Weighted Fourier Inequalities via Rearrangements

Abstract

The method of using rearrangements to give sufficient conditions for Fourier inequalities between weighted Lebesgue spaces is revisited, a comparison between two known sufficient conditions is completed, and the method is extended to provide sufficient conditions for a new range of indices. When $1<q<p<\infty$, simple conditions on weights ensure that the Fourier transform will map a weighted $L^p$ space into a weighted $L^q$ space. These are established in Theorems 1 and 4 of Benedetto and Heinig (J Fourier Anal Appl 9(1):1–37, 2003). The proofs apply when $2<q<p$ and $1<q<p<2$ but not in the remaining case, $1<q<2<p$. Here, counterexamples are given to show that these simple conditions are no longer sufficient when $1<q<2<p$. Also, various additional conditions are presented, any of which will restore sufficiency in that case.

1 Introduction

Fix a positive integer n. For which indices p, q and which weights U and W does there exist a constant $C<\infty$ such that the Fourier inequality

$$\begin{aligned} \bigg (\int _{{\mathbb {R}}^n} |{\hat{f}}(y)|^qU(y)\,dy\bigg )^{1/q} \le C\bigg (\int _{{\mathbb {R}}^n}|f(x)|^pW(x)\,dx\bigg )^{1/p} \end{aligned}$$

(1.1)

holds for all $f\in L^1({\mathbb {R}}^n)$? If there is such a constant then the Fourier transform is a bounded operator from a dense subspace of $L^p_W({\mathbb {R}}^n)$ into $L^q_U({\mathbb {R}}^n)$. Here the Fourier transform of a function in $L^1({\mathbb {R}}^n)$ is defined as an integral operator by

$$\begin{aligned} {\hat{f}}(x)=\int _{{\mathbb {R}}^n}e^{-2\pi ix\cdot y}f(y)\,dy. \end{aligned}$$

Using the inequality above, if C is finite, a standard argument will extend the integral operator to a linear operator on the whole space $L^p_W({\mathbb {R}}^n)$. So for our purposes it will suffice to restrict attention to functions in $L^1({\mathbb {R}}^n)$.

In the case $1<p\le q<\infty$, the question was addressed by three nearly simultaneous but different approaches appearing in 1983–1984, all involving rearrangements of the weights. See [2, 9, 12, 13, 17, 18]. The case $1< p<\infty , 0<q<p$ has been considered as well, in [4, 10] and later in [3], but clarification and improvement of this case is still possible. This is the object of the present paper.

The non-increasing rearrangement of $f^*$ of a $\mu$-measurable function f is defined for $t>0$ by,

$$\begin{aligned} f^*(t)=\inf \{\alpha>0:\mu _f(\alpha )\le t\},\quad \text {where}\quad \mu _f(\alpha )=\mu \{x:|f(x)|>\alpha \}. \end{aligned}$$

We refer to [5] for standard properties of the non-increasing rearrangement, just mentioning two: Hardy’s Lemma states that if $f_1$ and $f_2$ are non-negative measurable functions on $(0,\infty )$ then

$$\begin{aligned} \int _0^\infty f_1g\le \int _0^\infty f_2g \end{aligned}$$

holds for all non-negative non-decreasing g if and only if it holds with $g=\chi _{(t,\infty )}$ for each $t>0$. The Hardy–Littlewood–Polya Inequality shows that for any non-negative measurable f, g defined on ${\mathbb {R}}^n$,

$$\begin{aligned} \int _{{\mathbb {R}}^n}fg\le \int _0^\infty f^*g^*. \end{aligned}$$

Given non-negative, Lebesgue measurable functions U and W on ${\mathbb {R}}^n$, define u and w by $u=U^*$ and $1/w=(1/W)^*$. The functions u and w are defined on $(0,\infty )$, they take values in $[0,\infty ], u$ is non-increasing and w is non-decreasing. The rearranged Fourier inequality below, expressed in terms of the weights u and w, gives a sufficient condition for (1.1). For a proof see, for example, the proof of [3, Theorem 1].

Proposition 1.1

If $p,q\in (0,\infty )$ and

$$\begin{aligned} \bigg (\int _0^\infty ({\hat{f}})^*(t)^qu(t)\,dt\bigg )^{1/q} \le C\bigg (\int _0^\infty f^*(s)^pw(s)\,ds\bigg )^{1/p}, \quad f\in L^1({\mathbb {R}}^n),\quad \quad \end{aligned}$$

(1.2)

then (1.1) holds with the same constant C.

Working with (1.2) instead of (1.1) is the essence of the “rearrangement” approach to Fourier inequalities in weighted Lebesgue spaces. It has proven to be a powerful method but it is not the only approach. See, for example [25, 27].

The strategy we adopt for proving inequality (1.1) via inequality (1.2) with monotone u and w begins by using the mapping properties of the Fourier transform to find weighted Hardy-type inequalities that imply (1.2). This is done in Sect. 2. Then, known weight characterizations for the Hardy-type inequalities are employed to find sufficient conditions on the weights and indices for (1.2) and hence for (1.1) to hold. This is carried out in Sect. 3.

The mapping properties we use here are universally known: The Fourier transform maps $L^1$ to $L^\infty$ and maps $L^2$ to $L^2$. In [11], Jodiet and Torchinsky showed that for any operator with these mapping properties there exists a constant D such that

$$\begin{aligned} \int _0^x({\hat{f}})^*(t)^2\,dt\le D\int _0^x\bigg (\int _0^{1/t}f^*\bigg )^2\,dt \end{aligned}$$

(1.3)

for all $f\in L^1+L^2$.

The main Hardy inequality we use is the same one employed, either explicitly or implicitly, in the various 1983–1984 papers [2, 9, 12, 13, 17, 18]:

$$\begin{aligned} \bigg (\int _0^\infty \bigg (\int _0^{1/t} f\bigg )^qu(t)\,dt\bigg )^{1/q} \le C_1\bigg (\int _0^\infty f^pw\bigg )^{1/p},\qquad f\ge 0. \end{aligned}$$

(1.4)

A characterization of weights for this inequality is easily derived from known results; details are in Proposition 3.1 below. The case $p\le q$ illustrates the simple sufficient conditions that this method gives for the rearranged Fourier transform inequality (1.2). The following result was (essentially) given in each of [2, 3, 9, 13, 18].

Proposition 1.2

Let $u=U^*$ and $1/w=(1/W)^*$, and suppose $1<p\le q<\infty$. If

$$\begin{aligned} \sup _{x>0}\bigg (\int _0^{1/x}u\bigg )^{1/q}\bigg (\int _0^x w^{1-p'}\bigg )^{1/p'}<\infty \end{aligned}$$

(1.5)

then (1.2) holds, and hence (1.1) holds.

Throughout, conjugate indices are denoted with a prime, so $1/p+1/p'=1$. See [3] for estimates of the best constant C in (1.1), depending only on the indices p, q, the constant D from (1.3), and the value of the above supremum.

In Sect. 4 we complete a comparison begun in [3] of two theorems proved there. Both give sufficient conditions for (1.1), but the form of the conditions differ. We show that one of the two includes the results of the other.

When we are using (1.2) only to get to (1.1) we may assume that u is non-increasing and w is non-decreasing. Indeed, the monotonicity of the weights will figure prominently in our analysis. But the inequality (1.2) is of interest in its own right, without the monotonicity restrictions on u and w. It expresses the boundedness of the Fourier transform between weighted Lorentz $\Lambda$-spaces and has been studied in [3]. Work on the corresponding inequality for Lorentz $\Gamma$-spaces in [23, 24] resulted in necessary and sufficient conditions and was extended to the Fourier coefficient map in [21, 22]. We introduce the Lorentz $\Lambda$- and $\Gamma$-spaces, and improve (weaken) the known sufficient conditions for (1.2), using $\Gamma$-space techniques, in Sect. 5.

An example is given in Sect. 6 to show that the expected sufficient condition for (1.1), known to be valid when $2<q<p$ and $1<q<p<2$, fails to be sufficient when $1<q<2<p$.

Although, in (1.1), we consider all positive values of the index q, we restrict our attention to $p>1$. In fact, the case $p=1$ of (1.1) is rarely discussed, because of the following simple argument. With $C=(1/{\text {ess\,inf}} W)(\int _{{\mathbb {R}}^n}U)^q$, the trivial estimate $|{\hat{f}}|\le \int _{{\mathbb {R}}^n}|f|$ yields (1.1). On the other hand, for $L^1$ functions f approaching a point mass at x (in a suitable weak sense) $|{\hat{f}}|$ approaches a constant function. Applying (1.1) to such f yields $(\int _{{\mathbb {R}}^n}U)^q\le C w(x)$. Taking the essential infimum over x now shows that $C=(1/{\text {ess\,inf}} W)(\int _{{\mathbb {R}}^n}U)^q$ is best possible in (1.1).

A variation of this argument shows that if $0<p<1$ then (1.1) holds only if either U is almost everywhere zero or W is almost everywhere infinite.

2 Reduction to Hardy-Type Inequalities

The success of the rearrangement approach to (1.1) in the case $1<p\le q<\infty$ is well known. But methods and expectations change when $q<p$, because arguments based on the Fourier transform being of type $(p,q)=(1,\infty )$ and $(p,q)=(2,2)$ lend themselves most naturally to the case $p\le q$. Most authors were content to consider only that case. However, in [4, 10] a general estimate due to Calderón, based on weak-type mapping properties instead of the strong-type mapping properties mentioned above, was successfully used to give sufficient conditions for the Fourier inequality (1.1) when $1<q<p$.

Later, in [3], the strong-type conditions, combined with a duality argument, provided improved sufficient conditions using a reduction to the Hardy inequality (1.4). However, the argument that provided these improved conditions is only applicable under the restriction $\max (p',q)\ge 2$. This restriction is used implicitly in the proofs but, unfortunately, was not included in the statements of [3, Theorems 1 and 4]. In Sect. 6 we demonstrate that the restriction is essential by giving an example to show that the statements of [3, Theorems 1 and 4] may fail in the case $\max (p',q)<2$.

In our first main result, we give simple a priori conditions on indices and weights that enable the reduction to (1.4) go through even when the restriction $\max (p',q)\ge 2$ does not hold. Note that parts (a) and (b) are contained in the proof of [3, Theorem 1]. Parts (e) and (h) were suggested by [12, Corollary 3]. For parts (d) and (g) we need to define the weight condition $B_p$ for $p>0$: A non-negative function f is in $B_p$ provided there exists a constant $\beta <\infty$

$$\begin{aligned} y^p\int _y^\infty f(x)\,\frac{dx}{x^p}\le \beta \int _0^y f(x)\,dx,\quad y>0. \end{aligned}$$

Theorem 2.1

Let $u=U^*$ and $1/w=(1/W)^*$. Suppose $1<p<\infty$ and $0<q<p$, and suppose that the inequality (1.4) holds for some constant $C_1$. If any one of the following conditions holds, then so does the Fourier inequality (1.1).

• (a) $q\ge 2$;

• (b) $q>1$ and $p\le 2$;

• (c) there exists a constant $\beta$ such that for all $y>0$;

  $$\begin{aligned} y^{q/2}\int _y^\infty u(x)\,\frac{dx}{x^{q/2}} \le \beta \bigg (\int _0^y u(x)\,dx+y^{\max (1,q)}\int _y^\infty u(x)\,\frac{dx}{x^{\max (1,q)}}\bigg ); \end{aligned}$$

• (d) $u\in B_{q/2}$;

• (e) $q>1$ and $t^{2-q}u(t)$ is a decreasing function of t;

• (f) $q>1$ and there exists a constant $\beta$ such that for all $y>0$,

  $$\begin{aligned} y^{p'/2}\int _y^\infty w(x)^{1-p'}\,\frac{dx}{x^{p'/2}} \le \beta \bigg (\int _0^y w(x)^{1-p'}\,dx+y^{p'}\int _y^\infty w(x)^{1-p'}\,\frac{dx}{x^{p'}}\bigg ); \end{aligned}$$

• (g) $q>1$ and $w^{1-p'}\in B_{p'/2}$;

• (h) $q>1$ and $t^{2-p'}w(t)^{1-p'}$ is a decreasing function of t.

Explicit estimates of the constant C in (1.1) may be given in terms of the indices p and q, and the constants $D, \beta$, and $C_1$. We omit the details.

If none of these a priori conditions holds, then some additional conditions are needed to ensure the validity of (1.1). We approach the problem by adding another Hardy-type inequality to (1.4) so that together the two imply (1.2). This has already been done, in [4, 10], but it was based on weaker mapping properties of the Fourier transform than we use here and, predictably, gives a more restrictive sufficient condition than we obtain using the strong mapping properties. It is included, as (2.1), because it leads to a more tractable weight condition.

Theorem 2.2

Let $u=U^*$ and $1/w=(1/W)^*$. Suppose $1<p<\infty$ and $0<q<p$, and suppose that the inequality (1.4) holds. If either of the following inequalities also holds, for all $f\in L^1({\mathbb {R}}^n)$, then so does the rearranged Fourier inequality (1.2) and hence also the Fourier inequality (1.1):

$$\begin{aligned} \bigg (\int _0^\infty \bigg (x^{-1/2}\int _{1/x}^\infty t^{-1/2}f^*(t)\,dt\bigg )^qu(x)\,dx\bigg )^{1/q}\le & {} C_2\bigg (\int _0^\infty (f^*)^pw\bigg )^{1/p}.\quad \quad \quad \end{aligned}$$

(2.1)

$$\begin{aligned} \bigg (\int _0^\infty \bigg (\frac{1}{x}\int _{1/x}^\infty (f^*)^2\bigg )^{q/2}u(x)\,dx\bigg )^{1/q}\le & {} C_3\bigg (\int _0^\infty (f^*)^pw\bigg )^{1/p}.\quad \quad \end{aligned}$$

(2.2)

Once again, we omit the details of the available estimates for the constant C in (1.1) in terms of the indices p and q and the constants $D, C_1$, and $C_2$ or $C_3$.

Before proceeding to the proofs of these two theorems a discussion of duality is needed. We show that, when both $p>1$ and $q>1$, each of the inequalities (1.1) and (1.4) holds if and only if its counterpart, obtained by the replacements $p\mapsto q', q\mapsto p', U\mapsto W^{1-p'}$ and $W\mapsto U^{1-q'}$, also holds. Observe that positive exponents commute with the rearrangement, so

$$\begin{aligned} u=U^*\mapsto (W^{1-p'})^*=((1/W)^*)^{p'-1}=(1/w)^{p'-1}=w^{1-p'} \end{aligned}$$

and

$$\begin{aligned} 1/w=(1/W)^*\mapsto (1/U^{1-q'})^*=(U^*)^{q'-1}=1/u^{1-q'}. \end{aligned}$$

Thus, the replacements above imply $u\mapsto w^{1-p'}$ and $w\mapsto u^{1-q'}$.

Lemma 2.3

If $1<q<\infty$ and $1<p<\infty$ then for any weights U and W, and any constant C, (1.1) holds for all $f\in L^1({\mathbb {R}}^n)$ if and only if

$$\begin{aligned} \bigg (\int _{{\mathbb {R}}^n}|{\hat{g}}|^{p'}W^{1-p'}\bigg )^{1/p'}\le C\bigg (\int _{{\mathbb {R}}^n}|g|^{q'}U^{1-q'}\bigg )^{1/q'} \end{aligned}$$

(2.3)

for all $g\in L^1({\mathbb {R}}^n)$.

Proof

Suppose (1.1) holds with constant C and fix $g\in L^1({\mathbb {R}}^n)$. For any $f\in L^1({\mathbb {R}}^n)$ with $\int _{{\mathbb {R}}^n} |f|^pW\le 1$, we have $(\int _{{\mathbb {R}}^n}|{\hat{f}}|^q U)^{1/q}\le C$ so Hölder’s inequality yields,

$$\begin{aligned} \bigg |\int _{{\mathbb {R}}^n}f{\hat{g}}\bigg | =\bigg |\int _{{\mathbb {R}}^n}{\hat{f}}g\bigg | \le C\bigg (\int _{{\mathbb {R}}^n}|g|^{q'}U^{1-q'}\bigg )^{1/q'}. \end{aligned}$$

Taking the supremum over all such f, and using the density of $L^1({\mathbb {R}}^n)$ in $L^p_W({\mathbb {R}}^n)$, gives (2.3). The reverse implication is proved similarly.$\square$

Lemma 2.4

If $1<q<\infty$ and $1<p<\infty$ then for any weights u and w, and any constant $C_1$, (1.4) holds for all non-negative measurable f on $(0,\infty )$ if and only if

$$\begin{aligned} \bigg (\int _0^\infty \bigg (\int _0^{1/x} g\bigg )^{p'}w(x)^{1-p'}\,dx\bigg )^{1/p'} \le C_1\bigg (\int _0^\infty g^{q'}u^{1-q'}\bigg )^{1/q'} \end{aligned}$$

(2.4)

for all non-negative measurable g on $(0,\infty )$.

Proof

Suppose (1.4) holds with constant $C_1$ and fix a measurable $g\ge 0$ on $(0,\infty )$. For any $f\ge 0$ with $\int _0^\infty f^pw\le 1$, we have $(\int _0^\infty (\int _0^{1/t} f)^qu(t)\,dt)^{1/q}\le C_1$ so Hölder’s inequality yields,

$$\begin{aligned} \int _0^\infty f(x)\int _0^{1/x}g(t)\,dt\,dx= & {} \int _0^\infty \int _0^{1/t}f(x)\,dxg(t)\,dt\\\le & {} C_1\bigg (\int _0^\infty g^{q'}u^{1-q'}\bigg )^{1/q'}. \end{aligned}$$

Taking the supremum over all such f gives (2.4). The reverse implication is proved similarly.$\square$

It is worth pointing out that (1.4) requires the inequality to hold for all non-negative functions but, as we shall soon see, we will only apply it to non-increasing functions. Nothing is lost, however, for suppose (1.4) were known to hold for non-increasing functions. Then, for any f,

$$\begin{aligned} \bigg (\int _0^\infty \bigg (\int _0^{1/t} |f|\bigg )^qu(t)\,dt\bigg )^{1/q}&\le \bigg (\int _0^\infty \bigg (\int _0^{1/t} f^*\bigg )^qu(t)\,dt\bigg )^{1/q}\\&\le C\bigg (\int _0^\infty (f^*)^pw\bigg )^{1/p}\\&\le C\bigg (\int _0^\infty |f|^pw\bigg )^{1/p}. \end{aligned}$$

The last inequality is an exercise, using the Hardy–Littlewood–Polya inequality and the fact that w is non-decreasing. Thus (1.4) holds for all functions.

In the next lemma we isolate an estimate that will be used in the proofs of both main theorems.

Lemma 2.5

Let $0<q<\infty$ and $1<p<\infty$. If

$$\begin{aligned} \bigg (\int _0^\infty \bigg (\frac{1}{x}\int _0^x\bigg (\int _0^{1/t}f^*\bigg )^2 \,dt\bigg )^{q/2}u(x)\,dx\bigg )^{1/q}\le C_4\bigg (\int _0^\infty (f^*)^pw\bigg )^{1/p}\quad \quad \quad \end{aligned}$$

(2.5)

for all $f\in L^1({\mathbb {R}}^n)$ then (1.2) holds with $C=D^{1/2}C_4$. Here D is the constant of (1.3).

Proof

For each $f\in L^1({\mathbb {R}}^n), (({\hat{f}})^*)^2$ is decreasing, so by (1.3),

$$\begin{aligned} ({\hat{f}})^*(x)^2\le \frac{1}{x}\int _0^x(({\hat{f}})^*)^2 \le \frac{D}{x}\int _0^x\bigg (\int _0^{1/t}f^*\bigg )^2\,dt. \end{aligned}$$

Thus,

$$\begin{aligned} \bigg (\int _0^\infty (({\hat{f}})^*)(x)^qu(x)\,dx\bigg )^{1/q} \le D^{1/2}\bigg (\int _0^\infty \bigg (\frac{1}{x}\int _0^x \bigg (\int _0^{1/t}f^*\bigg )^2dt\bigg )^{q/2}u(x)\,dx\bigg )^{1/q}. \end{aligned}$$

The hypothesis completes the proof. $\square$

We now turn to the proof of Theorem 2.1.

Proof

To see part (a), let $q\ge 2$ and apply [11, Theorem 4.7] to get

$$\begin{aligned} \int _0^\infty (({\hat{f}})^*)^qu\le D^{q/2}\int _0^\infty \bigg (\int _0^{1/t}f^*\bigg )^qu(t)\,dt, \end{aligned}$$

where D is the constant from (1.3). Since (1.4) is assumed to hold, we also have (1.2) and hence (1.1).

For Part (b), the indices p and q are greater than 1 so the assumption (1.4) and Lemma 2.4 implies that (2.4) also holds. But since $1<q<p\le 2$, we have $2\le p'<q'$ so part (a) may be applied to conclude that (2.3) holds. Lemma 2.3 shows that (1.1) holds, as required.

Similar arguments show that, (f), (g), and (h) follow from parts (c), (d), and (e), respectively.

In view of part (a) it suffices to establish (c), (d), and (e) in the case $q<2$. A trivial estimate shows that part (d) follows from part (c). To see that part (e) follows from part (d), suppose $q>1$ and $t^{2-q}u(t)$ is decreasing. In this case,

$$\begin{aligned} y^{q/2}\int _y^\infty u(x)\,\frac{dx}{x^{q/2}} \le y^{q/2}y^{2-q}u(y)\int _y^\infty x^{q/2-2}\,dx =\frac{yu(y)}{1-q/2} \end{aligned}$$

and

$$\begin{aligned} \int _0^y u(x)\,dx\ge y^{2-q}u(y)\int _0^yx^{q-2}\,dx=\frac{yu(y)}{q-1}. \end{aligned}$$

It follows that $u\in B_{q/2}$.

It remains to establish (c) when $q<2$. By Lemma 2.5 it is enough to show that the condition of part (c) implies inequality (2.5). Let $\alpha =\max (1,q)$. Since $f^*$ is non-increasing and right continuous, it is an increasing pointwise limit of continuous non-increasing functions. (For instance, convolve $f^*$ with $n\chi _{(e^{-1/n},1)}$ in $((0,\infty ),dt/t)$ for $n=1,2,\dots$.) Therefore, we may assume without loss of generality that $f^*$ is continuous. For such an f, let $F(t)=-(\int _0^{1/t}f^*)^q$. We show that $t^{\alpha +1}F'(t)$ is non-decreasing in t by considering two cases. When $0<q\le 1, \int _0^{1/t}f^*$ decreases with t so

$$\begin{aligned} t^{1+\alpha }F'(t)=q\bigg (\int _0^{1/t}f^*\bigg )^{q-1}f^*(1/t) \end{aligned}$$

is a non-decreasing function of t. When $1<q<2, t\int _0^{1/t}f^*$ increases with t so

$$\begin{aligned} t^{1+\alpha }F'(t)=q\bigg (t\int _0^{1/t}f^*\bigg )^{q-1}f^*(1/t) \end{aligned}$$

is a non-decreasing function of t. Using the hypothesis of part(c), we get

$$\begin{aligned} \int _y^\infty t^{q/2}\int _t^\infty u(x)\,\frac{dx}{x^{q/2}}\,\frac{dt}{t^{1+\alpha }}&\quad =\int _y^\infty \int _y^xt^{q/2}\,\frac{dt}{t^{1+\alpha }}u(x)\,\frac{dx}{x^{q/2}}\\&\quad \le \frac{y^{q/2-\alpha }}{\alpha -q/2}\int _y^\infty u(x)\,\frac{dx}{x^{q/2}}\\&\quad \le \frac{\beta y^{-\alpha }}{\alpha -q/2}\bigg (\int _0^yu(x)\,dx+y^\alpha \int _y^\infty u(x)\,\frac{dx}{x^{\alpha }}\bigg )\\&\quad =\frac{\beta \alpha }{\alpha -q/2}\int _y^\infty \int _0^tu(x)\,dx\,\frac{dt}{t^{1+\alpha }}. \end{aligned}$$

Since this holds for all y, and $t^{1+\alpha }F'(t)$ is non-decreasing, Hardy’s lemma implies that

$$\begin{aligned} \int _0^\infty t^{1+\alpha }F'(t)t^{q/2}\int _t^\infty u(x)\,\frac{dx}{x^{q/2}}\,\frac{dt}{t^{1+\alpha }} \le \frac{\beta \alpha }{\alpha -q/2}\int _0^\infty t^{1+\alpha }F'(t)\int _0^tu(x)\,dx\,\frac{dt}{t^{1+\alpha }}. \end{aligned}$$

This simplifies to,

$$\begin{aligned} \int _0^\infty \int _0^xt^{q/2}F'(t)\,dt u(x)\,\frac{dx}{x^{q/2}} \le \frac{\beta \alpha }{\alpha -q/2}\int _0^\infty \int _x^\infty F'(t)\,dt\,u(x)\,dx. \end{aligned}$$

Now we apply Minkowski’s integral inequality with index 2 / q to get,

$$\begin{aligned}&\int _0^\infty \bigg (\frac{1}{x}\int _0^x\bigg (\int _0^{1/s}f^*\bigg )^2\,ds\bigg )^{q/2}u(x)\,dx\\&\quad =\int _0^\infty \bigg (\frac{1}{x}\int _0^x\bigg (\int _s^\infty F'(t)\,dt\bigg )^{2/q}\,ds\bigg )^{q/2} u(x)\,dx\\&\quad \le \int _0^\infty \int _0^\infty \bigg (\frac{1}{x}\int _0^{\min (x,t)}ds\bigg )^{q/2}F'(t)\,dtu(x)\,dx\\&\quad =\int _0^\infty \int _0^xt^{q/2}F'(t)\,dtu(x)\,\frac{dx}{x^{q/2}} +\int _0^\infty \int _x^\infty F'(t)\,dtu(x)\,dx\\&\quad \le \bigg (1+\frac{\beta \alpha }{\alpha -q/2}\bigg )\int _0^\infty \int _x^\infty F'(t)\,dtu(x)\,dx\\&\quad =\bigg (1+\frac{\beta \alpha }{\alpha -q/2}\bigg )\int _0^\infty \bigg (\int _0^{1/x}f^*\bigg )^qu(x)\,dx. \end{aligned}$$

This reduces the proof of (2.5) to our assumption that (1.4) holds.$\square$

Proof of Theorem 2.2.

Proof

The first part of the theorem is contained in the proof of [4, Theorem 1.1(ii)], but it may also be deduced from the second part as follows. By Minkowski’s integral inequality,

$$\begin{aligned} \bigg (\frac{1}{x}\int _{1/x}^\infty f^*(s)^2\,ds\bigg )^{1/2}&\quad \le \bigg (\frac{1}{x}\int _{1/x}^\infty \bigg (\frac{1}{s}\int _0^sf^*(t)\,dt\bigg )^2\,ds\bigg )^{1/2}\\&\quad \le \int _0^\infty \bigg (\frac{1}{x}\int _{\max (1/x,t)}^\infty \,\frac{ds}{s^2}\bigg )^{1/2}f^*(t)\,dt\\&\quad =\int _0^{1/x} f^*(t)\,dt +x^{-1/2}\int _{1/x}^\infty t^{-1/2}f^*(t)\,dt. \end{aligned}$$

This estimate and the extended Minkowski inequality show that (1.4) and (2.1) together imply (2.2).

To prove the second part we will apply Lemma 2.5. To begin, break the inner integral at $t=1/x$ and use the triangle inequality in the $L^2$ norm to get,

$$\begin{aligned}&\bigg (\frac{1}{x}\int _0^x\bigg (\int _0^{1/t}f^*\bigg )^2\,dt\bigg )^{1/2}\\&\quad \le \bigg (\frac{1}{x}\int _0^x\bigg (\int _0^{1/x}f^*\bigg )^2\,dt\bigg )^{1/2} +\bigg (\frac{1}{x}\int _0^x\bigg (\int _{1/x}^{1/t}f^*\bigg )^2\,dt\bigg )^{1/2}\\&\quad =\int _0^{1/x}f^* +\bigg (\frac{1}{x}\int _{1/x}^\infty \bigg (\frac{1}{t}\int _{1/x}^tf^*\bigg )^2\,dt\bigg )^{1/2}\\&\quad \le \int _0^{1/x}f^* +\bigg (\frac{1}{x}\int _{1/x}^\infty \bigg (\frac{1}{t-1/x}\int _{1/x}^tf^*\bigg )^2\,dt\bigg )^{1/2}\\&\quad \le \int _0^{1/x}f^* +2\bigg (\frac{1}{x}\int _{1/x}^\infty f^*(t)^2\,dt\bigg )^{1/2}. \end{aligned}$$

The last inequality above is the Hardy inequality on the interval $(1/x,\infty )$. Taking $c=\max (2^{1/q-1},1)$, and using the (extended) Minkowski inequality gives

$$\begin{aligned}&\bigg (\int _0^\infty \bigg (\frac{1}{x}\int _0^x \bigg (\int _0^{1/t}f^*\bigg )^2\,dt\bigg )^{q/2}u(x)\,dx\bigg )^{1/q}\\&\quad \le c\bigg (\int _0^\infty \bigg (\int _0^{1/x} f^*\bigg )^qu(x)\,dx\bigg )^{1/q}\\&\qquad +c\bigg (\int _0^\infty \bigg (\frac{1}{x}\int _{1/x}^\infty (f^*)^2\bigg )^{q/2}u(x)\,dx\bigg )^{1/q}. \end{aligned}$$

These estimates show that (2.5) holds whenever both (1.4) and (2.2) do. Now Lemma 2.5 completes the proof. $\square$

In the above proof we showed that inequalities (1.4) and (2.2) imply (2.5). To see that nothing is lost by this decomposition, we observe that the other implication also holds. Since the square of $\int _0^{1/x} f^*(t)\,dt$ is a decreasing function of x,

$$\begin{aligned} \int _0^{1/x} f^*(t)\,dt\le \bigg (\frac{1}{x}\int _0^x\bigg (\int _0^{1/t}f^*\bigg )^2\,dt\bigg )^{1/2}. \end{aligned}$$

Thus, if (2.5) holds then (1.4) holds for decreasing functions. It follows from the remark after Lemma 2.4, that (1.4) holds for all non-negative functions. Inequality (2.2) also follows from (2.5): Since $f^*$ is decreasing,

$$\begin{aligned} \frac{1}{x}\int _{1/x}^\infty f^*(t)^2\,dt \le \frac{1}{x}\int _{1/x}^\infty \bigg (\frac{1}{t}\int _0^tf^*\bigg )^2\,dt =\frac{1}{x}\int _0^x\bigg (\int _0^{1/t}f^*\bigg )^2\,dt. \end{aligned}$$

3 Explicit Weight Conditions

In the previous section, our approach was to find Hardy-type inequalities, depending on the weights u and w that imply the Fourier inequality (1.1) for the weights U and W. This puts us in a position to use known weight characterizations for Hardy-type operators to give conditions that ensure the validity of (1.1). In this section we do exactly that, beginning with the inequality (1.4).

Proposition 3.1

Suppose $0<q<\infty , 1< p<\infty$ and $1/r=1/q-1/p$. The inequality (1.4), that is,

$$\begin{aligned} \bigg (\int _0^\infty \bigg (\int _0^{1/x}f\bigg )^q u(x)\,dx\bigg )^{1/q} \le C_1\bigg (\int _0^\infty f^pw\bigg )^{1/p},\quad f\ge 0, \end{aligned}$$

holds (for some finite constant $C_1$) if and only if:

• (a) $1<p\le q$ and

  $$\begin{aligned} \sup _{x>0}\bigg (\int _0^{1/x} u\bigg )^{1/q}\bigg (\int _0^x w^{1-p'}\bigg )^{1/p'}<\infty ;\quad \text {or} \end{aligned}$$

• (b) $0<q<p, 1<p$, and

  $$\begin{aligned} \bigg (\int _0^\infty \bigg (\int _0^x u\bigg )^{r/p}\bigg ( \int _0^{1/x}w^{1-p'}\bigg )^{r/p'}u(x)\,dx\bigg )^{1/r}<\infty . \end{aligned}$$

  (3.1)

Proof

The change of variable $x\mapsto 1/x$ converts (1.4) to the standard form of the Hardy inequality found in [6, Theorem 1] and [26, Theorem 2.4]. The same change of variable is used to convert the weight conditions given in those results to the ones above. $\square$

It is pointed out in [26, p. 93] that if $q>1$ or if $w^{1-p'}$ is locally integrable, then the condition in Case (b) may be replaced by,

$$\begin{aligned} \bigg (\int _0^\infty \bigg (\int _0^{1/x} u\bigg )^{r/q}\bigg ( \int _0^xw^{1-p'}\bigg )^{r/q'}w(x)^{1-p'}\,dx\bigg )^{1/r}<\infty . \end{aligned}$$

(3.2)

The equivalence of these two forms (essentially using integration by parts) may be needed to reconcile previous results with those given here.

The next result applies to inequality (2.1) because (2.1) is obtained from (3.3) by taking $f=f^*$. Since (2.1) only requires that (3.3) hold for non-increasing functions, the weight condition (3.4) is sufficient for (2.1) but may be stronger than necessary. The condition (3.4) may be compared, via integration by parts, to [4, Condition (1.9)] when $q>1$.

Proposition 3.2

Suppose $0<q<p, 1< p<\infty$ and $1/r=1/q-1/p$. The inequality

$$\begin{aligned}&\bigg (\int _0^\infty \bigg (x^{-1/2}\int _{1/x}^\infty t^{-1/2}f(t)\,dt\bigg )^q u(x)\,dx\bigg )^{1/q} \le C_5\bigg (\int _0^\infty f^pw\bigg )^{1/p},\nonumber \\&\quad f\ge 0, \end{aligned}$$

(3.3)

holds (for some finite constant $C_5$) if and only if

$$\begin{aligned}&\bigg (\int _0^\infty \bigg (\int _x^\infty t^{-q/2} u(t)\,dt\bigg )^{r/p}\bigg (\int _{1/x}^\infty t^{-p'/2}w(t)^{1-p'}\,dt\bigg )^{r/p'}x^{-q/2}u(x)\,dx\bigg )^{1/r}\nonumber \\&\quad \ <\infty . \end{aligned}$$

(3.4)

Proof

This time we make the substitution $t\mapsto 1/t$ on both sides of the inequality and replace f(1 / t) by $t^{3/2}g(t)$. This puts it in the standard form of [26, Theorem 2.4], but with weights $x^{-q/2} u(x)$ and $t^{(3p-4)/2}w(1/t)$. As before, the same substitution is used to re-write the weight condition in the above form. $\square$

Necessary and sufficient conditions on weights for which (2.2) holds are known, but are substantially more complicated than for the other two inequalities.

Proposition 3.3

Suppose $0<q<p, 1<p<\infty$ and $1/r=1/q-1/p$. The inequality (2.2), that is,

$$\begin{aligned} \bigg (\int _0^\infty \bigg (\frac{1}{x}\int _{1/x}^\infty (f^*)^2\bigg )^{q/2} u(x)\,dx\bigg )^{1/q} \le C_3\bigg (\int _0^\infty (f^*)^pw\bigg )^{1/p},\quad f\in L^1({\mathbb {R}}^n), \end{aligned}$$

holds (for some finite constant $C_3$) whenever:

• (a) $0<q<p\le 2$ and

  $$\begin{aligned} \sup _{x_k}\bigg (\sum _{k\in {\mathbb {Z}}}\bigg (\int _{1/x_{k+1}}^{1/x_k}(x_{k+1}t-1)^{q/2}u(t)\,\frac{dt}{t^q}\bigg )^{r/q}\bigg (\int _0^{x_{k+1}}w\bigg )^{-r/p}\bigg )^{1/r}<\infty \quad \quad \quad \end{aligned}$$

  (3.5)

  where the supremum is taken over all increasing sequences $x_k, k\in {\mathbb {Z}}$; or

• (b) $0<q<2<p$, (3.5), and

  $$\begin{aligned}&\bigg (\int _0^\infty \bigg (\int _{1/x}^\infty \bigg (\frac{1}{t-1/x}\int _0^tw\bigg )^{-p/(p-2)}w(t)\,dt\bigg )^{(r/2)(p-2)/p}\nonumber \\&\quad \times \bigg (\int _x^\infty t^{-q/2}u(t)\,dt\bigg )^{r/p}x^{-q/2}u(x) \,dx\bigg )^{1/r}<\infty . \end{aligned}$$

  (3.6)

Proof

Every non-negative, decreasing function on $(0,\infty )$ can be represented as a limit of functions $(f^*)^2$ for $f\in L^1({\mathbb {R}}^n)$. So, letting $x\mapsto 1/x$ puts (2.2) in the form of [7, Theorem 5.1], with indices p / 2 and q / 2. Cases (ii) and (vi) of that theorem yield the results above, after letting $x\mapsto 1/x$ again. $\square$

We conclude this section with a summary of sufficient conditions for the Fourier inequality (1.1) in the case $q<p$.

Theorem 3.4

Suppose $0<q<p$ and $1<p<\infty$. Let U and W be non-negative, measurable function on ${\mathbb {R}}^n$ and set $u=U^*$ and $1/w=(1/W)^*$. Inequality (1.1), that is,

$$\begin{aligned} \bigg (\int _{{\mathbb {R}}^n} |{\hat{f}}(y)|^qU(y)\,dy\bigg )^{1/q} \le C\bigg (\int _{{\mathbb {R}}^n}|f(x)|^pW(x)\,dx\bigg )^{1/p},\quad f\in L^1({\mathbb {R}}^n),\quad \quad \end{aligned}$$

(3.7)

holds (for some finite constant C) provided (3.1) holds and (i), (ii) or (iii) is satisfied:

• (i) $2\le q< p$ or $1<q<p\le 2$.

• (ii) $1<q<2<p$ and

  • (a) one or more of (c)–(h) from Theorem 2.1,

  • (b) (3.4), or

  • (c) (3.5) and (3.6).

• (iii) $0<q<1<p$ and

  • (a)  (c) or (d) from Theorem 2.1,

  • (b) (3.4),

  • (c) $p\le 2$ and (3.5), or

  • (d) $p>2$ and (3.5) and (3.6).

Necessary and sufficient conditions for weighted Hardy inequalities may be expressed in a wide variety of different, but equivalent, forms. See, for example [19, 20, 28]. The form of the weight conditions in Propositions 3.1, 3.2, and 3.3 represent one choice but Theorems 2.1 and 2.2 may be combined with any of the various forms to give sufficient conditions for the Fourier inequality (1.1). For more information on Hardy inequalities, see [14, 15].

4 Comparison of Theorems 1 and 4 in [3]

In [3, Theorem 4], under the a priori assumption that $w\in B_p$ or $u^{1-q'}\in B_{q'}$, sufficient conditions for (1.1) are given that appear different than those of [3, Theorem 1]. The latter appear here in Proposition 1.2, see condition (1.5), and part (i) of Theorem 3.4, see the equivalent conditions (3.1) and (3.2). The weight conditions from the two theorems are compared in [3, Remark 5] but neither theorem is shown to directly imply the other. The next lemma completes the comparison, showing that the two theorems give equivalent weight conditions whenever [3, Theorem 4] applies. We conclude that [3, Theorem 1] is the stronger result because it does not require any a priori assumption.

Lemma 4.1

If $1<p<\infty , w\in B_p$ and w is increasing, then there is a constant $C_6$ such that

$$\begin{aligned} x\bigg (\int _0^x w\bigg )^{-1/p}\le \bigg (\int _0^x w^{1-p'}\bigg )^{1/p'}\le C_6x\bigg (\int _0^x w\bigg )^{-1/p} \end{aligned}$$

(4.1)

for all $x>0$.

Proof

By Hölder’s inequality,

$$\begin{aligned} x=\int _0^x w^{1/p}w^{-1/p}\le \bigg (\int _0^x w\bigg )^{1/p}\bigg (\int _0^xw^{1-p'}\bigg )^{1/p'}. \end{aligned}$$

This proves the first inequality of (4.1). For the second, recall [1, Theorem 1], which shows that since $w\in B_p$ there exists a constant $C_6<\infty$ such that the Hardy inequality

$$\begin{aligned} \bigg (\int _0^\infty \bigg (\frac{1}{t}\int _0^tf\bigg )^pw(t)\,dt\bigg )^{1/p} \le C_6\bigg (\int _0^\infty f^pw\bigg )^{1/p} \end{aligned}$$

holds for all non-increasing functions $f\ge 0$. Fix $x>0$ and, for $k>1/x$, let $f_k=\max (w(1/k),w)^{1-p'}\chi _{(0,x)}$. Since $f_k$ is non-increasing, so is its moving average. Therefore,

$$\begin{aligned} \bigg (\int _0^x w\bigg )^{1/p}\frac{1}{x}\int _0^x f_k \le \bigg (\int _0^\infty \bigg (\frac{1}{t}\int _0^t f_k\bigg )^p w(t)\,dt\bigg )^{1/p} \le C_6\bigg (\int _0^x f_k^pw\bigg )^{1/p}. \end{aligned}$$

Since $f_k^pw\le f_k$, and $f_k$ is bounded above, we have

$$\begin{aligned} \bigg (\int _0^x f_k\bigg )^{1/p'}\le C_6x\bigg (\int _0^x w\bigg )^{-1/p}. \end{aligned}$$

Letting $k\rightarrow \infty$ gives the second inequality of (4.1). $\square$

Using this lemma we show, in four cases, that if the appropriate a priori condition holds, then the Fourier inequalities that follow from [3, Theorem 4], are the same as those that follow from [3, Theorem 1].

Case 1 Suppose $1<p\le q, q\ge 2$, and $w\in B_p$. Then the weight pair (u, w) satisfies the weight condition of [3, Theorem 4(i)] if and only if it satisfies the weight condition of [3, Theorem 1(i)]. That is,

$$\begin{aligned} \sup _{x>0}x\bigg (\int _0^{1/x}u\bigg )^{1/q}\bigg (\int _0^x w\bigg )^{-1/p}<\infty \end{aligned}$$

if and only if (1.5) holds. This follows directly from Lemma 4.1.

Case 2 Suppose $2\le q<p, 1/r=1/q-1/p$, and $w\in B_p$. Then the weight pair (u, w) satisfies the weight condition of [3, Theorem 4(ii)] if and only if it satisfies the weight condition of [3, Theorem 1(ii)]. That is,

$$\begin{aligned} \bigg (\int _0^\infty \bigg (\int _0^xu\bigg )^{r/p}\bigg (\int _0^{1/x} w\bigg )^{-r/p}x^{-r}u(x)\,dx\bigg )^{1/r}<\infty \end{aligned}$$

(4.2)

if and only if (3.2) holds. Lemma 4.1 gives the equivalence of (4.2) and (3.1); the remark following Proposition 3.1 completes the argument.

Case 3 Suppose $1<p\le q<2$, and $u^{1-q'}\in B_{q'}$. Then the weight pair (u, w) satisfies the weight condition of [3, Theorem 4(iii)] if and only if it satisfies the weight condition of [3, Theorem 1(i)]. That is,

$$\begin{aligned} \sup _{x>0}\frac{1}{x}\bigg (\int _0^{1/x}u^{1-q'}\bigg )^{-1/q'}\bigg (\int _0^x w^{1-p'}\bigg )^{1/p'}<\infty \end{aligned}$$

if and only if (1.5) holds. This follows from Lemma 4.1, with p and w replaced by $q'$ and $u^{1-q'}$.

The last case does not include the index range $1<q<2<p$, which should have been excluded in the statements of [3, Theorems 1 and 4]. We also fix a typographic error in the weight condition of [3, Theorem 4(iv)].

Case 4 Suppose $1<q<p\le 2, 1/r=1/q-1/p$, and $u^{1-q'}\in B_{q'}$. Then the weight pair (u, w) satisfies the weight condition of [3, Theorem 4(iv)] if and only if it satisfies the weight condition of [3, Theorem 1(ii)]. That is,

$$\begin{aligned} \bigg (\int _0^\infty \bigg (\int _0^xw^{1-p'}\bigg )^{r/q'}\bigg (\int _0^{1/x} u^{1-q'}\bigg )^{-r/q'}x^{-r}w(x)^{1-p'}\,dx\bigg )^{1/r}<\infty \end{aligned}$$

if and only if (3.2) holds. This follows from Lemma 4.1, with p and w replaced by $q'$ and $u^{1-q'}$.

These four cases together show that [3, Theorem 1] is stronger than [3, Theorem 4] because it applies without the a priori conditions of [3, Theorem 4].

5 Lorentz Space Fourier Inequalities

In this section we return briefly to inequality (1.2), but without the monotonicity restrictions on the weights u and w. Let $0<p<\infty$ and w be a weight, and define the Lorentz $\Lambda$- and $\Gamma$-“norms” of f by,

$$\begin{aligned} \Vert f\Vert _{\Lambda _p(w)}= & {} \bigg (\int _0^\infty (f^*)^pw\bigg )^{1/p}\quad \text {and}\\ \Vert f\Vert _{\Gamma _p(w)}= & {} \bigg (\int _0^\infty \bigg (\frac{1}{t}\int _0^tf^*\bigg )^pw(t)\,dt\bigg )^{1/p}. \end{aligned}$$

Since $f^*$ is non-increasing, $\Vert f\Vert _{\Lambda _p(w)}\le \Vert f\Vert _{\Gamma _p(w)}$ for any weight w. If $w\in B_p$, then [1, Theorem 1] shows that the two are equivalent.

With this notation, inequality (1.2) becomes

$$\begin{aligned} \Vert {\hat{f}}\Vert _{\Lambda _q(u)}\le C\Vert f\Vert _{\Lambda _p(w)}. \end{aligned}$$

(5.1)

This inequality, which expresses the boundedness of the Fourier transform between Lorentz $\Lambda$-spaces, was considered in [3, Theorems 2 and 3]. Restricting the domain to the smaller $\Gamma$-space leads to the inequality

$$\begin{aligned} \Vert {\hat{f}}\Vert _{\Lambda _q(u)}\le C\Vert f\Vert _{\Gamma _p(w)}. \end{aligned}$$

(5.2)

These were studied in [23], where it was shown in [23, Theorem 3.4] that for $0<p\le q<\infty$, inequality (5.2) holds whenever

$$\begin{aligned} \sup _{x<y}\bigg (\frac{x}{y}\int _0^y u\bigg )^{1/q}\bigg (x^p\int _0^{1/x}w(t)\,dt+\int _{1/x}^\infty w(t)\,\frac{dt}{t^p}\bigg )^{-1/p}<\infty . \end{aligned}$$

(5.3)

Observe that this result includes [3, Theorem 2(i)]; for if $w\in B_p$, inequalities (5.1) and (5.2) are equivalent. Also, if u is non-increasing and

$$\begin{aligned} \sup _{x>0}\frac{1}{x}\bigg (\int _0^x u\bigg )^{1/q}\bigg (\int _0^{1/x}w\bigg )^{-1/p}<\infty , \end{aligned}$$

then straightforward estimates show that (5.3) also holds.

But, as we see next, results for the $\Gamma$-space inequality (5.2) may be used to further weaken sufficient conditions for the $\Lambda$-space inequality (5.1). This result may be viewed as a generalization of Proposition 1.2, without monotonicity conditions on the weights. We will need the level function $u^o$ of u defined by requiring that the function $x\mapsto \int _0^x u^o$ is the least concave majorant of $x\mapsto \int _0^x u$. Note that $u^o$ is non-increasing and it coincides with u when u is non-increasing. For properties of the level function, see [16] and the references therein.

Theorem 5.1

Let $1<p\le q <\infty$ and $q\ge 2$. Assume u and w are weight functions on $(0,\infty )$. If

$$\begin{aligned} \sup _{x>0} \bigg (\int _0^{1/x} u^o \bigg )^{1/q} \bigg ( \int _0^x w^{1-p'} \bigg )^{1/p'} < \infty , \end{aligned}$$

(5.4)

then there exists $C>0$ such that (5.1) holds for all $f\in L^1({\mathbb {R}}^n)$.

Proof

If $\int _0^{1/x} u^o$ is infinite for some $x>0$, concavity shows that it is infinite for all $x>0$. But then the finiteness of (5.4) implies w is infinite almost everywhere so the inequality (5.1) holds trivially. Henceforth, we assume that $\int _0^{1/x} u^o (t)\, dt<\infty$ for $x>0$.

Since the concave function $t \mapsto \int _0^t u^o$ is absolutely continuous, we may set

$$\begin{aligned} \sigma (t) = t^{q-2} u^o(1/t) = - t^q \, \dfrac{d}{dt} \bigg (\int _0^{1/t} u^o\bigg ) \end{aligned}$$

to get

$$\begin{aligned} \int _x^{\infty } \dfrac{\sigma (t)}{t^q} \, dt =\int _0^{1/x} u^o. \end{aligned}$$

(5.5)

Thus, for any $x<y$,

$$\begin{aligned} \frac{1}{y}\int _0^y u\le \frac{1}{y}\int _0^y u^o\le \frac{1}{x}\int _0^x u^o =\frac{1}{x}\int _{1/x}^{\infty } \dfrac{\sigma (t)}{t^q} \, dt. \end{aligned}$$

It follows that

$$\begin{aligned} \sup _{x<y}\bigg (\frac{x}{y}\int _0^y u\bigg )^{1/q}\bigg (x^q\int _0^{1/x}\sigma (t)\,dt+\int _{1/x}^\infty \sigma (t)\,\frac{dt}{t^q}\bigg )^{-1/q}\le 1<\infty . \end{aligned}$$

So, as we have seen above, [23, Theorem 3.4] shows there is a constant $C_7$ such that

$$\begin{aligned} \Vert {\hat{f}}\Vert _{\Lambda _q(u)}\le C_7\Vert f\Vert _{\Gamma _q(\sigma )}. \end{aligned}$$

Combining (5.5) with the hypothesis (5.4) shows that

$$\begin{aligned} \sup _{x>0} \bigg (\int _x^{\infty } \dfrac{\sigma (t)}{t^q} \, dt\bigg )^{1/q} \bigg ( \int _0^x w(t)^{1-p'}\,dt \bigg )^{1/p'} < \infty , \end{aligned}$$

which, by [6, Theorem 1], implies there exists a constant $C_8$ such that

$$\begin{aligned} \Vert f\Vert _{\Gamma _q(\sigma )}\le C_8\Vert f\Vert _{\Lambda _p(w)}. \end{aligned}$$

Taking $C=C_7C_8$ completes the proof. $\square$

6 Examples

In this section we produce explicit weights U and W, depending on indices p and q with $1<q<2<p<\infty$, for which the Hardy inequality (1.4) holds but the Fourier inequality (1.1) fails. This is Example 6.3. For simplicity, only the case $n=1$ is considered.

The existence of such an example shows that the conclusion of Theorems 2.1 and 2.2 does not follow from assumption (1.4) alone. It explains why one of (a)–(h) is assumed in Theorem 2.1 and why either (2.1) or (2.2) is assumed in Theorem 2.2.

So far we have worked only with the Fourier transform of functions on ${\mathbb {R}}^n$. But it is not difficult to adapt the proofs of Propositions 1.1 and 1.2, and Theorems 2.1, 2.2, and 3.4 to the case of Fourier series, or to the Fourier transform over any locally compact abelian group.

Example 6.1, suggested by [8, Exercise 3.1.6], looks at the case of the Fourier transform on ${\mathbb {T}}$, where the compactness of ${\mathbb {T}}$ permits a straightforward argument. This example is included because the simplicity of the construction illustrates the main idea of Example 6.3 but avoids much of the technical detail. Example 6.3 then takes on the case of the Fourier transform. In the proof, a Gaussian function is introduced to ensure convergence in the absence of compactness in the domain space.

Note that, by Proposition 3.1, proving (3.1) is enough to show that (1.4) holds. This tactic is used in both examples.

Example 6.1

Suppose p, q and r satisfy $1<q<2<p<\infty$ and $1/r=1/q-1/p$. Take $\alpha ,\beta \in (0,1)$ so that

$$\begin{aligned} \frac{1}{2}<\frac{\beta }{q}<\frac{\alpha }{p'}<\frac{1}{q}<1. \end{aligned}$$

Let $U(k)=(|k|+1)^{\beta -1}, k\in {\mathbb {Z}}$, and $W(x)=x^{(1-\alpha )(p-1)}, 0\le x<1$. Let u be the rearrangement of U with respect to counting measure on ${\mathbb {Z}}$ and define w by $1/w=(1/W)^*$ with respect to Lebesgue measure on ${\mathbb {T}}=[0,1]$. Then condition (3.1) holds but the Fourier series inequality

$$\begin{aligned} \bigg (\sum _{k\in {\mathbb {Z}}} |{\hat{f}}(k)|^qU(k)\bigg )^{1/q} \le C\bigg (\int _0^1|f(x)|^pW(x)\,dx\bigg )^{1/p}, \quad f\in L^1({\mathbb {T}}), \end{aligned}$$

fails to hold for any C.

Proof

Observe that $w(t)=t^{(1-\alpha )(p-1)}$ for $0\le t<1$ and $w(t)=\infty$ for $t>1$. Also, $u(t)=1$ for $0\le t<1$, and $u(t)=(k+1)^{\beta -1}$ for $2k-1\le t<2k+1, k=1,2,\dots$. It follows that $u(t)\le ((t+1)/2)^{\beta -1}$ and easy estimates show that (3.1) holds.

On the other hand, completing [8, Exercise 3.1.6] shows that the sum

$$\begin{aligned} g(x)=\sum _{k=2}^\infty k^{-1/2}(\log k)^{-2}e^{ik\log k}e^{2\pi ikx} \end{aligned}$$

defines a continuous function on ${\mathbb {T}}$. In particular, g is bounded on [0, 1] and thus $\Vert g\Vert _{L^p(W)}<\infty$ because W is integrable. However, ${\hat{g}}(k)=k^{-1/2}(\log k)^{-2}e^{ik\log k}$ for $k=2,3,\dots$, so

$$\begin{aligned} \Vert {\hat{g}}\Vert _{\ell ^q(U)}^q\ge \sum _{k=2}^\infty k^{-q/2}(\log k)^{-2q}(|k|+1)^{\beta -1}=\infty . \end{aligned}$$

Thus, the Fourier series inequality fails with $f=g$. $\square$

Before beginning with our second example, we set up to use van der Corput’s lemma several times.

Lemma 6.2

Suppose $1<a<b, x\in {\mathbb {R}}$, and $f\ge 0$ is continuously differentiable. If f is decreasing on [a, b] then

$$\begin{aligned} \bigg |\int _a^b f(y)e^{iy\log (y)}e^{2\pi ixy}\,dy\bigg |\le 24a^{1/2}f(a)+12\int _a^by^{-1/2}f(y)\,dy. \end{aligned}$$

If f is increasing on [a, b] then

$$\begin{aligned} \bigg |\int _a^b f(y)e^{iy\log (y)}e^{2\pi ixy}\,dy\bigg |\le 24b^{1/2}f(b). \end{aligned}$$

Proof

Suppose $[{\bar{a}},{\bar{b}}]\subseteq [a,b]$. Letting $y\mapsto {\bar{b}}y$, we have

$$\begin{aligned} \int _{{\bar{a}}}^{{\bar{b}}} e^{iy\log (y)}e^{2\pi ixy}\,dy ={\bar{b}}\int _{{\bar{a}}/{\bar{b}}}^1 e^{i{\bar{b}}y(\log (y)+\log ({\bar{b}})+2\pi x)}\,dy. \end{aligned}$$

The second derivative of $y(\log (y)+\log ({\bar{b}})+2\pi x)$ is 1 / y, which is at least 1 when $0<y\le 1$. So the van der Corput lemma [8, Proposition 2.6.7(b)] (with $k=2$) shows that

$$\begin{aligned} \bigg |\int _{{\bar{a}}}^{{\bar{b}}} E\,dy\bigg |\le {\bar{b}}(24{\bar{b}}^{-1/2})=24{\bar{b}}^{1/2}, \end{aligned}$$

where we have written $E=e^{iy\log (y)}e^{2\pi ixy}$ for simplicity.

First suppose that f is decreasing. Then $-f'(z)=|f'(z)|$ for $a<z<b$ so

$$\begin{aligned} \bigg |\int _a^b f(y)E\,dy\bigg |&=\bigg |\int _a^b \bigg (f(b)+\int _y^b |f'(z)|\,dz\bigg )E\,dy\bigg |\\&\le f(b)\bigg |\int _a^b E\,dy\bigg |+\int _a^b \bigg |\int _a^z E\,dy\bigg | |f'(z)|\,dz\\&\le 24b^{1/2}f(b)+24\int _a^b z^{1/2}|f'(z)|\,dz\\&= 24b^{1/2}f(b)+24\bigg ( a^{1/2}f(a)-b^{1/2}f(b)+\frac{1}{2}\int _a^b y^{-1/2}f(y)\,dy\bigg )\\&= 24a^{1/2}f(a)+12\int _a^b y^{-1/2}f(y)\,dy. \end{aligned}$$

Next suppose f is increasing. Then $f'(z)\ge 0$ for $a<z<b$ so

$$\begin{aligned} \bigg |\int _a^b f(y)E\,dy\bigg |&=\bigg |\int _a^b \bigg (f(a)+\int _a^y f'(z)\,dz\bigg )E\,dy\bigg |\\&\le f(a)\bigg |\int _a^b E\,dy\bigg |+\int _a^b \bigg |\int _z^b E\,dy\bigg | f'(z)\,dz\\&\le 24b^{1/2}f(a)+24\int _a^b b^{1/2}f'(z)\,dz\\&= 24b^{1/2}f(b). \end{aligned}$$

$\square$

Example 6.3

Suppose p, q and r satisfy $1<q<2<p<\infty$ and $1/r=1/q-1/p$. Take $\alpha ,\beta \in (0,1)$ so that

$$\begin{aligned} \frac{1}{2}<\frac{\beta }{q}<\frac{\alpha }{p'}<\frac{1}{q}<1. \end{aligned}$$

Let $U(y)=(|y|+1)^{\beta -1}$ for $y\in {\mathbb {R}}$ and $W(x)=|x|^{(1-\alpha )(p-1)}$ for $x\in {\mathbb {R}}$. Define u and w by $u=U^*$ and $1/w=(1/W)^*$. Then condition (3.1) holds but the Fourier inequality (1.1) fails to hold for any constant C.

Proof

Observe that $w(t)=(t/2)^{(1-\alpha )(p-1)}$ and $u(t)=((t/2)+1)^{\beta -1}$. With these in hand, easy estimates show that (3.1) holds.

For each $K>2+e$, let $\gamma =(2/3)(1+\log K)^{3/2}$ and define $m,n\in (0,1)$ by requiring that $(1+\log K)m=\frac{\pi }{3}$ and $(1+\log K)n=\frac{\pi }{2}$. Take

$$\begin{aligned} h(x)=\frac{\sqrt{2\pi }}{\gamma }e^{-2\pi ^2(x/\gamma )^2}\quad \text {to get}\quad {\hat{h}}(y)=e^{-\frac{1}{2}y^2\gamma ^2}. \end{aligned}$$

Also, take

$$\begin{aligned} g(x)=h(x)\int _e^Ky^{-1/2}\log (y)^{-2}e^{iy\log (y)}e^{2\pi ixy}\,dy. \end{aligned}$$

Clearly $g\in L^1$ and

$$\begin{aligned} {\hat{g}}(z)&=\int _{-\infty }^\infty g(x)e^{-2\pi ixz}\,dx\\&=\int _e^Ky^{-1/2}\log (y)^{-2}e^{iy\log (y)}{\hat{h}}(y-z)\,dy\\&=\int _e^Ky^{-1/2}\log (y)^{-2}e^{iy\log (y)}e^{-\frac{1}{2}(y-z)^2\gamma ^2}\,dy. \end{aligned}$$

First we estimate $\Vert g\Vert _{L^p(W)}$. Since $y^{-1/2}\log (y)^{-2}$ is decreasing for $y\ge e$, Lemma 6.2 gives,

$$\begin{aligned} \bigg |\int _e^Ky^{-1/2}\log (y)^{-2}e^{iy\log (y)}e^{2\pi ixy}\,dy\bigg | \le 24+12\int _e^K\log (y)^{-2}\,\frac{dy}{y}\le 36. \end{aligned}$$

Using this estimate, and making the substitution $x\mapsto \gamma x$, yields

$$\begin{aligned} \Vert g\Vert _{L^p(W)}\le 36\Vert h\Vert _{L^p(W)}=\gamma ^{-\alpha /p'}36\sqrt{2\pi }\bigg (\int _{-\infty }^\infty e^{-2\pi ^2px^2}|x|^{(1-\alpha )(p-1)}\,dx\bigg )^{1/p}. \end{aligned}$$

Since $\alpha /p'>0$ and $(1-\alpha )(p-1)>-1$, this norm goes to zero as $K\rightarrow \infty$.

Estimating $\Vert {\hat{g}}\Vert _{L^q(U)}$ is next. Suppose $e+1<z<K-1$. Then,

$$\begin{aligned} |{\hat{g}}(z)|&\ge \bigg |\int _{z-n}^{z+n}y^{-1/2}\log (y)^{-2}e^{-\frac{1}{2}(y-z)^2\gamma ^2}e^{iy\log (y)}\,dy\bigg |\\&-\bigg |\int _e^{z-n}y^{-1/2}\log (y)^{-2}e^{-\frac{1}{2}(y-z)^2\gamma ^2}e^{iy\log (y)}\,dy\bigg |\\&-\bigg |\int _{z+n}^Ky^{-1/2}\log (y)^{-2}e^{-\frac{1}{2}(y-z)^2\gamma ^2}e^{iy\log (y)}\,dy\bigg |\\&\equiv A-B_1-B_2. \end{aligned}$$

To estimate A, we will multiply by $1=|e^{-iz\log (z)}|$ and then reduce the modulus to its real part. This is justified by the mean value theorem: If $z-n<y<z+n$ then for some ${\bar{y}}\in (z-n,z+n)\subseteq (e,K)$,

$$\begin{aligned} |y\log (y)-z\log (z)|=(1+\log {\bar{y}})|y-z|\le (1+\log K)n=\pi /2 \end{aligned}$$

so $\cos (y\log (y)-z\log (z))\ge 0$. Similarly, if $z-m<y<z+m$ then $|y\log (y)-z\log (z)|\le \pi /3$ so $\cos (y\log (y)-z\log (z))\ge 1/2$. We have,

$$\begin{aligned} A&=\bigg |\int _{z-n}^{z+n}y^{-1/2}\log (y)^{-2}e^{-\frac{1}{2}(y-z)^2\gamma ^2}e^{iy\log (y)-iz\log (z)}\,dy\bigg |\\&\ge \int _{z-n}^{z+n}y^{-1/2}\log (y)^{-2}e^{-\frac{1}{2}(y-z)^2\gamma ^2}\cos (y\log (y)-z\log (z))\,dy \\&\ge \frac{1}{2}\int _{z-m}^zy^{-1/2}\log (y)^{-2}e^{-\frac{1}{2}(y-z)^2\gamma ^2}\,dy\\&\ge \frac{1}{2}z^{-1/2}(\log z)^{-2}\int _{z-m}^ze^{-\frac{1}{2}(y-z)^2\gamma ^2}\,dy\\&=\frac{1}{2}z^{-1/2}(\log z)^{-2}\gamma ^{-1}\int _0^{m\gamma }e^{-\frac{1}{2}y^2}\,dy\\&\ge \frac{1}{4}z^{-1/2}(\log z)^{-2}\gamma ^{-1}. \end{aligned}$$

The last estimate uses the fact that $m\gamma =(\pi /3)(2/3)(1+\log K)^{1/2}>1$ to show that

$$\begin{aligned} \int _0^{m\gamma }e^{-\frac{1}{2}y^2}\,dy\ge \int _0^1e^{-\frac{1}{2}}\,dy\ge \frac{1}{2}. \end{aligned}$$

To estimate $B_1$ we apply Lemma 6.2 with $x=0$. Since $n\gamma ^2=(\pi /2)(2/3)^2(1+\log K)^2\ge 5/(2e)$ we can show that $y^{-1/2}\log (y)^{-2}e^{-\frac{1}{2}(y-z)^2\gamma ^2}$ is an increasing function of y on $[e,z-n]$ by estimating the derivative of its logarithm to get

$$\begin{aligned} -\frac{1}{2y}-\frac{2}{y\log (y)}+(z-y)\gamma ^2\ge -\frac{1}{2e}-\frac{2}{e}+n\gamma ^2\ge 0. \end{aligned}$$

Thus,

$$\begin{aligned} B_1\le & {} 24 \log (z-n)^{-2}e^{-\frac{1}{2}n^2\gamma ^2}\le 24e^{-\frac{1}{2}n^2\gamma ^2}\\= & {} 24e^{-\frac{1}{2}\left( \frac{\pi }{2}\right) ^2\left( \frac{2}{3}\right) ^2(1+\log K)} \le 24K^{-\pi ^2/18}. \end{aligned}$$

For $y>z$ it is clear that the function $y^{-1/2}\log (y)^{-2}e^{-\frac{1}{2}(y-z)^2\gamma ^2}$ is decreasing, so we may use Lemma 6.2, with $x=0$, to estimate $B_2$ as well. We have

$$\begin{aligned} B_2&\le 24\log (z+n)^{-2}e^{-\frac{1}{2}n^2\gamma ^2}+12\int _{z+n}^K\log (y)^{-2}e^{-\frac{1}{2}(y-z)^2\gamma ^2}\,\frac{dy}{y}\\&\le 24\log (z+n)^{-2}e^{-\frac{1}{2}n^2\gamma ^2} +12e^{-\frac{1}{2}n^2\gamma ^2}\log (z+n)^{-1}\le 36K^{-\pi ^2/18}. \end{aligned}$$

Since $\pi ^2/18>1/2$,

$$\begin{aligned} \frac{1}{2}A\ge \frac{1}{8}K^{-1/2}(\log K)^{-2}\frac{3}{2}(1+\log K)^{-3/2}\ge 60 K^{-\pi ^2/18}=B_1+B_2. \end{aligned}$$

for K sufficiently large. It follows that for such K, and $z\in (e+1,K-1)$,

$$\begin{aligned} |{\hat{g}}(z)|\ge A-B_1-B_2\ge \frac{1}{2}A\ge \frac{1}{8}z^{-1/2}(\log z)^{-2}\gamma ^{-1}. \end{aligned}$$

Now

$$\begin{aligned} \Vert {\hat{g}}\Vert _{L^q(U)}&\ge \frac{1}{8}\gamma ^{-1}\bigg (\int _{e+1}^{K-1} z^{-q/2}(\log z)^{-2q}(z+1)^{\beta -1}\,dz\bigg )^{1/q}\\&\ge \frac{3}{16}(1+\log K)^{-3/2}(1+\log K)^{-2}\bigg (\int _{e+1}^{K-1} (z+1)^{-q/2+\beta -1}\,dz\bigg )^{1/q}\\&= \frac{3}{16}(1+\log K)^{-7/2}\Big (\beta -\frac{q}{2}\Big )^{-1/q} (K^{\beta -q/2}-(e+2)^{\beta -q/2})^{1/q}. \end{aligned}$$

This goes to infinity as $K\rightarrow \infty$. We conclude that there is no finite constant C such that $\Vert {\hat{f}}\Vert _{L^q(U)}\le C\Vert f\Vert _{L^p(W)}$ for all $f\in L^1$. $\square$

As we have pointed out earlier, the weight conditions (3.1) and (3.2) are equivalent when $q>1$. Thus the weight condition of [3, Theorem 1(ii)] holds in the example above, yet the Fourier inequality fails.

The weight condition of [3, Theorem 4(iv)] holds as well because, as we have seen in Sect. 4, the weight conditions of [3, Theorems 1 and 4] are equivalent whenever the appropriate a priori condition holds. To see that the weight u of Example 6.3 satisfies $u^{1-q'}\in B_{q'}$, observe that the continuous function of x given by

$$\begin{aligned} \frac{x^{q'}\int _x^\infty \big (\frac{t}{2}+1\big )^{(1-\beta )(q'-1)}t^{-q'}\,dt}{\int _0^x\big (\frac{t}{2}+1\big )^{(1-\beta )(q'-1)}\,dt} \end{aligned}$$

is bounded as $x\rightarrow 0$ and as $x\rightarrow \infty$. The details are omitted.