Abstract
In this chapter, we study several important operators in Gaussian harmonic analysis. First, we consider Riesz and Bessel potentials with respect to the Ornstein–Uhlenbeck operator L, and then, Riesz and Bessel fractional derivatives. We study their regularity on Gaussian Lipschitz spaces, on Gaussian Besov–Lipschitz spaces, and on Gaussian Triebel–Lizorkin spaces. The results obtained are essentially similar to the classical results, as mentioned before, the methods of proofs are completely different. The boundedness results for Gaussian Besov–Lipschitz and Triebel–Lizorkin spaces were obtained by A. E. Gatto, E. Pineda, and W. Urbina, and appeared initially in [110] and [111]. These results can be extended to the case of Laguerre and Jacobi expansions by analogous arguments.
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In this chapter, we study several important operators in Gaussian harmonic analysis. First, we consider Riesz and Bessel potentials with respect to the Ornstein–Uhlenbeck operator L, and then, Riesz and Bessel fractional derivatives. We study their regularity on Gaussian Lipschitz spaces, on Gaussian Besov–Lipschitz spaces, and on Gaussian Triebel–Lizorkin spaces. The results obtained are essentially similar to the classical results, as mentioned before, the methods of proofs are completely different. The boundedness results for Gaussian Besov–Lipschitz and Triebel–Lizorkin spaces were obtained by A. E. Gatto, E. Pineda, and W. Urbina, and appeared initially in [110] and [111]. These results can be extended to the case of Laguerre and Jacobi expansions by analogous arguments.
8.1 Riesz and Bessel Potentials with Respect to the Gaussian Measure
8.1.1 Gaussian Riesz Potentials
In the classical case, the Riesz potential of order β > 0 is defined as the negative fractional powers of − Δ,
which means, using Fourier transform, that
For more details, see [118, 252].
The Gaussian fractional integrals or Gaussian Riesz potentials can also be defined as negative fractional powers of (−L). However, because the Ornstein–Uhlenbeck operator has eigenvalue 0, the negative powers are not defined on all of L 2(γ d); thus, we need to be more careful with the definition. Let us consider \(\varPi _{0}f=f-\displaystyle \int _{\mathbb {R}^{d}}f(y)\gamma _{d}(dy)\) the L 2(γ d) for f ∈ L 2(γ d), the orthogonal projection on the orthogonal complement of the eigenspace corresponding to the eigenvalue 0.
Definition 8.1
The Gaussian fractional integral or Riesz potential of order β > 0, I β is defined spectrally as
which means that for any multi-index ν, |ν| > 0 its action on the Hermite polynomial H ν is given by
and for ν = 0 = (0, …, 0), I β(H 0) = 0.
By linearity, using the fact that the Hermite polynomials are an algebraic basis of \(\mathscr {P}(\mathbb {R}^d),\) I β can be defined for any polynomial function \(f(x) = \sum _{\nu } \widehat {f}_\gamma (\nu ) {\mathbf {H}}_\nu \) as
and similarly for f ∈ L 2(γ d).
From (8.4), it is clear that the Gaussian Riesz potentials I β are the simplest Meyer’s multipliers, because in this case
with h(x) = x the identity function.
Proposition 8.2
The Gaussian Riesz potential I β, β > 0, has the following integral representations, for \(f \in \mathscr (\mathbb {R}^d)\) is a polynomial or \(f \in C^2_b(\mathbb {R}^d),\)
with respect to the Ornstein–Uhlenbeck semigroup, and
with respect to the Poisson–Hermite semigroup,
Proof
It is enough to prove that (8.6) holds for the Hermite polynomials. By the change of variables u = |ν|t
Then, again as the Hermite polynomials are an algebraic base of the set of polynomials \(\mathscr {P}({\mathbb {R}^d}),\) the formula holds for any polynomial. It can be proved that (8.6) also holds for \(f \in C^2_b(\mathbb {R}^d).\)
Observe that the integral representation (8.7) only means a change of scale, as I β = [(−L)1∕2]−β. Taking the change of variables \(u=t \sqrt {|\nu |},\)
again using that the Hermite polynomials are an algebraic base of the set of polynomials \(\mathscr {P}({\mathbb {R}^d}),\) □
Following the classical case, in general, we prefer to use the representation of I β (8.7), using the Poisson–Hermite semigroup. This representation will be crucial later to get several boundedness results to operators associated with L.
On the other hand, let us recall that in the classical case (see [252, Chapter V §1]), Riesz potentials have the following integral representation:
In the Gaussian case, we can also get an integral representation, as follows:
Theorem 8.3
The Gaussian Riesz potential I β, β > 0, has an integral representation,
where the kernel N β∕2(x, y) is defined as
Proof
To find the integral representation of I β, because the negative powers of L do not exist in all of L 2(γ d), we add a small multiple of the identity. Hence, let us consider the operator (𝜖I d − L), where I d is the identity in \(\mathbb {R}^d\) and 𝜖 > 0, and let us take its negative powers. The advantage of this trick is that it can be represented as a Laplace transform and this allows us to use the expression for Mehler’s kernel M t(x, y). More precisely, for 𝜖 > 0 and β > 0,
therefore, the kernel of (𝜖I − L)−β∕2 is
because, if f ∈ L 1(γ d),
As Π 0 is the orthogonal projection of the orthogonal complement of the eigenspace corresponding to the eigenvalue 0, then J 0 = I − Π 0, where J 0 is the orthogonal projection on the subspace generated by H 0 ≡ 1 (that is, the constants), and then we have
The kernel of J 0 is clearly \(\pi ^{-d/2} e^{-|y|{ }^2}\) and trivially \( \epsilon ^{-\beta } = \int _0^{\infty } t^{\beta - 1} e^{-\epsilon t} dt\), then the kernel of (𝜖I − L)−β∕2Π 0 is
We can take 𝜖 → 0 in the integral above without problems, then
Therefore, the kernel of I β is given by
taking r = e −t. Thus
□
In [102], it is proven that these operators are not of weak type (1, 1) with respect to γ. On the other hand, the strong type (p, p), for 1 < p < ∞,
follows either directly, from the hypercontractivity property of the Ornstein–Uhlenbeck semigroup, or by applying P. A. Meyer’s multiplier theorem, Theorem 6.2.
The classical Riesz potentials are homogeneous (see E. Stein [252, Chapter V (10)]), but it is easy to see that this is not the case for the Gaussian Riesz potentials I β.
Moreover, it is well-known that the classical Riesz potentials are of strong type (p, q) with \(\frac {1}{q}=\frac {1}{p}-\frac {\beta }{d}\), that is to say, the classical Riesz potentials “improve” in the sense that \(I_{\beta }: L^p(\mathbb R^d) \rightarrow L^q(\mathbb R^d)\) continuously, with \(\frac {1}{q} = \frac {1}{p} - \frac {\beta }{d}\). The Gaussian Riesz potentials, however, do not improve integrability. More formally, for any β > 0 for the Gaussian Riesz potential I β, there is no q > p such that it sends L p(γ d) → L q(γ d) continuously. This can be proved using the following counterexample, due to L. Forzani and W. Urbina, [87]. For every a > 0, let us split I β as,
where the kernel (8.11) is split into the sum of two parts,
The operator
can be written as
where T t = e Lt is the Ornstein–Uhlenbeck semigroup (see Chapter 2). Taking into account that T t is a hypercontractive semigroup, I 1 turns out to be of strong type (p, q), with q = 1 + (p − 1)e 4t.
Additionally, I 2 is an operator defined for every function on L p(γ d). To prove that it does not improve integrability, it would be enough to show that for every q > p there is a function f ∈ L p(dγ) such that I 2f∉L q(γ d). Let us take \(\frac {1}{q}<c<\frac {1}{p}\) and \(f(y)=e^{c|y|{ }^2}\chi _{|y|\ge 1}\in L^p(d\gamma )\).Footnote 1 It can be proved (see [86]), that the kernel \( N^2_\beta (x,y) \ge C e^{|x|{ }^2} \frac {e^{|y|{ }^2}}{|y|} \) in the region \( \{(x,y): \ \ |x| \ge 1, \ \ \frac {1}{4} |y|{ }^2 + 1 < |x|{ }^2 < \frac {3}{4} |y|{ }^2 \} \). Hence, \( I_2 (e^{c|x|{ }^2}) \ge \frac {e^{c|x|{ }^2}}{|x|{ }^2} \) for |x|≥ 1; therefore, \( I_2 (e^{c|x|{ }^2}) \notin L^q (\gamma _d)\).
The reason why Gaussian Riesz potentials do not improve integrability is the fact that L satisfies a logarithmic Sobolev inequality and not a Sobolev inequality. Nevertheless, a \(L^p\log L(\gamma _d)\) inequality can still be pulled out. Following E. Fabes’ suggestion, applying certain techniques used by L. Gross in [119], to prove that hypercontractivity implies a Sobolev logarithmic inequality, we can prove the following result:
Proposition 8.4
For any β > 0 the Gaussian Riesz potential I β maps L p(γ d) into \(L^p\log L(\gamma _d)\) continuously; in other words, the following inequality holds
for each f ∈ L p(γ d).
Proof
Indeed, for β > 0, consider the generalized Poisson–Hermite semigroup \(P_t^{\beta }=e^{-(-L)^{\beta }t}\), defined in (3.38). Let f be a polynomial, such that \( \int _{\mathbb {R}^d}f\ d\gamma =0\), I βf ≠ 0, and set \(F(t)=P_t^{\beta }(I_{\beta } f)\), then for every t > 0,
where the above inequality is a consequence of the hypercontractivity of \(P_t^{\beta }.\) In (8.14) we let t → 0+ to get
Using a lemma proved in [119],
But F′(0) = (−L)βI βf = f. Now, combining (8.15) and (8.16) we get
By applying Hölder’s inequality to the second term of the sum appearing on the right-hand side of the above inequality, and then the L p(dγ) continuity of I β, we get inequality (8.13). □
Thus, although I β do not improve in the L p(γ d) “scale,” they do improve in the “logarithmic scale” \( L^p(\gamma _d) \log L(\gamma _d).\)
8.1.2 Gaussian Bessel Potentials
Definition 8.5
The Gaussian Bessel potential of order β > 0, \(\mathscr {J}_\beta \), is defined spectrally as
meaning that for the Hermite polynomials we have,
Again, by linearity, \(\mathscr {J}_\beta \) can be extended to any polynomial; thus, if f =∑kJ kf, then
From (8.18), it is clear that the Gaussian Bessel potentials \(\mathscr {J}_\beta \) are not Meyer’s multipliers, but a composition of two Meyer’s multipliers, because in this case
with h 1(x) = (1 + x)−β and h 2(x) = x.
Using a similar argument to that above (8.7), the Bessel potentials can be represented as
P. A. Meyer’s multiplier theorem, Theorem 6.2, shows that \(\mathscr {J}_\beta \) is a bounded operator on L p(γ d), 1 < p < ∞, and again (8.20) can be extended to L p(γ d), using the density of the polynomials there.
On the other hand, again using P. A. Meyer’s multiplier theorem, Theorem 6.2, we get that the operators
are bounded on every L p(γ d), 1 < p < ∞; because, for instance, for any multi-index ν, |ν| > 0
with h(x) = (x + 1)β. These give the relation between the Riesz and Bessel potentials, similar to those in the classical case (see [252, Chapter V. Lemma. 2]).
It is easy to see, from the fact that \(\mathscr {J}_\beta \) is a multiplier, that it is also a bijection over the set of polynomials \({\mathscr P}\). Additionally, the Gaussian Sobolev spaces can be characterized in terms of Gaussian Bessel potentials,
Proposition 8.6
For β ≥ 0 and 1 ≤ p < ∞
Proof
First of all, observe that \( \mathscr {J}_\beta \) maps the family of polynomials \(\mathscr {P}(\mathbb {R}^d)\) into itself injectively. Then, as we already know \( \mathscr {J}_\beta \) is continuous in L p(γ d), then we conclude \( \mathscr {J}_\beta : L^{p}(\gamma _d) \to L^{p}_\alpha (\gamma _d)\) is bijective. □
Moreover, considering the family \(\{\mathscr {J}_\beta \}_\beta \) it is easy to see that it is a strongly continuous semigroup on L p(γ), 1 ≤ p < ∞, having as infinitesimal generator \(\frac {1}{2}\log (I-L).\)
8.2 Fractional Derivatives with Respect to the Gaussian Measure
8.2.1 Gaussian Riesz Fractional Derivate
In the classical case, fractional derivates for the Laplacian operator are defined as,
for 0 < β < 2, \(c_\beta =\frac {2^{\beta }\varGamma (d+\beta /2)}{\pi ^{d/2}\varGamma (-\beta /2)}\), see [255].
For the case of doubling measures, and more recently for s-dimensional non- doubling measures, this has been generalized by A. E. Gatto, C. Segovia, and S. Vàgi in [108].
On the other hand, observe that
where P t is the classical Poisson semigroup. Then, following the classical case:
Definition 8.7
The Gaussian Riesz fractional derivative of order β > 0, D β is defined spectrally as
meaning that for the Hermite polynomials, we have
Thus, by linearity, D β can be extended to any polynomial (see [164] and [224]).
Now, if f is a polynomial, by the linearity of the operators I β and D β, (8.3) and (8.24), we get
In the case of 0 < β < 1 we have the following integral representation for f a polynomial,
where \( c_\beta =\int _0^\infty u^{-\beta -1}(1-e^{-u})du:\) because for the Hermite polynomials we have, by the change of variables \(u=\sqrt {\left |\nu \right |}t\),
The identity (8.26) is very important in the development of a version of A. P. Calderón’s reproduction formula (see Theorem 8.31 below).
Now, if β ≥ 1, let k be the smallest integer greater than β i.e. k − 1 ≤ β < k, then the fractional derivative D β can be represented as
where \(c^k_{\beta } = \int _0^{\infty } u^{-\beta -1} ( 1-e^{-u} )^k \, du\) and f a polynomial function (see [239]).
As was mentioned earlier, fractional derivatives D β can be used to characterize the Gaussian Sobolev spaces \(L^p_\beta (\gamma _d)\). First, we need to extend the fractional derivative operator D β to all the Gaussian Sobolev spaces \( L^p_{\beta }(\gamma _d), \; 1<p<\infty .\) The union of these spaces
is a natural domain of D β. Observe that the definition of D β in all the spaces \( L_{\beta }^p(\gamma _d), \; 1<p<\infty ,\) is based on an application of Meyer’s multiplier theorem, Theorem 6.2.
Theorem 8.8
Let β > 0 and 1 < p < ∞.
-
i)
If {P n}n is a sequence of polynomials such that limn→∞P n = f in \( L_{\beta }^p(\gamma _d)\), then limnD βP n exists in \(L_{\beta }^p(\gamma _d)\) and does not depend on the choice of a sequence {P n}n. If \(f \in L_{\beta }^p(\gamma _d) \cap L_{\beta }^r(\gamma _d),\) then the limit does not depend on the choice of p or r. Thus, the fractional derivative is well defined by
$$\displaystyle \begin{aligned}D^{\beta} f = \lim_{n \to \infty } D^{\beta} P_n \;\; \mathrm{in}\;\; L_{\beta}^p(\gamma_d),\quad \mathit{\mbox{as}} \quad \lim_{n \to \infty} P_n= f \;\; \mathrm{in} \;\; L_{\beta}^p(\gamma_d),\end{aligned}$$f ∈ L β(γ d), is well defined.
-
ii)
\(f \in L_{\beta }^p(\gamma _d)\) if and only if D βf ∈ L p(γ d). Moreover,
$$\displaystyle \begin{aligned} B_{p,\beta}\left\| f \right\|{}_{p,\beta} \leq \left\| D^{\beta} f \right\|{}_{p,\gamma_d} \leq A_{p,\beta}\left\| f \right\|{}_{p,\beta}. \end{aligned} $$(8.28)
Proof
ii) Let f be a polynomial. Then
where g = (1 − L)−β∕2f. Note that g is also a polynomial. Observe that by construction,
Using Meyer’s multiplier theorem, Theorem 6.2, with the holomorphic function h(z) = (1 + z)−β∕2, we get
To prove the converse inequality, observe that the polynomial g can be rewritten as
and using Meyer’s multiplier theorem again we obtain,
Thus, we get (8.28) for polynomials.
i) The completeness of \(L_{\beta }^p(\gamma _d)\) can be proved using (8.28), and the fact that for r ≥ p the embedding \(L_{\beta }^r(\gamma _d) \subset L_{\beta }^p(\gamma _d)\) is continuous. Finally, from there, we can obtain (8.28) for any \(f \in L_{\beta }^p(\gamma _d)\). □
From the previous result and Proposition 7.3, we can immediately obtain a characterization of the Gaussian Sobolev spaces.
Corollary 8.9
Assume that 1 < p < ∞ and β > 0. Then
If \(\beta = k \in \mathbb {N},\) then
This characterization of Sobolev spaces is the most common one in the classical case.
8.2.2 Gaussian Bessel Fractional Derivates
We can also define the Gaussian Bessel fractional derivatives, \({\mathscr D}^\beta \).
Definition 8.10
The Gaussian Bessel fractional derivatives of order β, \({\mathscr D}^\beta ,\) are defined spectrally as
which means that for the Hermite polynomials, we have
thus, by linearity, it can be extended to any polynomial (see [ 224 ]).
In the case of 0 < β < 1, we have the following integral representation,
where, as before, \(c_\beta =\int _0^\infty u^{-\beta -1}(1-e^{-u})du\) and f is a polynomial.
Moreover, if β ≥ 1, let k be the smallest integer greater than β, i.e.. k − 1 ≤ β < k, then we have the following representation of \({\mathscr D}^\beta f\)
where \(c^k_{\beta } = \int _0^{\infty } u^{-\beta -1} (1-e^{-u} )^k du\) and f is a polynomial (see [239]).
8.3 Boundedness of Fractional Integrals and Fractional Derivatives on Gaussian Lipschitz Spaces
The boundedness results in the case of Gaussian Lipschitz spaces initially appeared in A. E. Gatto and W. Urbina’s article [109]. First, observe that the Gaussian Riesz potentials are not bounded operators on L ∞(γ d) and, therefore, not on Lip α(γ) either. Then, to make sense of Riesz potentials on L ∞, we consider, for β > 0, the truncated Gaussian Riesz potentials,
We want to study the truncated Gaussian Riesz potentials \(I_{\beta }^{T}\) on the Gaussian Lipschitz spaces Lip α(γ d),
Theorem 8.11
For 0 < β < 1 and α > 0, the Riesz potential of order β, \( I_{\beta }^{T}: Lip_{\alpha }(\gamma _d) \rightarrow Lip_{\alpha +\beta }(\gamma _d)\) is bounded.
Proof
Let f ∈ Lip α(γ d), i.e., f ∈ L ∞ such that \(\Big \|\frac {\partial P_t f}{\partial t}\Big \|{ }_{\infty ,\gamma _d}\leq At^{-1+\alpha }\). First, observe that
that is, P tf ∈ L ∞ and then
Therefore, \(I^{T}_{\beta } f \in L^{\infty }\). Now, using the semigroup property and Fubini’s theorem,
If α + β < 1, then for 0 ≤ s ≤ 1
Now, for (I), because t < s
and, for (II), as t > s
Thus,
which implies I βf ∈ Lip α+β(γ d). The general case follows in a similar manner. □
Now, we study the action of the Bessel potentials on the Gaussian Lipschitz spaces Lip α(γ), which is much better than the case of the Riesz potentials:
Theorem 8.12
Let α, β > 0 then \(\mathscr {J}_{\beta }\) is bounded from Lip α(γ) to Lip α+β(γ).
Proof
Let f ∈ Lip α(γ) and consider a fixed integer n > α + β, then
Using (8.20), the fact that f ∈ L ∞, and consequently P t+sf ∈ L ∞, we obtain
therefore,
i.e. \(P_{t}(\mathscr {J}_{\beta }f) \in L^{\infty }\).
Now, we want to verify the Lipschitz condition. Differentiating (8.35), we get
and this implies
Because β > 0 as t + s > t,
On the other hand, because n > α + β, as t + s > s
Therefore,
Thus, \(\mathscr {J}_{\beta }f\in Lip_{\alpha +\beta }(\gamma ),\) and moreover
□
Finally, let us study the action of the fractional derivative D β on the Gaussian Lipschitz spaces.
Theorem 8.13
For 0 < β < α < 1, the fractional derivate of order β, D β : Lip α(γ d) → Lip α−β(γ d) is bounded.
Proof
Let f ∈ Lip α(γ d), i.e., f ∈ L ∞ such that \(\Big \|\frac {\partial P_t f}{\partial t}\Big \|{ }_{\infty ,\gamma }\leq At^{-1+\alpha }\). Observe that using (7.44) and Proposition 7.23, we get
Thus, D βf ∈ L ∞(γ d). Now, using (8.26), and fixing s, we have
Using Proposition 7.27, we have
and then, using the fundamental theorem of calculus, we get
Then, as t < s,
On the other hand,
Therefore,
which implies D βf ∈ Lip α−β(γ d). □
8.4 Boundedness of Fractional Integrals and Fractional Derivatives on Gaussian Besov–Lipschitz Spaces
As we discussed in the previous section, in the case of the Lipschitz spaces only a truncated version of the Riesz potentials is bounded from Lip α(γ d) to Lip α+β(γ d). Now, we study the boundedness properties of the Riesz potentials on Besov–Lipschitz spaces, and we see that in this case, the results are better.
Theorem 8.14
Let α ≥ 0, β > 0, 1 < p < ∞, 1 ≤ q ≤∞ then I β is bounded from \(B_{p,q}^{\alpha }(\gamma _{d})\) into \(B_{p,q}^{\alpha +\beta }(\gamma _{d})\).
Proof
Let k > α + β a fixed integer, \(f\in B_{p,q}^{\alpha }(\gamma _{d})\), using the integral representation of Riesz potentials (8.7), the semigroup property of {P t}t≥0 and the fact that P ∞f(x) is a constant and the semigroup is conservative, we get
Using the fact that P ∞f(x) is a constant again, and the chain rule,
Then, using Minkowski’s integral inequality
Hence, if 1 ≤ q < ∞,
Now, as β > 0 using Lemma 3.5, and as t + s > t,
because \(f\in B_{p}^{\alpha ,q}(\gamma _{d})\).
On the other hand, as k > α + β using Lemma 3.5 again, because t + s > s, and Hardy’s inequality (10.101), we obtain
because \(f\in B_{p,q}^{\alpha }(\gamma _{d})\). Therefore, \(I_{\beta }f\in B_{p,q}^{\alpha +\beta }(\gamma _{d})\) and, moreover,
Now, if q = ∞, (8.39) can be written as
Using that β > 0, Lemma 3.5, as t + s > t, and because \(f\in B_{p,\infty }^{\alpha }(\gamma _{d})\),
Now, because k > α + β, using Lemma 3.5, as t + s > s, and because \(f\in B_{p,\infty }^{\alpha }(\gamma _{d})\), we get
Therefore,
and this implies that \(I_{\beta }f \in B_{p,\infty }^{\alpha +\beta }(\gamma _{d})\) and A k(I βf) ≤ CA k(f).
Moreover, as I β is a bounded operator on L p(γ d), 1 < p < ∞,
□
Now, we are going to study the boundedness properties of the Bessel potentials on Besov–Lipschitz spaces.
Theorem 8.15
Let α ≥ 0, 1 ≤ p, q < ∞, then for β > 0,
-
i)
\(\mathscr {J}_\beta \) is bounded on \(B_{p,q}^{\alpha }(\gamma _{d})\).
-
ii)
Moreover, \(\mathscr {J}_{\beta }\) is bounded from \(B_{p,q}^{\alpha }(\gamma _{d})\) to \(B_{p,q}^{\alpha +\beta }(\gamma _{d})\).
-
iii)
Finally, for q = ∞, \(\mathscr {J}_{\beta }\) is bounded from \(B_{p,\infty }^{\alpha }(\gamma _{d})\) into \(B_{p,\infty }^{\alpha +\beta }(\gamma _{d})\).
Proof
-
i)
Let us see that \(\mathscr {J}_\beta \) is bounded on \(B_{p,q}^{\alpha }(\gamma _{d})\). Using Lebesgue’s dominated convergence theorem, Minkowski’s integral inequality, and Jensen’s inequality, we have
$$\displaystyle \begin{aligned} \begin{array}{rcl} \Big\|\frac{\partial^{k}P_{t}}{\partial t^{k}}\Big(\mathscr{J}_\beta f \Big)\Big\|{}_{p,\gamma_{d}}^{q} &\displaystyle =&\displaystyle (\int_{\mathbb{R}^{d}}|\frac{\partial^{k} P_{t}}{\partial t^{k}}(\frac{1}{\varGamma(\beta)}\int_{0}^{+\infty}s^{\beta} e^{-s}P_{s}f(x)\frac{ds}{s})|{}^{p}\gamma_{d}(dx))^{\frac{q}{p}}\\ &\displaystyle \leq&\displaystyle (\frac{1}{\varGamma(\beta)}\int_{0}^{+\infty}s^{\beta}e^{-s}(\int_{\mathbb{R}^{d}}|\frac{\partial^{k} P_{t}P_{s}f(x)}{\partial t^{k}}|{}^{p}\gamma_{d}(dx))^{\frac{1}{p}}\frac{ds}{s})^q\\ &\displaystyle \leq&\displaystyle \frac{1}{\varGamma(\beta)}\int_{0}^{+\infty}s^{\beta}e^{-s}\Big\|\frac{\partial^{k}P_{t}P_{s}f}{\partial t^{k}}\Big\|{}_{p,\gamma_{d}}^{q}\frac{ds}{s}, \end{array} \end{aligned} $$and then, using Tonelli’s theorem,
Therefore,
$$\displaystyle \begin{aligned} \begin{array}{rcl} \| \mathscr{J}_\beta f\|{}_{B_{p,q}^{\alpha}}&\displaystyle =&\displaystyle \| \mathscr{J}_\beta f\|{}_{p,\gamma_{d}}+\int_{0}^{+\infty}\Big(t^{k-\alpha}\Big\|\frac{\partial^{k}P_{t}}{\partial t^{k}}\Big(\mathscr{J}_\beta f \Big)\Big\|{}_{p,\gamma_{d}}\Big)^{q}\frac{dt}{t}\\ &\displaystyle \leq&\displaystyle \|f\|{}_{p,\gamma_{d}}+\int_{0}^{+\infty}\Big(t^{k-\alpha}\Big\|\frac{\partial^{k} P_{t}f}{\partial t^{k}}\Big\|{}_{p,\gamma_{d}}\Big)^{q}\frac{dt}{t}=\|f\|{}_{B_{p,q}^{\alpha}}. \end{array} \end{aligned} $$ -
ii)
We use the notation u(x, t) = P tf(x) and \(U(x,t)=P_{t}\mathscr {J}_{\beta }f(x)\), using the representation (3.8) of P t we have,
$$\displaystyle \begin{aligned}U(x,t)=\displaystyle\int_{0}^{+\infty}T_{s}(\mathscr{J}_{\beta}f)(x)\mu_{t}^{(1/2)}(ds).\end{aligned}$$Therefore,
$$\displaystyle \begin{aligned}U(x,t_{1}+t_{2})=P_{t_{1}}(P_{t_{2}}(\mathscr{J}_{\beta}f))(x)= \displaystyle\int_{0}^{+\infty}T_{s}(P_{t_{2}}(\mathscr{J}_{\beta}f))(x)\mu_{t_{1}}^{(1/2)}(ds).\end{aligned}$$Now, let k, l be integers greater than α, β respectively, by differentiating k times with respect to t 2 and l times with respect to t 1,
$$\displaystyle \begin{aligned}\displaystyle\frac{\partial^{k+l}U(x,t_{1}+t_{2})}{\partial (t_{1}+t_{2})^{k+l}}=\int_{0}^{+\infty}T_{s} (\frac{\partial^{k}P_{t_{2}}}{\partial t_{2}^{k}}(\mathscr{J}_{\beta}f))(x)\frac{\partial^{l}}{\partial t_{1}^{l}}\mu_{t_{1}}^{(1/2)}(ds).\end{aligned}$$Thus,
$$\displaystyle \begin{aligned}\displaystyle\frac{\partial^{k+l} U(x,t)}{\partial t^{k+l}}=\int_{0}^{+\infty}T_{s}(\frac{\partial^{k}P_{t_{2}}}{\partial t_{2}^{k}}(\mathscr{J}_{\beta}f))(x)\frac{\partial^{l}}{\partial t_{1}^{l}}\mu_{t_{1}}^{(1/2)}(ds),\end{aligned}$$if t = t 1 + t 2 and therefore, using the L p continuity of T s and (3.21)
$$\displaystyle \begin{aligned} \begin{array}{rcl} {} \Big\|\frac{\partial^{k+l} U(\cdot,t)}{\partial t^{k+l}}\Big\|{}_{p,\gamma}&\displaystyle \leq&\displaystyle \int_{0}^{+\infty}\|T_{s} (\frac{\partial^{k}P_{t_{2}}}{\partial t_{2}^{k}}(\mathscr{J}_{\beta}f))\Big\|{}_{p,\gamma} |\frac{\partial^{l}}{\partial t_{1}^{l}} \mu_{t_{1}}^{(1/2)}(ds)| \\ &\displaystyle \leq&\displaystyle \int_{0}^{+\infty}\| \frac{\partial^{k}P_{t_{2}}}{\partial t_{2}^{k}}(\mathscr{J}_{\beta}f)\Big\|{}_{p,\gamma} |\frac{\partial^{l}}{\partial t_{1}^{l}} \mu_{t_{1}}^{(1/2)}(ds)| \\ &\displaystyle =&\displaystyle \Big\|\frac{\partial^{k}P_{t_{2}}}{\partial t_{2}^{k}}(\mathscr{J}_{\beta}f)\Big\|{}_{p,\gamma}\int_{0}^{+\infty}|\frac{\partial^{l}}{\partial t_{1}^{l}} \mu_{t_{1}}^{(1/2)}(ds)| \\ &\displaystyle \leq&\displaystyle C t_{1}^{-l} \Big\|\frac{\partial^{k}}{\partial t_{2}^{k}}P_{t_{2}}\mathscr{J}_{\beta}f\Big\|{}_{p,\gamma}. \end{array} \end{aligned} $$(8.40)On the other hand, using the representation of Bessel potential (8.20), we have
$$\displaystyle \begin{aligned}P_{t}(\mathscr{J}_{\beta}f)(x)=\displaystyle\frac{1}{\varGamma(\beta)}\int_{0}^{+\infty}s^{\beta}e^{-s}P_{t+s}f(x)\frac{ds}{s}\end{aligned}$$then
$$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{\partial^{k}P_{t}}{\partial t^{k}}(\mathscr{J}_{\beta}f)(x)&\displaystyle =&\displaystyle \frac{1}{\varGamma(\beta)}\int_{0}^{+\infty}s^{\beta}e^{-s}\frac{\partial^{k} P_{t+s}f(x)}{\partial t^{k}}\frac{ds}{s}\\ &\displaystyle =&\displaystyle \frac{1}{\varGamma(\beta)}\int_{0}^{+\infty}s^{\beta}e^{-s}\frac{\partial^{k} P_{t+s}f(x)}{\partial (t+s)^{k}}\frac{ds}{s}, \end{array} \end{aligned} $$and this implies that
$$\displaystyle \begin{aligned} \begin{array}{rcl} \Big\|\frac{\partial^{k} P_{t}}{\partial t^{k}}(\mathscr{J}_{\beta}f)\Big\|{}_{p,\gamma}&\displaystyle \leq&\displaystyle \frac{1}{\varGamma(\beta)}\int_{0}^{+\infty}s^{\beta}e^{-s}\Big\|\frac{\partial^{k}P_{t+s}f}{\partial (t+s)^{k}}\Big\|{}_{p,\gamma}\frac{ds}{s}< \infty, \end{array} \end{aligned} $$because \(f\in B_{p,q}^{\alpha }(\gamma _{d})\). Now, because the definition of \(B_{p,q}^{\alpha }(\gamma _{d})\) is independent on the integer k > α that we can choose, let us take k > α + β and l > β, then k + l > α + 2β > α + β; thus, k + l is an integer greater than α + β. Let us now see that
$$\displaystyle \begin{aligned}\displaystyle\Big(\int_{0}^{+\infty}\Big(t^{k+l-(\alpha+\beta)}\Big\|\frac{\partial^{k+l} U(\cdot,t)}{\partial t^{k+l}}\Big\|{}_{p,\gamma}\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}}<+\infty. \end{aligned}$$In fact, taking t 1 = t 2 = t∕2 in (8.40), we get
$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \displaystyle\Big(\int_{0}^{+\infty}\Big(t^{k+l-(\alpha+\beta)}\Big\|\frac{\partial^{k+l} U(\cdot,t)}{\partial t^{k+l}}\Big\|{}_{p,\gamma}\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}} \quad \quad \quad \quad \quad \\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \leq C \Big(\int_{0}^{+\infty}\Big(t^{k+l-(\alpha+\beta)}\Big\|\frac{\partial^{k}P_{\frac{t}{2}}}{\partial (\frac{t}{2})^{k}}(\mathscr{J}_{\beta}f)\Big\|{}_{p,\gamma}(\frac{t}{2})^{-l}\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}}\\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \leq \frac{C}{\varGamma(\beta)}\Big(\int_{0}^{+\infty}\Big(t^{k-(\alpha+\beta)}\Big(\int_{0}^{+\infty}s^{\beta}e^{-s} \Big\|\frac{\partial^{k}P_{s+\frac{t}{2}}f}{\partial(\frac{t}{2})^{k}}\Big\|{}_{p,\gamma}\frac{ds}{s}\Big)\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}}\\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \leq\frac{C}{\varGamma(\beta)}\Big[\int_{0}^{+\infty}t^{(k-(\alpha+\beta))q}\Big(\int_{0}^{t}s^{\beta} \Big\|\frac{\partial^{k}P_{s+\frac{t}{2}}f}{\partial(s+\frac{t}{2})^{k}}\Big\|{}_{p,\gamma}\frac{ds}{s}\Big)^{q}\\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + \Big(\int_{t}^{+\infty}s^{\beta} \Big\|\frac{\partial^{k}P_{s+\frac{t}{2}}f}{\partial(s+\frac{t}{2})^{k}}\Big\|{}_{p,\gamma}\frac{ds}{s}\Big)^{q}\frac{dt}{t}\Big]^{\frac{1}{q}}. \end{array} \end{aligned} $$Again using that (a + b)q ≤ C q(a q + b q) if a, b ≥ 0, q ≥ 1, but because (a + b)1∕q ≤ a 1∕q + b 1∕q if a, b ≥ 0, q ≥ 1,
$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \frac{C}{\varGamma(\beta)}\Big[\int_{0}^{+\infty}t^{(k-(\alpha+\beta))q}\Big(\int_{0}^{t}s^{\beta} \Big\|\frac{\partial^{k}P_{s+\frac{t}{2}}f}{\partial(s+\frac{t}{2})^{k}}\Big\|{}_{p,\gamma}\frac{ds}{s}\Big)^{q}\quad \quad \quad \quad \quad \\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \quad \quad + \Big(\int_{t}^{+\infty}s^{\beta} \Big\|\frac{\partial^{k}P_{s+\frac{t}{2}}f}{\partial(s+\frac{t}{2})^{k}}\Big\|{}_{p,\gamma}\frac{ds}{s}\Big)^{q}\frac{dt}{t}\Big]^{\frac{1}{q}}\\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \leq\frac{C}{\varGamma(\beta)}\Big[\int_{0}^{+\infty}t^{(k-(\alpha+\beta))q}\Big(\int_{0}^{t}s^{\beta} \Big\|\frac{\partial^{k}P_{s+\frac{t}{2}}f}{\partial(s+\frac{t}{2})^{k}}\Big\|{}_{p,\gamma}\frac{ds}{s}\Big)^{q} \frac{dt}{t}\Big]^{1/q}\\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \quad \quad + \frac{C}{\varGamma(\beta)}\Big[ \int_{0}^{+\infty}t^{(k-(\alpha+\beta))q}\Big(\int_{t}^{+\infty}s^{\beta} \Big\|\frac{\partial^{k}P_{s+\frac{t}{2}}f}{\partial(s+\frac{t}{2})^{k}}\Big\|{}_{p,\gamma}\frac{ds}{s}\Big)^{q}\frac{dt}{t}\Big]^{\frac{1}{q}}\\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad =(I)+(II). \end{array} \end{aligned} $$Now, using Lemma 3.5 and because β > 0
$$\displaystyle \begin{aligned} \begin{array}{rcl} (I) &\displaystyle =&\displaystyle \frac{C}{\varGamma(\beta)}\Big[\int_{0}^{+\infty}t^{(k-(\alpha+\beta))q}\Big(\int_{0}^{t}s^{\beta} \Big\|\frac{\partial^{k}P_{s+\frac{t}{2}}f}{\partial(s+\frac{t}{2})^{k}}\Big\|{}_{p,\gamma}\frac{ds}{s}\Big)^{q}\frac{dt}{t}\Big]^{\frac{1}{q}}\\ &\displaystyle \leq&\displaystyle \frac{C}{\varGamma(\beta)}\Big[\int_{0}^{+\infty}t^{(k-(\alpha+\beta))q}\Big(\int_{0}^{t}s^{\beta} \Big\|\frac{\partial^{k}P_{\frac{t}{2}}f}{\partial(\frac{t}{2})^{k}}\Big\|{}_{p,\gamma}\frac{ds}{s}\Big)^{q}\frac{dt}{t}\Big]^{\frac{1}{q}}\\ &\displaystyle =&\displaystyle \frac{C}{\beta\varGamma(\beta)}\Big(\int_{0}^{+\infty}\Big(t^{k-\alpha}\Big\|\frac{\partial^{k}P_{\frac{t}{2}}f}{\partial(\frac{t}{2})^{k}}\Big\|{}_{p,\gamma}\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}}\\ &\displaystyle =&\displaystyle C_{\alpha,\beta}\Big(\int_{0}^{+\infty}\Big(u^{k-\alpha}\Big\|\frac{\partial^{k}P_{u}f}{\partial u^{k}}\Big\|{}_{p,\gamma}\Big)^{q}\frac{du}{u}\Big)^{\frac{1}{q}}<+\infty, \end{array} \end{aligned} $$because \(f\in B_{p}^{\alpha ,q}(\gamma _{d})\).
On the other hand, using Hardy inequality, because k > α + β and Lemma 3.5, we get
$$\displaystyle \begin{aligned} \begin{array}{rcl} (II) &\displaystyle =&\displaystyle \frac{C}{\varGamma(\beta)}\Big(\int_{0}^{+\infty}t^{(k-(\alpha+\beta))q}\Big(\int_{t}^{+\infty}s^{\beta} \Big\|\frac{\partial^{k}P_{s+\frac{t}{2}}f}{\partial(s+\frac{t}{2})^{k}}\Big\|{}_{p,\gamma}\frac{ds}{s}\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}}\\ &\displaystyle \leq&\displaystyle \frac{C}{\varGamma(\beta)}\Big(\int_{0}^{+\infty}t^{(k-(\alpha+\beta))q}\Big(\int_{t}^{+\infty}s^{\beta} \Big\|\frac{\partial^{k}P_{s}f}{\partial s^{k}}\Big\|{}_{p,\gamma}\frac{ds}{s}\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}}\\ &\displaystyle \leq&\displaystyle \frac{C}{\varGamma(\beta)}\frac{1}{k-(\alpha+\beta)}\int_{0}^{+\infty}\Big(s^{k-\alpha} \Big\|\frac{\partial^{k}}{\partial s^{k}}P_{s}f\Big\|{}_{p,\gamma}\Big)^{q}\frac{ds}{s}\Big)^{\frac{1}{q}}<+\infty \end{array} \end{aligned} $$because \(f\in B_{p,q}^{\alpha }(\gamma _{d})\). Thus, \(\mathscr {J}_{\beta }f\in B_{p,q}^{\alpha +\beta }(\gamma _{d})\) and, moreover,
$$\displaystyle \begin{aligned}\displaystyle\|\mathscr{J}_{\beta}f\|{}_{B_{p,q}^{\alpha+\beta}}\leq C _{\alpha,\beta}\|f\|{}_{B_{p,q}^{\beta}}. \end{aligned}$$ -
iii)
Let k > α + β a fixed integer, \(f\in B_{p,\infty }^{\alpha }(\gamma _{d})\), by using the representation of Bessel potential (8.20), we get
$$\displaystyle \begin{aligned}P_{t}(\mathscr{J}_{\beta}f)(x)=\displaystyle\frac{1}{\varGamma(\beta)}\int_{0}^{+\infty}s^{\beta}e^{-s}P_{t+s}f(x)\frac{ds}{s};\end{aligned}$$thus, using the chain rule, we obtain
$$\displaystyle \begin{aligned} \begin{array}{rcl} \frac{\partial^{k}}{\partial t^{k}}P_{t}(\mathscr{J}_{\beta}f)(x)&\displaystyle =&\displaystyle \frac{1}{\varGamma(\beta)}\int_{0}^{+\infty}s^{\beta}e^{-s}u^{(k)}(x,t+s)\frac{ds}{s}, \end{array} \end{aligned} $$which implies, using Minkowski’s integral inequality,
$$\displaystyle \begin{aligned} \begin{array}{rcl} \Big\|\frac{\partial^{k}}{\partial t^{k}} P_{t}(\mathscr{J}_{\beta}f)\Big\|{}_{p,\gamma}&\displaystyle \leq&\displaystyle \frac{1}{\varGamma(\beta)}\int_{0}^{+\infty}s^{\beta}e^{-s}\|u^{(k)}(\cdot,t+s)\|{}_{p,\gamma}\frac{ds}{s}\\ &\displaystyle =&\displaystyle \frac{1}{\varGamma(\beta)}\int_{0}^{t}s^{\beta}e^{-s}\|u^{(k)}(\cdot,t+s)\|{}_{p,\gamma}\frac{ds}{s}\\ &\displaystyle &\displaystyle \quad \quad \quad + \frac{1}{\varGamma(\beta)}\int_{t}^{\infty}s^{\beta}e^{-s}\|u^{(k)}(\cdot,t+s)\|{}_{p,\gamma}\frac{ds}{s}=(I)+(II). \end{array} \end{aligned} $$Now, as β > 0, using Lemma 3.5, as t + s > t, and because \(f\in B_{p,\infty }^{\beta }(\gamma _{d})\),
$$\displaystyle \begin{aligned} \begin{array}{rcl} (I) &\displaystyle \leq&\displaystyle \frac{1}{\varGamma(\beta)}\Big\|\frac{\partial^{k}P_{t}f}{\partial t^{k}}\Big\|{}_{p,\gamma}\int_{0}^{t}s^{\beta}e^{-s}\frac{ds}{s} \leq \frac{1}{\varGamma(\beta)}\Big\|\frac{\partial^{k}P_{t}f}{\partial t^{k}}\Big\|{}_{p,\gamma}\int_{0}^{t}s^{\beta-1}ds\\ &\displaystyle \leq&\displaystyle \frac{1}{\varGamma(\beta)}\frac{t^{\beta}}{\beta}A_{k}(f)t^{-k+\alpha} =C_{\beta} A_{k}(f)t^{-k+\alpha+\beta}. \end{array} \end{aligned} $$On the other hand, as k > α + β using Lemma 3.5, as t + s > s, and because \(f\in B_{p,\infty }^{\alpha }(\gamma _{d})\)
$$\displaystyle \begin{aligned} \begin{array}{rcl} (II) &\displaystyle \leq&\displaystyle \frac{1}{\varGamma(\beta)}\int_{t}^{\infty}s^{\beta}e^{-s}\Big\|\frac{\partial^{k}P_{s}f}{\partial s^{k}}\Big\|{}_{p,\gamma}\frac{ds}{s} \leq \frac{A_{k}(f)}{\varGamma(\beta)}\int_{t}^{\infty}s^{\beta}e^{-s} s^{-k+\alpha}\frac{ds}{s}\\ &\displaystyle \leq&\displaystyle \frac{A_{k}(f)}{\varGamma(\beta)}\int_{t}^{\infty}s^{-k+\alpha+\beta-1}ds =\frac{A_{k}(f)}{\varGamma(\beta)}\frac{t^{-k+\alpha+\beta}}{k-(\alpha+\beta)} = C_{k,\alpha,\beta}A_{k}(f) t^{-k+\alpha+\beta}. \end{array} \end{aligned} $$Therefore,
$$\displaystyle \begin{aligned} \begin{array}{rcl} \Big\|\frac{\partial^{k}}{\partial t^{k}} P_{t}({\mathscr{J}}_{\beta}f)\Big\|{}_{p,\gamma}&\displaystyle \leq&\displaystyle C A_{k}(f) t^{-k+\alpha+\beta}, \end{array} \end{aligned} $$then \(\mathscr {J}_{\beta }f\in B_{p,\infty }^{\alpha +\beta }(\gamma _{d})\) and \(A_{k}(\mathscr {J}_{\beta }f)\leq CA_{k}(f)\). Thus,
$$\displaystyle \begin{aligned} \begin{array}{rcl} \Big\|{\mathscr{J}}_{\beta}f\Big\|{}_{B_{p,\infty}^{\alpha+\beta}}&\displaystyle =&\displaystyle \Big\|{\mathscr{J}}_{\beta}f\Big\|{}_{p,\gamma}+A_{k}({\mathscr{J}}_{\beta}f)\leq\|f\|{}_{p,\gamma}+C A_{k}(f) \leq C\|f\|{}_{B_{p,\infty}^{\alpha}}. \end{array} \end{aligned} $$□
Now, we study the boundedness of the Riesz fractional derivatives and of the Bessel fractional derivatives on Besov–Lipschitz spaces. We use the representation (8.24) of the fractional derivative and Hardy’s inequalities. Because they require different techniques, we consider two cases:
-
The bounded case, 0 < β < α < 1.
-
The unbounded case 0 < β < α.
Let us start with the bounded case for the Riesz derivative:
Theorem 8.16
Let 0 < β < α < 1, 1 ≤ p < ∞ and 1 ≤ q ≤∞ then D β is bounded from \(B_{p,q}^{\alpha }(\gamma _{d})\) into \(B_{p,q}^{\alpha -\beta }(\gamma _{d})\).
Proof
Let \(f\in B_{p,q}^{\alpha }(\gamma _{d})\), using Hardy’s inequality (10.100), with p = 1, and the fundamental theorem of calculus,
Thus, using Minkowski’s integral inequality
because \(f\in B_{p,q}^{\alpha }(\gamma _{d})\subset B_{p,1}^{\beta }(\gamma _{d}) \), 1 ≤ q ≤∞ as α > β, i.e., D βf ∈ L p(γ d).
Now, by analogous argument
and again, using Minkowski’s integral inequality
Then, if 1 ≤ q < ∞, by (8.43)
Now, because r > t using Lemma 3.5 and the fact that 0 < β < 1,
On the other hand, as r > t using Hardy’s inequality (10.101), because (1 − α)q > 0, we get
Thus,
as \(f\in B_{p,q}^{\alpha }(\gamma _{d}).\) Then, \(D_{\beta }f\in B_{p,q}^{\alpha -\beta }(\gamma _{d})\) and
Therefore, \(D_{\beta }f:B_{p,q}^{\alpha }\rightarrow B_{p,q}^{\alpha -\beta }\) is bounded.
Now if q = ∞, inequality (8.43) can be written as
Now, using Lemma 3.5, because r > t,
and by Lemma 3.5, because r > t, and the fact that \(f \in B_{p,\infty }^\alpha ,\)
Thus,
Hence, \(D_{\beta }f \in B_{p,\infty }^{\alpha -\beta }(\gamma _{d})\) then A(D βf) ≤ CA(f), and
Therefore, \(D_{\beta }:B_{p,\infty }^{\alpha }\rightarrow B_{p,\infty }^{\alpha -\beta }\) is bounded. □
Next, we study the boundedness of the Bessel fractional derivative on Besov–Lipschitz spaces for the bounded case 0 < β < α < 1 :
Theorem 8.17
Let 0 < β < α < 1, 1 ≤ p < ∞ and 1 ≤ q ≤∞, then \({\mathscr D}_{\beta }\) is bounded from \(B_{p,q}^{\alpha }(\gamma _{d})\) into \(B_{p,q}^{\alpha -\beta }(\gamma _{d})\).
Proof
Let f ∈ L p(γ d), using the fundamental theorem of calculus we can write,
Now, using Hardy’s inequality (10.100), with p = 1 in both integrals, we have
Therefore, according to Minkowski’s integral inequality
because \(f\in B_{p,q}^{\alpha }(\gamma _{d})\subset B_{p,1}^{\beta }(\gamma _{d}) \), 1 ≤ q ≤∞ as α > β, i.e. D βf ∈ L p(γ d).
On the other hand, using the fundamental theorem of calculus and, Hardy’s inequality (10.100) again, with p = 1 in the second integral, we have
Thus, using Minkowski’s integral inequality,
Then, if 1 ≤ q < ∞, using (8.44) and Minkowski’s integral inequality, we get
For the first term, the argument is the same as that considered in the second part of the proof of Theorem 8.16; thus,
because \(f\in B_{p,q}^{\alpha }(\gamma _{d})\), and for the second term trivially
because α > α − β and the inclusion relation given in Proposition 7.36.
Hence, if 1 ≤ q < ∞,
so \(\mathscr {D}_{\beta }f\in B_{p,q}^{\alpha -\beta }(\gamma _{d})\) and, moreover,
If q = ∞, using the same argument as in Theorem 8.16, inequality (8.44) can be written as
for t > 0, then, \(\mathscr {D}_{\beta }f\in B_{p,\infty }^{\alpha -\beta }(\gamma _{d})\) and \(A(\mathscr {D}_{\beta }f)\leq C_{\alpha ,\beta }A(f)\); thus,
□
We consider now the unbounded case for fractional derivatives (removing the condition that the indexes must be less than 1). To do this, we need to consider forward differences. Remember that for a given function f, the k-th order forward difference of f starting at t with increment s is defined as
The forward differences have the following properties (see Appendix Lemma 10.30), which will be needed in what follows. For any positive integer k
-
i)
\(\varDelta _{s}^{k}(f,t)=\varDelta _{s}^{k-1}(\varDelta _{s}(f,\cdot ),t)=\varDelta _{s}(\varDelta _{s}^{k-1}(f,\cdot ),t).\)
-
ii)
\(\varDelta _{s}^{k}(f,t)=\displaystyle \int _{t}^{t+s}\int _{v_{1}}^{v_{1}+s}\ldots \int _{v_{k-2}}^{v_{k-2}+s}\int _{v_{k-1}}^{v_{k-1}+s}f^{(k)}(v_{k})dv_{k}dv_{k-1}\ldots dv_{2}dv_{1}.\)
-
iii)
For any positive integer k
$$\displaystyle \begin{aligned} \frac{\partial }{\partial s}(\varDelta_s^k (f,t) ) = k \,\varDelta_s^{k-1} (f',t+s), \end{aligned} $$(8.45)and for any integer j > 0,
$$\displaystyle \begin{aligned} \frac{\partial^j }{\partial t^j}(\varDelta_s^k (f,t) ) =\varDelta_s^{k} (f^{(j)},t). \end{aligned} $$(8.46)
Observe that using the binomial theorem and the semigroup property of {P t}, we have
where as usual, u(x, t) = P tf(x). Additionally, we need the following result:
Lemma 8.18
Let f ∈ L p(γ d), 1 ≤ p < ∞ and \(k,n\in \mathbb {N}\) then
Proof
From property ii) of forward differences (see Lemma 10.30), we have
then, using Minkowski’s integral inequality k-times and Lemma 3.5,
□
Let us start studying the boundedness of the Riesz fractional derivative in \(B_{p,q}^{\alpha }(\gamma _{d})\)
Theorem 8.19
Let 0 < β < α, 1 ≤ p < ∞ and 1 ≤ q ≤∞ then
D β is bounded from \(B_{p,q}^{\alpha }(\gamma _{d})\) into \(B_{p,q}^{\alpha -\beta }(\gamma _{d})\).
Proof
Let \(f\in B_{p,q}^{\alpha }(\gamma _{d})\), using (8.47), Hardy’s inequality (10.100), p = 1, the fundamental theorem of calculus, and property iii) of forward differences (see Lemma 10.30), we get
Now, using Minkowski’s integral inequality and Lemma 8.18
because \(f\in B_{p,q}^{\alpha }(\gamma _{d})\subset B_{p,1}^{\beta }(\gamma _{d})\), as α > β. Therefore, D βf ∈ L p(γ d).
On the other hand,
Thus, if n is the smaller integer greater than α, i.e., n − 1 ≤ α < n, then according to Lemma 10.30 iv),
therefore, using Minkowski’s integral inequality
Now, if 1 ≤ q < ∞, by (8.49),
Then, using Lemma 8.18,
because \(f\in B_{p,q}^{\alpha }(\gamma _{d})\), and using Lemma 3.5
because \(f\in B_{p,q}^{\alpha }(\gamma _{d})\). Therefore, if 1 ≤ q < ∞, \(D_{\beta }f\in B_{p,q}^{\alpha -\beta }(\gamma _{d})\); moreover,
Thus, \(D_{\beta }f:B_{p,q}^{\alpha }\rightarrow B_{p,q}^{\alpha -\beta }\) is bounded.
If q = ∞, inequality (8.49) can be written as
and then as \(f \in B^\beta _{p, \infty }\), by Lemma 8.18,
and as above, using Lemma 3.5,
□
There is an alternative proof of the fact that D βf ∈ L p(γ d) without using Hardy’s inequality following the same scheme as in the proof of i) [109, Theorem 3.5], and using the inclusion \(B^\alpha _{p,q} \subset B^{\beta +\epsilon }_{p, \infty }\) with β + 𝜖 < k.
Now, let us consider the case of the Bessel derivative.
Theorem 8.20
Let 0 < β < α, 1 ≤ p < ∞ and 1 ≤ q ≤∞ then
\(\mathscr {D}_{\beta }\) is bounded from \(B_{p,q}^{\alpha }(\gamma _{d})\) into \(B_{p,q}^{\alpha -\beta }(\gamma _{d})\).
Proof
Let \(f\in B_{p,q}^{\alpha }(\gamma _{d})\), and set v(x, t) = e −tu(x, t). Then, using Hardy’s inequality (10.100), the fundamental theorem of calculus, and property iii) of forward differences (see Lemma 10.30),
and this implies, using Minkowski’s integral inequality,
Now, using property ii) of forward differences (see Lemma 10.30),
\(\|\varDelta _{r}^{k-1}(v',r)\|{ }_{p,\gamma }\leq \displaystyle \int _{r}^{2r}\int _{v_{1}}^{v_{1}+r}\ldots \int _{v_{k-2}}^{v_{k-2}+r}\|v^{(k)}(\cdot ,v_{k-1})\|{ }_{p,\gamma }dv_{k-1}dv_{k-2}\ldots dv_{2}dv_{1}\) and using Leibniz’s differentiation rule for the product
Then
Therefore,
Thus,
because \(f\in B_{p,q}^{\alpha }(\gamma _{d})\subset B_{p,1}^{\beta -j}(\gamma _d) \) as α > β > β − j ≥ 0, for j ∈{0, …, k − 1}, then \(\mathscr {D}_{\beta }f\in L^{p}(\gamma _d)\).
On the other hand,
Let n be the smaller integer greater than β, i.e., n − 1 ≤ β < n, we have
where w(x, t) = e −tu (n)(x, t). Now, using the fundamental theorem of calculus,
Then, using Hardy’s inequality (10.100) and property iii) of forward differences (see Lemma 10.30),
and according to Minkowski’s integral inequality, we get
Now, using an analogous argument to that above, Lemma 10.30 and Leibniz’s product rule give us
and this implies that
Thus,
Now, if 1 ≤ q < ∞, using (8.50) we have,
For each 1 ≤ j ≤ k, 0 < α − β + k − j ≤ β and using Lemma 3.5
as \(f\in B_{p,q}^{\alpha }(\gamma _{d})\subset B_{p,q}^{\alpha -\beta +(k-j)}(\gamma _{d})\) for any 0 ≤ j ≤ k.
Now, for the case j = 0,
Using Lemma 3.5, and k > β,
because \(f\in B_{p,q}^{\alpha }(\gamma _{d})\) and n + k > α. For the second term, using Lemma 3.5 and Hardy’s inequality (10.101)
because \(f\in B_{p,q}^{\alpha }(\gamma _{d})\).
Therefore, \(\mathscr {D}_{\beta }f\in B_{p,q}^{\alpha -\beta }(\gamma _{d})\). Moreover,
Finally, if q = ∞, from the inequality (8.50)
and then, the argument is essentially similar to the previous case, as in the last part of the proof of Theorem 8.19. □
8.5 Boundedness of Fractional Integrals and Fractional Derivatives on Gaussian Triebel–Lizorkin Spaces
First, we study the boundedness of the Riesz potentials I β on Gaussian Triebel–Lizorkin spaces.
Theorem 8.21
Let α ≥ 0, β > 0, 1 < p < ∞, 1 ≤ q < ∞ then I β is bounded from \(F_{p,q}^{\alpha }(\gamma _{d})\) into \(F_{p,q}^{\alpha +\beta }(\gamma _{d})\).
Proof
Let k > α + β + 1 be an integer fixed and \(f \in F_{p,q}^{\alpha }(\gamma _{d}).\) Using the integral representation of Riesz potentials (8.7), the semigroup property of {P t}t≥0, and the fact that P ∞f(x) is a constant, we get
Then, again using that P ∞f(x) is a constant and the chain rule,
-
i)
Case β ≥ 1: Using (8.51), the change of variables r = t + s, dr = ds, and Hardy’s inequality (10.101), we have
$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \Big(\int_{0}^{+\infty}\Big(t^{k-(\alpha+\beta)}|\frac{\partial^{k}P_{t}(I_{\beta}f)(x)}{\partial t^{k}}|\Big)^{q}\frac{dt}{t}\Big)^{1/q} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \leq \frac{1}{\varGamma(\beta)}\Big(\int_{0}^{+\infty}t^{q(k-(\alpha+\beta))}\Big(\int_{0}^{+\infty}s^{\beta-1} |u^{(k)}(x, t+s) | ds \Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}}\\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad = \frac{1}{\varGamma(\beta)}\Big(\int_{0}^{+\infty}t^{q(k-(\alpha+\beta))}\Big(\int_{t}^{+\infty}(r-t)^{\beta-1} |u^{(k)}(x,r)|dr\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}}\\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \leq\frac{1}{\varGamma(\beta)}\Big(\int_{0}^{+\infty}t^{q(k-(\alpha+\beta))} \Big(\int_{t}^{+\infty}r^{\beta-1} |u^{(k)}(x,r)|dr\Big)^{q} \frac{dt}{t}\Big)^{\frac{1}{q}}\\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \leq \frac{1}{\varGamma(\beta)}\frac{1}{(k-(\alpha+\beta))^{1/q}} \Big(\int_{0}^{+\infty}\Big(r^{k-\alpha} |u^{(k)}(x,r)|\Big)^{q}\frac{dr}{r}\Big)^{\frac{1}{q}}, \end{array} \end{aligned} $$and, therefore,
$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \|\Big(\int_{0}^{+\infty}\Big(t^{k-(\alpha+\beta)}|\frac{\partial^{k}P_{t}(I_{\beta}f)}{\partial t^{k}}|\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}}\Big\|{}_{p,\gamma}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \leq C_{k,\alpha,\beta} \|\Big(\int_{0}^{+\infty}\Big(r^{k-\alpha} |\frac{\partial^{k}P_{r}f}{\partial r^{k}}|\Big)^{q}\frac{dr}{r}\Big)^{\frac{1}{q}}\Big\|{}_{p,\gamma} <\infty , \end{array} \end{aligned} $$because \(f\in F_{p,q}^\alpha \). By (8.12) and the previous estimate,
$$\displaystyle \begin{aligned}\| I_{\beta}f\Big\|{}_{F_{p,q}^{\alpha+\beta}} \leq C \| f\Big\|{}_{F_{p,q}^{\alpha}}.\end{aligned}$$ -
ii)
Case 0 < β < 1: again using (8.51),
$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \Big(\int_{0}^{+\infty}\Big(t^{k-(\alpha+\beta)}|\frac{\partial^{k}P_{t} (I_{\beta}f)(x)}{\partial t^{k}}|\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \leq\frac{1}{\varGamma(\beta)}\Big(\int_{0}^{+\infty}t^{q(k-(\alpha+\beta))}\Big(\int_{0}^{+\infty}s^{\beta} |u^{(k)}(x, t+s)|\frac{ds}{s}\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}} \\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \leq \frac{C}{\varGamma(\beta)}(\int_{0}^{+\infty}t^{q(k-(\alpha+\beta))-1}\Big(\int_{0}^{t}s^{\beta-1} |u^{(k)}(x, t+s)|ds\Big)^{q}dt\Big)^{\frac{1}{q}}\\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \quad +\frac{C}{\varGamma(\beta)} (\int_{0}^{+\infty}t^{q(k-(\alpha+\beta))-1}\Big(\int_{t}^{+\infty}s^{\beta-1} |u^{(k)}(x, t+s)|ds\Big)^{q}dt\Big)^{\frac{1}{q}}\\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad =(I) + (II) . \end{array} \end{aligned} $$Writing \(s^{\beta -1} = s^{\frac {\beta -1}{q}+ \frac {\beta -1}{q'}},\, \frac {1}{q}+\frac {1}{q'} =1,\) and using Hölder’s inequality in the internal integral,
$$\displaystyle \begin{aligned} \begin{array}{rcl} (I) &\displaystyle \leq&\displaystyle \frac{C_{\beta}}{\beta^{\frac{q-1}{q}}} \Big(\int_{0}^{+\infty}t^{q(k-\alpha)-\beta-1}\int_{0}^{t}s^{\beta-1} |u^{(k)}(x, t+s)|{}^{q}ds\, dt\Big)^{1/q}. \end{array} \end{aligned} $$Using the Fubini–Tonelli theorem and using that q(k − β) − β − 1 > 0, as k > β + β + 1, we get
$$\displaystyle \begin{aligned} \begin{array}{rcl} (I) &\displaystyle \leq&\displaystyle \frac{C_{\beta}}{\beta^{\frac{q-1}{q}}}\Big(\int_{0}^{+\infty}s^{\beta-1}\int_{s}^{+\infty}t^{q(k-\alpha)-\beta-1} |u^{(k)}(x, t+s)|{}^{q}dt \,ds\Big)^{1/q}\\ &\displaystyle \leq&\displaystyle \frac{C_{\beta}}{\beta^{\frac{q-1}{q}}}\Big(\int_{0}^{+\infty}s^{\beta-1} \int_{s}^{+\infty}(t+s)^{q(k-\alpha)-\beta-1} |u^{(k)}(x, t+s)|{}^{q}dt \,ds\Big)^{1/q}. \end{array} \end{aligned} $$Then, by the change of variables r = t + s and using Hardy’s inequality (10.101),
$$\displaystyle \begin{aligned} \begin{array}{rcl} (I)&\displaystyle \leq&\displaystyle \frac{C_{\beta}}{\beta^{\frac{q-1}{q}}} \Big(\int_{0}^{+\infty}s^{\beta-1} \int_{2s}^{+\infty}r^{q(k-\alpha)-\beta-1} |u^{(k)}(x, r)|{}^{q}dr \, ds\Big)^{1/q}\\ &\displaystyle \leq&\displaystyle \frac{C_{\beta}}{\beta^{\frac{q-1}{q}}}\Big( \int_{0}^{+\infty}s^{\beta-1} \int_{s}^{+\infty}r^{q(k-\alpha)-\beta-1} |u^{(k)}(x, r)|{}^{q}dr \, ds \Big)^{1/q}\\ &\displaystyle \leq&\displaystyle \frac{C_{\beta}}{\beta} \Big(\int_{0}^{+\infty}\Big(r^{k-\alpha}|u^{(k)}(x, r)|\Big)^{q}\frac{dr}{r}\Big)^{1/q} . \end{array} \end{aligned} $$On the other hand, because β < 1, then t < s implies that s β−1 < t β−1, and by the change of variables r = t + s and according to Hardy’s inequality (10.101), as k > α + β + 1 > α + 1, we obtain
$$\displaystyle \begin{aligned} \begin{array}{rcl} (II) &\displaystyle \leq&\displaystyle C_{\beta} \Big(\int_{0}^{+\infty}t^{q(k-\alpha-1)-1}\Big(\int_{t}^{+\infty} |u^{(k)}(x, t+s)|ds\Big)^{q}dt\Big)^{\frac{1}{q}}\\ &\displaystyle \leq&\displaystyle C_{\beta} \Big(\int_{0}^{+\infty}t^{q(k-\alpha-1)-1}\Big(\int_{2t}^{+\infty} |u^{(k)}(x, r)|dr\Big)^{q}dt\Big)^{\frac{1}{q}}\\ &\displaystyle \leq&\displaystyle C_{\beta} \frac{1}{(k-\alpha-1)^{1/q}}\Big(\int_{0}^{+\infty}\Big(r^{k-\alpha}|\frac{\partial^{k}P_{r}f(x)}{\partial r^{k}}|\Big)^{q}\frac{dr}{r}\Big)^{\frac{1}{q}}. \end{array} \end{aligned} $$Therefore,
$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \Big\|\Big(\int_{0}^{+\infty}\Big(t^{k-(\alpha+\beta)}|\frac{\partial^{k}P_{t}(I_{\beta}f)(x)}{\partial t^{k}}|\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}}\Big\|{}_{p,\gamma}\\ &\displaystyle &\displaystyle \quad \quad \quad \leq C_{k,\alpha,\beta }\Big\|\Big(\int_{0}^{+\infty}\Big(r^{k-\alpha}|\frac{\partial^{k}P_{r}f}{\partial r^{k}}|\Big)^{q}\frac{dr}{r} \Big)^{\frac{1}{q}}\Big\|{}_{p,\gamma} < \infty . \end{array} \end{aligned} $$as \(f\in F_{p,q}^{\alpha }\). Then, using (8.12) and the previous estimate, we get
$$\displaystyle \begin{aligned}\| I_{\beta}f\|{}_{F_{p,q}^{\alpha+\beta}} \leq C \| f\|{}_{F_{p,q}^{\alpha}}.\end{aligned}$$
□
Next, we study the boundedness properties of the Bessel potentials on Triebel–Lizorkin spaces.
Theorem 8.22
Let α ≥ 0, 1 ≤ p, q < ∞ then for every β > 0,
-
i)
\(\mathscr {J}_\beta \) is bounded on \(F_{p,q}^{\alpha }(\gamma _{d})\).
-
ii)
Moreover, \(\mathscr {J}_{\beta }\) is bounded from \(F_{p,q}^{\alpha }(\gamma _{d})\) to \(F_{p,q}^{\alpha +\beta }(\gamma _{d})\).
Proof
-
i)
Let us prove that \(\mathscr {J}_\beta \) is bounded on \(F_{p,q}^{\alpha }(\gamma _{d})\). Using Lebesgue’s dominated convergence theorem, Minkowski’s integral inequality, and iii), we have
$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \left(\int_0^{\infty}(s^{k-\alpha} \left|\frac{\partial^{k}P_{s}}{\partial s^{k}} \Big(\mathscr{J}_{\beta}g\Big)(x) \right |)^{q}\frac{ds}{s} \right)^{1/q} \quad \quad \quad \quad \quad \\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad = \left( \int_0^{\infty} (s^{k-\alpha} \left|\frac{\partial^{k}P_{s}}{\partial s^{k}}\Big(\frac{1}{\varGamma(\beta)} \int_{0}^{+\infty}t^{\beta}e^{-t}P_{t}g(x)\frac{dt}{t}\Big)\right|)^{q}\frac{ds}{s}\right)^{1/q}\\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \leq \frac{1}{\varGamma(\beta)}\int_{0}^{+\infty}t^{\beta}e^{-t}\left( \int_0^{\infty} (s^{k-\alpha} \left|\frac{\partial^{k}P_{s}(P_{t}g)}{\partial s^{k}}(x) \right|) ^{q}\frac{ds}{s} \right) ^{1/q}\frac{dt}{t}, \end{array} \end{aligned} $$then, again using Minkowski’s integral inequality, and iii)
$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \Big\|\left(\int_0^{\infty}\Big(s^{k-\alpha} \left|\frac{\partial^{k}P_{s}}{\partial s^{k}} \Big(\mathscr{J}_{\beta} g\Big) \right |\Big)^{q}\frac{ds}{s} \right)^{1/q}\Big\|{}_{p,\gamma} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \leq \Big\|\frac{1}{\varGamma(\beta)}\int_{0}^{+\infty}t^{\beta}e^{-t}\left( \int_0^{\infty} \Big(s^{k-\alpha} \left|\frac{\partial^{k} P_{s}(P_{t}g)}{\partial s^{k}} \right|\Big)^{q}\frac{ds}{s} \right) ^{1/q}\frac{dt}{t}\Big\|{}_{p,\gamma}\\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \leq \frac{1}{\varGamma(\beta)}\int_{0}^{+\infty} t^{\beta}e^{-t}\Big\|\left( \int_0^{\infty} \Big(s^{k-\alpha} \left|\frac{\partial^{k} P_{s}(P_{t}g)}{\partial s^{k}} \right|\Big)^{q}\frac{ds}{s} \right) ^{1/q}\Big\|{}_{p,\gamma}\frac{dt}{t}\\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad \leq\frac{1}{\varGamma(\beta)}\int_{0}^{+\infty} t^{\beta}e^{-t}\Big\|\left( \int_0^{\infty} \Big(s^{k-\alpha} \left|\frac{\partial^{k}P_{s} g}{\partial s^{k}} \right|\Big) ^{q}\frac{ds}{s} \right)^{1/q}\Big\|{}_{p,\gamma}\frac{dt}{t}\\ &\displaystyle &\displaystyle \quad \quad \quad \quad \quad =\Big\|\left( \int_0^{\infty} \Big(s^{k-\alpha} \left|\frac{\partial^{k} P_{s} g}{\partial s^{k}} \right|\Big) ^{q}\frac{ds}{s} \right) ^{1/q}\Big\|{}_{p,\gamma}. \end{array} \end{aligned} $$Thus,
$$\displaystyle \begin{aligned} \begin{array}{rcl} \|\mathscr{J}_{\beta} g\|{}_{F_{p,q}^{\beta}}&\displaystyle =&\displaystyle \|\mathscr{J}_{\beta}g\|{}_{p,\gamma}+ \Big\|\left(\int_0^{\infty}\Big(s^{k-\alpha} \left|\frac{\partial^{k}P_{s}}{\partial s^{k}} \Big(\mathscr{J}_{\beta} g\Big) \right |\Big)^{q}\frac{ds}{s} \right)^{1/q}\Big\|{}_{p,\gamma} \\ &\displaystyle \leq&\displaystyle \|g\|{}_{p,\gamma}+\Big\|\left( \int_0^{\infty} \Big(s^{k-\alpha} \left|\frac{\partial^{k}P_{s} g}{\partial s^{k}} \right|) ^{q}\frac{ds}{s} \right) ^{1/q}\Big\|{}_{p,\gamma}=\|g\|{}_{F_{p,q}^{\alpha}}. \end{array} \end{aligned} $$ -
ii)
Let k > α + β + 1 be a fixed integer, let \(f\in F_{p,q}^{\alpha }(\gamma _{d}),\) and let \(h={\mathscr {J}}_{\beta }f\). We consider two cases:
-
ii-1)
If β ≥ 1. Taking the change of variables u = t + s and using Hardy’s inequality, we get
$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \Big(\int_{0}^{+\infty}\Big(t^{k-(\alpha+\beta)}|\frac{\partial^{k}P_{t}h(x)}{\partial t^{k}}|\Big)^{q}\frac{dt}{t}\Big)^{1/q} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ &\displaystyle &\displaystyle \quad \quad \quad \leq \frac{1}{\varGamma(\beta)}\Big(\int_{0}^{+\infty}t^{q(k-(\alpha+\beta))}\Big(\int_{0}^{+\infty}s^{\beta}e^{-s} |\frac{\partial^{k}P_{t+s}f(x)}{\partial(t+s)^{k}}|\frac{ds}{s}\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}}\\ &\displaystyle &\displaystyle \quad \quad \quad \leq \frac{1}{\varGamma(\beta)}\Big(\int_{0}^{+\infty}t^{q(k-(\alpha+\beta))}\Big(\int_{t}^{+\infty}(u-t)^{\beta-1} |\frac{\partial^{k}P_{u}f(x)}{\partial u^{k}}|du\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}}\\ &\displaystyle &\displaystyle \quad \quad \quad \leq\frac{1}{\varGamma(\beta)}\Big(\int_{0}^{+\infty}\Big(\int_{t}^{+\infty}u^{\beta-1} |\frac{\partial^{k}P_{u}f(x)}{\partial u^{k}}|du\Big)^{q}t^{q(k-(\alpha+\beta))-1}dt\Big)^{\frac{1}{q}}\\ &\displaystyle &\displaystyle \quad \quad \quad \leq \frac{1}{\varGamma(\beta)}\frac{1}{k-(\alpha+\beta)} \Big(\int_{0}^{+\infty}\Big(u^{k-\alpha} |\frac{\partial^{k}P_{u}f(x)}{\partial u^{k}}|\Big)^{q}\frac{du}{u}\Big)^{\frac{1}{q}}. \end{array} \end{aligned} $$Therefore,
$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \Big\|\Big(\int_{0}^{+\infty}\Big(t^{k-(\alpha+\beta)}|\frac{\partial^{k}P_{t}h}{\partial t^{k}}|\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}}\Big\|{}_{p,\gamma}\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ &\displaystyle &\displaystyle \quad \quad \quad \quad \leq \frac{1}{\varGamma(\beta)(k-(\alpha+\beta))} \|\Big(\int_{0}^{+\infty}\Big(u^{k-\alpha} |\frac{\partial^{k}P_{u}f}{\partial u^{k}}|\Big)^{q}\frac{du}{u}\Big)^{\frac{1}{q}}\Big\|{}_{p,\gamma} <\infty, \end{array} \end{aligned} $$because \(f\in F_{p,q}^{\alpha }(\gamma _d)\). Thus \({\mathscr {J}}_{\beta }f\in F_{p,q}^{\alpha +\beta }(\gamma _d)\).
-
ii-2)
If 0 < β < 1.
$$\displaystyle \begin{aligned} \begin{array}{rcl} &\displaystyle &\displaystyle \Big(\int_{0}^{+\infty}\Big(t^{k-(\alpha+\beta)}|\frac{\partial^{k}P_{t} h(x)}{\partial t^{k}}|\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}} \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \\ &\displaystyle &\displaystyle \quad \quad \quad \leq\frac{1}{\varGamma(\beta)}\Big(\int_{0}^{+\infty}t^{q(k-(\alpha+\beta))}\Big(\int_{0}^{+\infty}s^{\beta}e^{-s} |\frac{\partial^{k}P_{t+s}f(x)}{\partial(t+s)^{k}}|\frac{ds}{s}\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}} \\ &\displaystyle &\displaystyle \quad \quad \quad \leq \frac{C}{\varGamma(\beta)}(\int_{0}^{+\infty}t^{q(k-(\alpha+\beta))-1}\Big(\int_{0}^{t}s^{\alpha-1}e^{-s} |\frac{\partial^{k}P_{t+s}f(x)}{\partial(t+s)^{k}}|ds\Big)^{q}dt\Big)^{\frac{1}{q}}\\ &\displaystyle &\displaystyle \quad \quad \quad +\frac{C}{\varGamma(\beta)} (\int_{0}^{+\infty}t^{q(k-(\alpha+\beta))-1}\Big(\int_{t}^{+\infty}s^{\alpha-1}e^{-s} |\frac{\partial^{k}P_{t+s}f(x)}{\partial(t+s)^{k}}|ds\Big)^{q}dt\Big)^{\frac{1}{q}}\\ &\displaystyle &\displaystyle \quad \quad \quad =I + II. \end{array} \end{aligned} $$Now, e −s < 1 and as β < 1, then s β−1 < t β−1 for t < s.
Hence, again by the change of variables u = t + s and using Hardy’s inequality, we get
$$\displaystyle \begin{aligned} \begin{array}{rcl} II &\displaystyle \leq&\displaystyle \frac{C}{\varGamma(\beta)} \Big(\int_{0}^{+\infty}t^{q(k-\beta-1)-1}\Big(\int_{t}^{+\infty} |\frac{\partial^{k}P_{t+s}f(x)}{\partial(t+s)^{k}}|ds\Big)^{q}dt\Big)^{\frac{1}{q}}\\ &\displaystyle \leq&\displaystyle \frac{C}{\varGamma(\beta)} \Big(\int_{0}^{+\infty}t^{q(k-\beta-1)-1}\Big(\int_{t}^{+\infty} |\frac{\partial^{k}P_{u}f(x)}{\partial u^{k}}|du\Big)^{q}dt\Big)^{\frac{1}{q}}\\ &\displaystyle \leq&\displaystyle \frac{C}{\varGamma(\beta)}\Big(\int_{0}^{+\infty}\Big(u^{k-\alpha}|\frac{\partial^{k}P_{u}f(x)}{\partial u^{k}}|\Big)^{q}\frac{du}{u}\Big)^{\frac{1}{q}} \end{array} \end{aligned} $$On the other hand, using e −s < 1 again,
$$\displaystyle \begin{aligned} \begin{array}{rcl} I^q &\displaystyle \leq&\displaystyle \frac{C}{\varGamma(\beta)}\int_{0}^{+\infty}t^{q(k-(\alpha+\beta))-1}\Big(\int_{0}^{t}s^{\beta-1} |\frac{\partial^{k}P_{t+s}f(x)}{\partial(t+s)^{k}}|ds\Big)^{q}dt\\ &\displaystyle =&\displaystyle \frac{C}{\varGamma(\beta)\beta^{q}}\int_{0}^{+\infty}t^{q(k-\alpha)-1}\Big(\frac{\beta }{ t^{\beta}}\int_{0}^{t}s^{\beta-1} |\frac{\partial^{k}P_{t+s}f(x)}{\partial(t+s)^{k}}|ds\Big)^{q}dt \end{array} \end{aligned} $$Now, as β > 0, \(\displaystyle \int _{0}^{t}s^{\beta -1}ds=\frac {t^{\beta }}{\beta },\) then using Jensen’s inequality for the probability measure \(\displaystyle \frac {\beta }{t^{\beta }}s^{\beta -1} ds\) and Fubini’s theorem, we get
$$\displaystyle \begin{aligned} \begin{array}{rcl} I^q&\displaystyle \leq&\displaystyle \frac{C}{\varGamma(\beta)\beta^{q}}\int_{0}^{+\infty}t^{q(k-\alpha)-1}\Big(\frac{\beta }{ t^{\beta}}\int_{0}^{t}s^{\beta-1} |\frac{\partial^{k}P_{t+s}f(x)}{\partial(t+s)^{k}}|{}^{q}ds\Big)dt\\ &\displaystyle =&\displaystyle \frac{C}{\varGamma(\beta)\beta^{q-1}}\int_{0}^{+\infty}s^{\beta-1}\Big(\int_{s}^{+\infty}t^{q(k-\alpha)-\beta-1} |\frac{\partial^{k}P_{t+s}f(x)}{\partial(t+s)^{k}}|{}^{q}dt\Big)ds\\ &\displaystyle \leq&\displaystyle \frac{C}{\varGamma(\beta)\beta^{q-1}}\int_{0}^{+\infty}s^{\beta-1}\Big(\int_{s}^{+\infty}(t+s)^{q(k-\alpha)-\beta-1} |\frac{\partial^{k}P_{t+s}f(x)}{\partial(t+s)^{k}}|{}^{q}dt\Big)ds \end{array} \end{aligned} $$as q(k − α) − β − 1 > 0, because 0 < β < 1. Finally, again taking the change of variables u = t + s and using Hardy’s inequality, we get
$$\displaystyle \begin{aligned} \begin{array}{rcl} I^q&\displaystyle \leq&\displaystyle \frac{C}{\varGamma(\beta)\beta^{q-1}}\int_{0}^{+\infty}s^{\beta-1}\Big(\int_{2s}^{+\infty}u^{q(k-\alpha)-\beta-1} |\frac{\partial^{k}P_{u}f(x)}{\partial u^{k}}|{}^{q}du\Big)ds\\ &\displaystyle \leq&\displaystyle \frac{C}{\varGamma(\beta)\beta^{q-1}}\int_{0}^{+\infty}s^{\beta-1}\Big(\int_{s}^{+\infty}u^{q(k-\alpha)-\beta-1} |\frac{\partial^{k}P_{u}f(x)}{\partial u^{k}}|{}^{q}du\Big)ds\\ &\displaystyle \leq&\displaystyle \frac{C}{\varGamma(\beta)\beta^{q-1}}\int_{0}^{+\infty}\Big(u^{k-\beta}|\frac{\partial^{k}P_{u}f(x)}{\partial u^{k}}|\Big)^{q}\frac{du}{u}. \end{array} \end{aligned} $$Hence,
$$\displaystyle \begin{aligned} \begin{array}{rcl} \Big\|\Big(\int_{0}^{+\infty}&\displaystyle &\displaystyle \Big(t^{k-(\alpha+\beta)}|\frac{\partial^{k}P_{t}h}{\partial t^{k}}|\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}}\Big\|{}_{p,\gamma}\\ &\displaystyle \leq&\displaystyle C_{k,\alpha,\beta }\Big\|\Big(\int_{0}^{+\infty}\Big(u^{k-\alpha}|\frac{\partial^{k}P_{u}f}{\partial u^{k}}|\Big)^{q}\frac{du}{u} \Big)^{\frac{1}{q}}\Big\|{}_{p,\gamma} < \infty. \end{array} \end{aligned} $$Thus, \({\mathscr {J}}_{\beta }f\in F_{p,q}^{\alpha +\beta }(\gamma _d)\), for 0 < β < 1.
Therefore, in both cases we have,
$$\displaystyle \begin{aligned} \begin{array}{rcl} \|{\mathscr{J}}_{\beta}f\|{}_{F_{p,q}^{\alpha+\beta}}&\displaystyle =&\displaystyle \|{\mathscr{J}}_{\beta}f\|{}_{p,\gamma}+ \|\Big(\int_{0}^{+\infty}\Big(t^{k-(\alpha+\beta)}|\frac{\partial^{k}P_{t}{\mathscr{J}}_{\beta}f}{\partial t^{k}}|\Big)^{q}\frac{dt}{t}\Big)^{\frac{1}{q}}\Big\|{}_{p,\gamma} \\ &\displaystyle \leq &\displaystyle C_{\beta} \|f\|{}_{p,\gamma}+C_{k,\alpha,\beta}\|\Big(\int_{0}^{+\infty}\Big(u^{k-\alpha}|\frac{\partial^{k}P_{u}}{\partial u^{k}}|\Big)^{q}\frac{du}{u} \Big)^{\frac{1}{q}}\Big\|{}_{p,\gamma}\\ &\displaystyle \leq&\displaystyle C_{k,\alpha,\beta }\|f\|{}_{F_{p,q}^{\alpha}}. \end{array} \end{aligned} $$
-
ii-1)
□
Now, we study the boundedness of the Riesz fractional derivatives and of the Bessel fractional derivatives on Triebel–Lizorkin spaces. Again, because they require different techniques, we consider two cases:
-
The bounded case, 0 < β < α < 1.
-
The unbounded case 0 < β < α.
Let us start with the bounded case for the Riesz derivative.
Theorem 8.23
Let 1 ≤ p, q < ∞, for 0 < β < α < 1, D β is bounded from \(F_{p,q}^{\alpha }(\gamma _{d})\) into \(F_{p,q}^{\alpha -\beta }(\gamma _{d})\).
Proof
Let \(f\in F_{p,q}^{\alpha }(\gamma _{d})\), using the fundamental theorem of calculus, and Hardy’s inequality (10.100) with p = 1,
Thus,
because \( F_{p,q}^{\alpha }(\gamma _{d})\subset F_{p,1}^{\beta }(\gamma _{d}) \) (α > β and q ≥ 1). Now, using an analogous argument using Hardy’s inequality (10.100) with p = 1,
This implies that
Writing \(r^{-\beta } = r^{\frac {-\beta }{q}+ \frac {-\beta }{q'}},\, \frac {1}{q}+\frac {1}{q'} =1,\) and using Hölder’s inequality in the internal integral, we have
Then, according to the Fubini–Tonelli theorem, we get
It is easy to prove that (2 − α)q + β − 2 > −1. We need to study two cases:
Case #1 – if (2 − α)q + β − 2 < 0: as r < t and taking the change of variables w = t + r, we have
Then using Hardy’s inequality (10.101) as (2 − α)q − 1 > 0
Case #2 – if (2 − β)q + β − 2 ≥ 0: taking the change of variables w = t + r, we get
and using Hardy’s inequality (10.101),
Therefore, in both cases we have
To estimate (II), observe that r −β < t −β, for r > t and β > 0, then using the same argument as before to estimate (I) case #1, taking the change of variables w = t + r, and using Hardy’s inequality (10.101), so that
Finally,
as \(f\in F_{p,q}^{\beta }(\gamma _{d})\). Using the previous estimate and (8.52)
□
In the following theorem, we study the boundedness of the Bessel fractional derivative on Triebel–Lizorkin spaces for the bounded case 0 < β < α < 1.
Theorem 8.24
Let 0 < β < α < 1, 1 ≤ p, q < ∞ then \({\mathscr D}^{\beta }\) is bounded from \(F_{p,q}^{\alpha }(\gamma _{d})\) into \(F_{p,q}^{\alpha -\beta }(\gamma _{d})\).
Proof
Let f ∈ L p(γ d), using the fundamental theorem of calculus, we can write
Now, using Hardy’s inequality (10.100) with p = 1 in both integrals, we have
Thus,
Therefore, if \(f\in F_{p,q}^{\alpha }(\gamma _{d})\), we get
because \( F_{p,q}^{\alpha }(\gamma _{d})\subset F_{p,1}^{\beta }(\gamma _{d}) \), as α > β, and q ≥ 1.
Using a similar argument to that above, the fundamental theorem of calculus and Hardy’s inequality (10.100) with p = 1, we get
Therefore,
Then, we have
Now, the first term can be estimated as in the proof of Theorem 3, estimates (8.53) and (8.54), so that
which is finite as \(f\in F_{p,q}^{\alpha }(\gamma _{d}).\) For the second term, we have
as \( F_{p,q}^{\alpha }(\gamma _{d})\subset F_{p,q}^{\alpha -\beta }(\gamma _{d}) \); thus,
Therefore, \(\mathscr {D}^{\beta }f\in F_{p,q}^{\alpha -\beta }(\gamma _{d})\) and moreover, using the previous estimate and (8.55)
□
To study the general case for fractional derivatives (removing the condition that the indexes must be less than 1), we need to consider forward differences again. Also, we need the generalized version of Hardy’s inequality (see Theorem 10.26 in the Appendix, and also the following technical results):
Lemma 8.25
For any positive integer k,
For the proof of this result, see Lemma 10.30 in the Appendix, or [109] Lemma 3.1, ii).
Lemma 8.26
Let t ≥ 0, β > 0 and let k be the smallest integer greater than β, and let f differentiable up to order k, then
where \(C_{\beta ,k}=\displaystyle \int _{0}^{1}\ldots \int _{0}^{1}(v_{1}+\ldots +v_{k})^{\beta -k}dv_{1}\ldots dv_{k}\)
Proof
Using Lemma10.26, with p = 1, and Lemma 8.25 we have,
taking r = s(v 1 + … + v k) then dr = (v 1 + … + v k)ds,
Therefore,
where \(C_{\beta ,k}=\displaystyle \int _{0}^{1}\ldots \int _{0}^{1}(v_{1}+\ldots +v_{k})^{\beta -k}dv_{1}\ldots dv_{k}<\infty .\) □
We need to use (8.47)
and (8.48)
Let us consider the unbounded case 0 < β < α for the Riesz derivative,
Theorem 8.27
Let 0 < β < α, 1 ≤ p, q < ∞, then D β is bounded from \(F_{p,q}^{\alpha }(\gamma _{d})\) into \(F_{p,q}^{\alpha -\beta }(\gamma _{d})\).
Proof
Let \(f\in F_{p,q}^{\alpha }(\gamma _{d})\), using (8.47), (8.48) and Lemma 8.26,
Then
because \( F_{p,q}^{\alpha }(\gamma _{d})\subset F_{p,1}^{\beta }(\gamma _{d})\), (α > β and 1 ≤ q < ∞).
Let \(n\in \mathbb {N}, n>\alpha \); using Lemma 10.30(8.46) and Lemma 8.26, we get
Therefore,
which is inequality (8.53) for n = k = 1. The rest of the proof follows the argument used in Theorem 8.23, so that
taking L p(γ)-norm both sides of the inequality, we get the result. □
Finally, the following result extends Theorem 8.24 to the general case 0 < β < α:
Theorem 8.28
Let 0 < β < α, 1 < p < ∞ and 1 ≤ q < ∞, then \(\mathscr {D}^{\beta }\) is bounded from \(F_{p,q}^{\alpha }(\gamma _{d})\) into \(F_{p,q}^{\alpha -\beta }(\gamma _{d})\).
Proof
Let \(f\in F_{p,q}^{\alpha }(\gamma _{d})\), k be an integer such that k − 1 ≤ β < k and v(x, r) = e −ru(x, r), using Lemma 8.26 and Leibniz’s differentiation rule for the product
then
because \(F_{p,q}^{\alpha }(\gamma _{d})\subset F_{p,1}^{\beta -j}(\gamma _{d})\), as α > β ≥ β − j ≥ 0, for j = 0, …, k − 1 and q ≥ 1.
Now, let \(n\in \mathbb {N}, n>\alpha \) and w(x, t) = e −tu (n)(x, t), using Lemma 8.26, we get
Now, using Leibniz’s rule, \(w^{(k)}(x,r)=\displaystyle \sum _{j=0}^{k}\binom {k}{j}(-1)^{j}e^{-r}u^{(k+n-j)}(x,r)\) and then
for all r > 0. Thus,
Therefore,
For 0 ≤ j ≤ k − 1, we have β − j ≥ β − (k − 1) ≥ 0, and taking into account that each term of the above sum is bounded by the left side of the inequality (8.56), with k replaced by k − j and β replaced by β − j, we get that
for 0 ≤ j ≤ k − 1. Unfortunately, the remaining case j = k requires a special argument that uses the following known inequality for the Poisson–Hermite semigroup:
(see Lemma 3.4; see also [226, Lemma 1], or [224]). Then
We first consider the case k ≤ β. The term (I) is estimated as term (I) in the proof of Theorem 8.23.
Because β ≥ k − 1, taking the change of variables v = t + s, we get
Therefore, using Hardy’s inequality (10.101),
Next, consider the case k > α. In this case, using inequality (8.57) and Hardy’s inequality (10.100), we have
On the other hand,
Using the usual argument the change of variables v = t + s and Hardy’s inequality (10.101), we get
Finally, using inequality (8.57) again, we get
Hence, in both cases, we get that
as \(f\in F_{p,q}^{\alpha }(\gamma _{d})\). Therefore, \(\mathscr {D}^{\beta }f\in F_{p,q}^{\alpha -\beta }(\gamma _{d})\) and moreover,
□
8.6 Notes and Further Results
-
1.
Observe that the arguments given in the proofs of theorems in this chapter are still valid in the classical case taking the Poisson integral; therefore, they are alternative proofs to those given in E. Stein’s book [252].
-
2.
Moreover, if instead of considering the Ornstein–Uhlenbeck operator and the Poisson–Hermite semigroup, we consider the Laguerre differential operator in \(\mathbb R^d_{+},\) for α = (α 1, ⋯ , α d) a multi-index,
$$\displaystyle \begin{aligned} \mathscr{L}^{\alpha} = \sum^{d}_{i=1} \Bigg[ x_i \frac{\partial^2}{\partial x^2_i} + (\alpha_i + 1 - x_i ) \frac{\partial}{\partial x_i} \Bigg] \end{aligned} $$(8.58)and the corresponding Poisson–Laguerre semigroup, or if we consider the Jacobi differential operator in (−1, 1)d,
$$\displaystyle \begin{aligned} \mathscr{L}^{\alpha,\beta} = - \sum^{d}_{i=1} \Bigg[ (1-x_i^2)\frac{\partial^2}{\partial x_i^2} + (\beta_i -\alpha_i-\left(\alpha_i +\beta_i +2\right)x_i) \frac{\partial}{\partial x_i} \Bigg], \end{aligned} $$(8.59)and the corresponding Poisson–Jacobi semigroup (for more details, we refer the reader to [279]), the arguments are completely analogous. To see this, it is more convenient to use the representation of P t in terms of the one-sided stable measure \(\mu ^{(1/2)}_t(ds)\) and to write Lemma 3.3 in those terms (see [225]). In other words, we can define in an analogous manner Laguerre–Lipschitz spaces and Jacobi–Lipschitz spaces, and prove the corresponding notions of fractional integrals and fractional derivatives (see [25, 117]).
-
3.
Following similar arguments to those given in Chapter 7, we can define in an analogous manner Laguerre–Besov–Lipschitz spaces and Jacobi–Besov–Lipschitz spaces, in addition to Laguerre–Triebel–Lizorkin spaces and Jacobi–Triebel–Lizorkin spaces, and then prove that the corresponding notions of fractional integrals and fractional derivatives of corresponding operators \({\mathscr L}^{\alpha ,\beta }\) and \({\mathscr L}^{\alpha }\) behave similarly.
-
4.
In [146], G. E. Karadzhov & M. Milman show that the Gaussian Riesz potentials I β maps \(L^p(\log L)_a\) continuously into \(L^p(\log L)_{a+\beta },\) for 1 < p < ∞ and \(a \in \mathbb {R}.\) The proof is using extrapolation in an abstract setting. Moreover, their proof is in fact valid for any hypercontractive semigroup (see [146, Theorem 5.7]).
-
5.
We can also consider alternative Riesz potentials, alternative Bessel potential, alternative Riesz and alternative Bessel fractional derivatives using the same formulas as before, but with respect to \(\overline {L},\) the alternative Ornstein–Uhlenbeck operator (2.14). This case is actually simpler, as 0 is not a eigenvalue of \(\overline {L}.\) For instance, for β > 0 the alternative Riesz potential \(\overline {I}_\beta \) can be defined as
$$\displaystyle \begin{aligned} \overline{I}_\beta=(-\overline{L})^{-\beta/2}, \end{aligned} $$(8.60)meaning that any multi-index ν such that |ν| > 0 its action on the Hermite polynomial H ν is
$$\displaystyle \begin{aligned} \overline{I}_\beta {\mathbf{H}}_\nu(x)=\frac 1{(\left| \nu \right|+d)^{\beta/2}}{\mathbf{H}}_\nu(x). \end{aligned} $$(8.61)\(\overline {I}_\beta \) has the following integral representation, using the fact that \(\overline {L}\) is the infinitesimal generator of the semigroup \(\{T^{(d)}_t\}_t = \{e^{-td}T_t\}_t\), the d-translated Ornstein–Uhlenbeck semigroup (2.78),
(8.62)The integral representation (8.62) is crucial to getting the L p(γ d)-boundedness results of some of the Gaussian singular integrals considered in Chapter 9.
Similar representations can be found for Bessel potentials and the fractional derivatives associated with \(\overline {L}.\)
-
6.
In [164], alternate representations of I β and D β are obtained.
Proposition 8.29
Suppose f \(\in C_B^2(\mathbb {R}^d)\) such that \(\int _{\mathbb {R}^d}f(y)\gamma _d(dy)=0\) , then
$$\displaystyle \begin{aligned} D^\beta f =\frac {1}{\beta c_{\beta}}\int_0^{\infty} t^{-\beta}\frac{\partial}{\partial t}P_t f dt, \ \ \ 0<\beta<1, \end{aligned} $$(8.63)$$\displaystyle \begin{aligned} I_\beta f =-\frac {1}{\beta \varGamma(\beta)}\int_0^{\infty} t^\beta \frac{\partial}{\partial t}P_t f dt,\ \ \ \beta >0. \end{aligned} $$(8.64)Proof
Let us start proving (8.63). Integrating by parts in (8.26) ,we get
$$\displaystyle \begin{aligned} \begin{array}{rcl} D_{\beta} f(x)&\displaystyle =&\displaystyle \frac{1}{c_\beta} \lim_{\substack{a \to 0^{+} \\ b \to \infty}} \int_a^b t^{-\beta-1}\left( P_t f(x)-f(x) \right) dt\\ &\displaystyle =&\displaystyle \frac{1}{c_\beta}\lim_{\substack{a \to 0^{+} \\ b \to \infty}} \Biggl\{\frac{t^{-\beta}}{-\beta}\left(P_t f(x)-f(x)\right)\Big|{}^{b}_{a}+\frac{1} {\beta}\int_a^b t^{-\beta}\frac{\partial}{\partial t}P_t f(x)dt \Biggr\}\\ &\displaystyle =&\displaystyle \frac{1}{\beta c_{\beta}} \int_0^{\infty} t^{-\beta}\frac{\partial}{\partial t}P_t f(x)dt \end{array} \end{aligned} $$because, using (3.28) and (3.29), we have
$$\displaystyle \begin{aligned} \lim_{ b \to \infty}\left( \frac{P_b f(x)-f(x)}{b^{\beta}}\right)=0 \end{aligned}$$and
$$\displaystyle \begin{aligned} \begin{array}{rcl} \lim_{ a \to 0^{+}}\left| \frac{P_a f(x)-f(x)}{a^{\beta}}\right|&\displaystyle \leq&\displaystyle \lim_{ a \to 0^{+}}\frac{1}{a^\beta} \int_0^a \left| \frac{\partial}{\partial s} P_s f(x)\right|ds\\ &\displaystyle \leq&\displaystyle C_{d,f}(d+|x|)\lim_{ a \to 0^{+}} \frac{1-e^{-a}}{a^\beta}=0. \end{array} \end{aligned} $$Let us prove now (8.64). Again, by integrating by parts, we have
$$\displaystyle \begin{aligned} \begin{array}{rcl} I_{\beta} f(x)&\displaystyle =&\displaystyle \frac{1}{\varGamma(\beta)} \lim_{\substack{a \to 0^{+} \\ b \to \infty}} \int_a^b t^{\beta-1} P_t f(x) dt\\ &\displaystyle =&\displaystyle \frac{1}{\varGamma(\beta)}\lim_{\substack{a \to 0^{+} \\ b \to \infty}} \Biggl\{\frac{t^{\beta}}{\beta}P_t f(x)\Big|{}^{b}_{a}-\frac{1}{\beta}\int_a^b t^{\beta}\frac{\partial}{\partial t}P_t f(x)dt \Biggr\} \\ &\displaystyle =&\displaystyle -\frac{1}{\beta \varGamma(\beta)} \int_0^{\infty} t^{\beta}\frac{\partial}{\partial t}P_t f(x)dt, \end{array} \end{aligned} $$because, using the previous result
$$\displaystyle \begin{aligned} \lim_{ b \to \infty}\left| P_b f(x)b^{\beta}\right|\leq C_{d,f}(d+|x|)\lim_{ b \to \infty}b^{\beta}e^{-b}=0 \end{aligned}$$and
$$\displaystyle \begin{aligned} \lim_{ a \to 0^{+}}\left| P_a f(x) a^{\beta}\right|=0. \end{aligned}$$□
Observe that because the previous proposition holds for f = H β, the Hermite polynomial of order β, |β| > 0, then it holds for any non-constant polynomial f such that \(\int _{\mathbb {R}^d}f(y)\gamma _d(dy)=0\).
By using (3.3) and (8.63), D β can be expressed explicitly as
$$\displaystyle \begin{aligned} D_\beta f(x)=\int_{\mathbb{R}^d}K_\beta (x,y)f(y)dy, \end{aligned}$$where,
$$\displaystyle \begin{aligned} \begin{array}{rcl} K_\beta (x,y)&\displaystyle =&\displaystyle C_d\int_0^{\infty}\int_0^1 t^{-\beta}e^{t^2/4log r}(-log r)^{1/2} \frac{e^{-\frac{\left| y-rx \right|{}^2}{1-r^2}}}{(1-r^2)^{d/2}} \\ &\displaystyle &\displaystyle \ \ \quad \quad \times \left( \frac{2r^2 \left|y-rx \right|{}^2-2r(1-r^2) \langle y-rx,x \rangle -dr^2(1-r^2)}{(1-r^2)^2} \right) \frac{dr}{r}dt. \end{array} \end{aligned} $$Now, let us write
$$\displaystyle \begin{aligned} q_t(x,y)=-t\frac{\partial}{\partial t}\left(\int_0^1t\frac{\exp \left( t^2/4\log r\right) }{(-\log r)^{3/2}}\frac{\exp \left( \frac{-\left| y-rx\right|{}^2}{1-r^2}\right) }{(1-r^2)^{d/2}}\frac{dr}r \right), \end{aligned} $$(8.65)and define the operator Q t as
$$\displaystyle \begin{aligned} Q_t f(x)=-t\frac{\partial}{\partial t}P_t f(x)=\int_{\mathbb{R}^d} q_t(x,y)f(y)dy. \end{aligned} $$(8.66)Following [108] we immediately get from (8.63) and (8.64) the following formulas:
Corollary 8.30
Suppose f \(\in C_B^2(\mathbb {R}^d)\) such that \(\int _{\mathbb {R}^d}f(y)\gamma _d(dy)=0\) . Then, we have
$$\displaystyle \begin{aligned} -\beta D_\beta f =\frac {1}{c_{\beta}}\int_0^{\infty} t^{-\beta-1}Q_t f dt, \, 0<\beta<1, \end{aligned} $$(8.67)$$\displaystyle \begin{aligned} \beta I_\beta =\frac {1}{\varGamma(\beta)}\int_0^{\infty} t^{\beta-1}Q_t f dt, \, \beta >0. \end{aligned} $$(8.68) -
7.
An interesting use of the family {Q t} is that it allows us to give a version of A. P. Calderón’s reproducing formula for the Gaussian measure; see [164].
Theorem 8.31
-
i)
Suppose f ∈ L 1(γ d) such that \(\int _{\mathbb {R}^d}f(y)\gamma _d(dy)=0\), then we have
$$\displaystyle \begin{aligned} f =\int_0^{\infty} Q_t f\frac{dt}{t}. \end{aligned} $$(8.69) -
ii)
Suppose f a polynomial such that \(\int _{\mathbb {R}^d}f(y)\gamma _d(dy)=0\) , then we have
$$\displaystyle \begin{aligned} f =C_\beta \int_0^{\infty} \int_0^{\infty} t^{-\beta}s^\beta Q_t\left( Q_s f \right)\frac{ds}{s}\frac{dt}{t} \ \ \ 0<\beta<1. \end{aligned} $$(8.70)Also,
$$\displaystyle \begin{aligned} \int_0^{\infty} \int_0^{\infty} t^{-\beta}s^\beta Q_t\left( Q_s f \right)\frac{ds}{s}\frac{dt}{t}=d_\beta \int_0^{\infty} u\frac{\partial^2}{\partial u^2}P_u f du. \end{aligned} $$(8.71)
Formula (8.70) is the aforementioned version of Calderón’s reproducing formula for the Gaussian measure.
Proof
-
i)
Using (3.28) and (3.29) we have,
$$\displaystyle \begin{aligned}\int_0^{\infty}Q_t f\frac{dt}{t}=\lim_{\substack{a \to 0^{+} \\ b \to \infty}}(-\int_a^b \frac{\partial}{\partial t}P_t f dt) =\lim_{\substack{a \to 0^{+} \\ b \to \infty}}(-P_t f) \Big|{}^{b}_{a} = f. \end{aligned}$$ -
ii)
Let us prove (8.70), given f, a polynomial such that \(\int _{\mathbb {R}^d}f(y)\gamma _d(dy)=0\), by Corollary 8.30, we have
$$\displaystyle \begin{aligned} D_{\beta} \left( I_{\beta}f \right)=\frac{1}{\beta c_{\beta}}\int_0^{\infty}t^{-\beta-1} Q_t\left(I_{\beta} f \right)dt. \end{aligned}$$Now, using the definition of Q t and Fubini’s theorem, we have
$$\displaystyle \begin{aligned} Q_t\left(I_{\beta}f \right) =\frac{1}{\beta \varGamma(\beta)}\int_{\mathbb{R}^d}\int_0^{\infty}s^{\beta-1}Q_s(f)(y)dsdy. \end{aligned}$$Again, using the definition of Q s, we obtain
$$\displaystyle \begin{aligned} f=D_{\beta} \left( I_{\beta}f \right) =d_\beta\int_0^{\infty}\int_0^{\infty} t^{-\beta-1} s^{\beta-1}Q_t\left(Q_s f \right) dsdt. \end{aligned}$$To show (8.71), we see that from (8.66)
$$\displaystyle \begin{aligned} Q_t\left(Q_s f\right)(x)=ts\frac{\partial}{\partial t}\frac{\partial}{\partial s}P_{t+s} f(x). \end{aligned}$$But
$$\displaystyle \begin{aligned} \frac{\partial}{\partial t}\frac{\partial}{\partial s}P_{t+s} f(x)=\frac{\partial^2}{\partial u^2}P_u f(x)\Big|{}_{u=t+s}, \end{aligned}$$then
$$\displaystyle \begin{aligned} \begin{array}{rcl} \int_0^{\infty}\int_0^{\infty} t^{-\beta-1} s^{\beta-1}Q_t\left(Q_s f\right)dsdt&\displaystyle =&\displaystyle \int_0^{\infty}\int_0^{\infty} t^{-\beta}s^{\beta}\frac{\partial^2}{\partial u^2}P_u f\Big|{}_{u=t+s}dsdt\\ &\displaystyle =&\displaystyle d_\beta \int_0^{\infty}u\frac{\partial^2}{\partial u^2}P_u fdu, \end{array} \end{aligned} $$where \(d_\beta =\frac {B(-\beta +1,\beta +1)}{a_\beta c_\beta }\), B(−β + 1, β + 1) being the beta function of parameter (−β + 1, β + 1). □
-
i)
-
8.
Also, in [117], P. Graczyk, J. J. Loeb, I. López, A. Nowak, and W. Urbina obtained an analog of A. P. Calderón’s reproducing formula for the Laguerre case .
-
9.
Using more abstract approaches to Besov and Triebel–Lizorkin spaces associated with a general differential operator, as in [154], many of the results contained in this chapter would follow from the functional calculus for the Ornstein–Uhlenbeck operator.
Change history
18 December 2019
This book was inadvertently published without updating the following corrections:
Notes
- 1.
For d = 1 the function f, defined above, is the same as that used by H. Pollard in his famous counterexample in [230].
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Urbina-Romero, W. (2019). Gaussian Fractional Integrals and Fractional Derivatives, and Their Boundedness on Gaussian Function Spaces. In: Gaussian Harmonic Analysis. Springer Monographs in Mathematics. Springer, Cham. https://doi.org/10.1007/978-3-030-05597-4_8
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