Abstract
In this paper, we introduce the concept of operator geometrically convex functions for positive linear operators and prove some Hermite–Hadamard type inequalities for these functions. As applications, we obtain trace inequalities for operators which give some refinements of previous results.
Similar content being viewed by others
Avoid common mistakes on your manuscript.
1 Introduction and preliminaries
Let \({\mathcal {A}}\) be a sub-algebra of B(H) stand for the commutative \(C^{*}\)-algebra of all bounded linear operators on a complex Hilbert space H with inner product \(\langle \cdot ,\cdot \rangle \). An operator \(A\in {\mathcal {A}}\) is positive and write \(A\ge 0\) if \(\langle Ax,x\rangle \ge 0\) for all \(x\in H\). Let \({\mathcal {A}}^{+}\) stand for all strictly positive operators in \({\mathcal {A}}\).
Let A be a self-adjoint operator in \({\mathcal {A}}\). The Gelfand map establishes a \(*\)-isometrically isomorphism \(\varPhi \) between the set \(C({{\mathrm{Sp}}}(A))\) of all continuous functions defined on the spectrum of A, denoted \({{\mathrm{Sp}}}(A)\), and the \(C^{*}\)-algebra \(C^{*}(A)\) generated by A and the identity operator \(1_{H}\) on H as follows:
For any \(f,g\in C({{\mathrm{Sp}}}(A)))\) and any \(\alpha , \beta \in \mathbb {C}\) we have:
-
\(\varPhi (\alpha f+\beta g)=\alpha \varPhi (f)+\beta \varPhi (g);\)
-
\(\varPhi (fg)=\varPhi (f)\varPhi (g)\) and \( \varPhi (\bar{f})=\varPhi (f)^{*};\)
-
\(\Vert \varPhi (f)\Vert =\Vert f\Vert :=\sup _{t\in {{\mathrm{Sp}}}(A)}|f(t)|;\)
-
\(\varPhi (f_{0})=1_{H}\) and \(\varPhi (f_{1})=A,\) where \(f_{0}(t)=1\) and \(f_{1}(t)=t\), for \(t\in {{\mathrm{Sp}}}(A)\).
with this notation we define
and we call it the continuous functional calculus for a self-adjoint operator A.
If A is a self-adjoint operator and f is a real valued continuous function on \({{\mathrm{Sp}}}(A)\), then \(f(t)\ge 0\) for any \(t\in {{\mathrm{Sp}}}(A)\) implies that \(f(A)\ge 0\), i.e. f(A) is a positive operator on H. Moreover, if both f and g are real valued functions on \({{\mathrm{Sp}}}(A)\) then the following important property holds:
in the operator order of B(H), see [17].
Let I be an interval in \({\mathbb {R}}\). Then \(f:I\rightarrow {\mathbb {R}}\) is said to be convex function if
for \(a,b\in I\) and \(\lambda \in [0,1]\).
The following inequality holds for any convex function f defined on \({\mathbb {R}}\)
It was firstly discovered by Hermite in 1881 in the journal Mathesis (see [9]). But this result was nowhere mentioned in the mathematical literature and was not widely known as Hermite’s result [13].
Beckenbach, a leading expert on the history and the theory of convex functions, wrote that this inequality was proven by Hadamard in 1893 [1]. In 1974, Mitrinovič found Hermites note in Mathesis [9]. Since (2) was known as Hadamards inequality, the inequality is now commonly referred as the Hermite–Hadamard inequality [13].
Definition 1
[11] A continuous function \(f:I\subset {\mathbb {R}}^{+}\rightarrow {\mathbb {R}}^{+}\) is said to be geometrically convex function (or multiplicatively convex function) if
for \(a,b\in I\) and \(\lambda \in [0,1]\).
The author of [7] established the Hermite–Hadamard type inequalities for geometrically convex functions as follows:
Theorem 1
Let \(f:I\subseteq {\mathbb {R}}^{+}\rightarrow {\mathbb {R}}^{+}\) be a geometrically convex function and \(a,b\in I\) with \(a< b\). If \(f\in L^{1}[a,b]\), then
By changing variables \(t=a^{\lambda }b^{1-\lambda }\) we have
Remark 1
It is well-known that for positive numbers a and b
The author of [8] mentioned the following inequality, but here we provide a short proof which gives a refinement for above theorem.
Theorem 2
Let f be a geometrically convex function defined on I a sub-interval of \({\mathbb {R}}^{+}\). Then, we have
for \(a,b\in I\).
Proof
Since f is geometrically convex function, we can write
for all \(\lambda \in [0,1]\).
So, we have
Integrate (3) over [0, 1], we have
\(\square \)
Lemma 1
[11, Page 156] Suppose that I is a subinterval of \({\mathbb {R}}^{+}\) and \(f:I\rightarrow (0,\infty )\) is a geometrically convex function. Then
is a convex function. Conversely, if J is an interval for which \(\exp (J)\) is a subinterval of \({\mathbb {R}}^{+}\) and \(F: J\rightarrow {\mathbb {R}}\) is a convex function, then
is geometrically convex function.
Theorem 3
Let f be a geometrically convex function defined on [a, b] such that \(0<a<b\). Then, we have
for \(a, b\in I\).
Proof
Let \(f:[a,b]\rightarrow {\mathbb {R}}\) be a geometrically convex function. So, by Lemma 1 we have
is convex.
Then, by [12, Remark 1.9.3]
By definition of F, we obtain
It follows that
Since \(\exp (x)\) is increasing, we have
Using change of variable \(t=\exp (x)\) to obtain the desired result. \(\square \)
The author of [11, p. 158] showed that every polynomial P(x) with non-negative coefficients is a geometrically convex function on \([0,\infty )\). More generally, every real analytic function \(f(x)=\sum _{n=0}^{\infty }c_{n}x^{n}\) with non-negative coefficients is geometrically convex function on (0, R) where R denotes the radius of convergence. This gives some different examples of geometrically convex function. It is easy to show that \(\exp (x)\) is geometrically convex function.
In this paper, we introduce the concept of operator geometrically convex functions and prove the Hermite–Hadamard type inequalities for these class of functions. These results lead us to obtain some inequalities for trace functional of operators.
2 Inequalities for operator geometrically convex functions
In this section, we prove Hermite–Hadamard type inequality for operator geometrically convex function.
In [4] Dragomir investigated the operator version of the Hermite–Hadamard inequality for operator convex functions. Let \(f:I\rightarrow {\mathbb {R}}\) be an operator convex function on the interval I then, for any self-adjoint operators A and B with spectra in I, the following inequalities holds
for the first inequality in above, see [15].
To give operator geometrically convex function definition, we need following lemmas.
Lemma 2
[10, Lemma 3] Let A and B be two operators in \({\mathcal {A}}^{+}\), and f a continuous function on \({{\mathrm{Sp}}}(A)\). Then, \(AB=BA\) implies that \(f(A)B=Bf(A)\).
Since \(f(t)=t^{\lambda }\) is continuous function for \(\lambda \in [0,1]\) and \({\mathcal {A}}\) is a commutative \(C^{*}\)-algebra, we have \(A^{\lambda }B=BA^{\lambda }\). Moreover, by applying above lemma for \(f(t)=t^{1-\lambda }\) again, we have \(A^{\lambda }B^{1-\lambda }=B^{1-\lambda }A^{\lambda }\), for operators A and B in \({\mathcal {A}}^{+}\). It means \(A^{\lambda }\) and \(B^{1-\lambda }\) commute together whenever A and B commute.
Lemma 3
Let A and B be two operators in \({\mathcal {A}}^{+}\). Then
is convex.
Proof
We know that \(\{\lambda A+(1-\lambda )B : 0\le \lambda \le 1\}\) is convex for arbitrary operator A and B. So, \(\{\lambda \log A+(1-\lambda )\log B : 0\le \lambda \le 1\}\) is convex. Since A and B are commutative and knowing that \(e^{f}\) is convex when f is convex, we have
So, \(A^{\lambda }B^{1-\lambda }\) is convex for \(0\le \lambda \le 1\). \(\square \)
Lemma 4
[17, Theorem 5.3] Let A and B be in a Banach algebra such that \(AB=BA\). Then
Let A and B be two positive operators in \({\mathcal {A}}\) with spectra in I. Now, Lemma 2 and functional calculus [17, Theorem 10.3(c)] imply that
for \(0\le \lambda \le 1\).
Definition 2
A continuous function \(f:I\subseteq {\mathbb {R}}^{+}\rightarrow {\mathbb {R}}^{+}\) is said to be operator geometrically convex if
for \(A, B\in {\mathcal {A}}^{+}\) such that \({{\mathrm{Sp}}}(A)\), \({{\mathrm{Sp}}}(B)\subseteq I\).
Now, we are ready to prove Hermite–Hadamard type inequality for operator geometrically convex functions.
Theorem 4
Let f be an operator geometrically convex function. Then, we have
for \(0\le t\le 1\) and \(A,B\in {\mathcal {A}}^{+}\) such that \({{\mathrm{Sp}}}(A), {{\mathrm{Sp}}}(B)\subseteq I\).
Proof
Since f is operator geometrically convex function, we have \(f(\sqrt{AB})\le \sqrt{f(A)f(B)}\). Let replace A and B by \(A^{t}B^{1-t}\) and \(A^{1-t}B^{t}\) respectively, we obtain
It is well-known that \(\log t\) is operator monotone function on \((0,\infty )\) (see [16]), i.e., \(\log t\) is operator monotone function if \(\log A\le \log B\) when \(A\le B\). So, by above inequality, we have
Therefore,
Integrate above inequality over [0, 1], we can write the following
The last above equality follows by knowing that
Hence, from (10), we have
This proved left inequality of (8).
On the other hand, we have \(f(A^{t}B^{1-t})\le f(A)^{t}f(B)^{1-t}\). It follows that
So,
Now, integrate of (11) on [0, 1], we have
This completes the proof. \(\square \)
We should mention, when f is operator geometrically convex function, then we have
So, we have
Integrate above inequality over [0, 1], we obtain
for \(0\le t\le 1\) and \(A,B\in {\mathcal {A}}^{+}\) such that \({{\mathrm{Sp}}}(A), {{\mathrm{Sp}}}(B)\subseteq I\).
Let \(A, B\in {\mathcal {A}}\) and \(A\le B\), by continuous functional calculus [17, Theorem 10.3(b)], we can easily obtain \(\exp (A)\le \exp (B)\). This means \(\exp (t)\) is operator monotone on \([0,\infty )\) for \(A,B\in {\mathcal {A}}\).
On the other hand, like the classical case, the arithmetic-geometric mean inequality holds for operators as following
with respect to operator order for positive non-commutative operator in B(H). Whenever, A and B commute together, then inequality (12) reduces to
Since \(\exp (t)\) is an operator monotone function, by above inequality we have
for \(A, B\in {\mathcal {A}}^{+}\) and \(\nu \in [0,1]\). So, in this case \(\exp (t)\) is an operator geometrically convex function on \([0,\infty )\).
Let replace f in Theorem 4 by \(\exp (t)\) as an operator geometrically convex function, we have
So,
for \(A, B\in {\mathcal {A}}^{+}\).
Here, we mention some remarks for operator geometrically convex functions.
Remark 2
\(f(x)=\Vert x\Vert \) is geometrically convex function for usual operator norms since the following hold
Above inequality is a special case of McIntosh inequality.
Remark 3
If f(t) is an operator geometrically convex function, then so is \(g(t)=tf(t)\)
for \(\alpha \in [0,1]\) and \(A, B\in {\mathcal {A}}^{+}\).
Remark 4
Operator geometrically convex functions is an algebra with some complication of operators spectra. To see this we make use of the following inequality
for \(A, B, C, D\in {\mathcal {A}}^{+}\).
Let f and g be operator geometrically convex functions.
First, we prove that \(f+g\) is an operator geometrically convex function
for \(A, B\in {\mathcal {A}}^{+}\). In the last inequality above we applied (15).
Second, we show that mf is an operator geometrically convex function for a scalar m
for \(A, B\in {\mathcal {A}}^{+}\).
Third, \(h=fg\) is an operator geometrically convex function
for \(A, B\in {\mathcal {A}}^{+}\).
Let \(\{e_{i}\}_{i\in I}\) be an orthonormal basis of H, we say that \(A\in B(H)\) is trace class if
The definition of \(\Vert A\Vert _{1}\) does not depend on the choice of the orthonormal basis \(\{e_{i}\}_{i\in I}\). We denote by \(B_{1}(H)\) the set of trace class operators in B(H).
We define the trace of a trace class operator \(A\in B_{1}(H)\) to be
where \(\{e_{i}\}_{i\in I}\) an orthonormal basis of H.
Note that this coincides with the usual definition of the trace if H is finite-dimensional. We observe that the series (17) converges absolutely.
The following result collects some properties of the trace:
Theorem 5
We have
-
1.
If \(A\in B_{1}(H)\) then \(A^{*}\in B_{1}(H)\) and
$$\begin{aligned} {{\mathrm{Tr}}}(A^{*})=\overline{{{\mathrm{Tr}}}(A)}; \end{aligned}$$(18) -
2.
If \(A\in B_{1}(H)\) and \(T\in B(H)\), then \(AT, TA\in B_{1}(H)\) and
$$\begin{aligned} {{\mathrm{Tr}}}(AT)={{\mathrm{Tr}}}(TA) \quad and \quad \Vert {{\mathrm{Tr}}}(AT)\Vert \le \Vert A\Vert _{1}\Vert T\Vert ; \end{aligned}$$(19) -
3.
\({{\mathrm{Tr}}}(\cdot )\) is a bounded linear functional on \(B_{1}(H)\) with \(\Vert {{\mathrm{Tr}}}\Vert =1\);
-
4.
If \(A, B\in B_{1}(H)\) then \({{\mathrm{Tr}}}(AB)={{\mathrm{Tr}}}(BA)\).
For the theory of trace functionals and their applications the reader is referred to [14].
For \(A,B\ge 0\) we have \({{\mathrm{Tr}}}(AB)\le {{\mathrm{Tr}}}(A){{\mathrm{Tr}}}(B)\). Also, since \(f(t)=t^{\frac{1}{2}}\) is monotone we have
for positive operator A and B in B(H).
We know that \(f(t)={{\mathrm{Tr}}}(t)\) is operator geometrically convex function [6, p. 513], i.e.
for \(0\le t\le 1\) and positive operators \(A,B\in B_{1}(H) \).
For commutative case, we have
since \(({{\mathrm{Tr}}}(AB))^{\frac{1}{2}}\le {{\mathrm{Tr}}}(AB)^{\frac{1}{2}}\).
Moreover, by Theorem 4 we can write
Let replace A and B by \(A^{2}\) and \(B^{2}\) in above inequality, respectively. By applying commutativity of algebra and knowing that \({{\mathrm{Tr}}}(A)^{2}\le ({{\mathrm{Tr}}}A)^{2}\) for positive operator A, we have
3 More results on trace functional class for product of operators
In this section we prove some trace functional class inequalities for operators which are not necessarily commutative.
We consider the wide class of unitarily invariant norms \(|||\cdot |||\). Each of these norms is defined on an ideal in B(H) and it will be implicitly understood that when we talk of |||T|||, then the operator T belongs to the norm ideal associated with \(|||\cdot |||\). Each unitarily invariant norm \(|||\cdot |||\) is characterized by the invariance property \(|||UTV|||=|||T|||\) for all operators T in the norm ideal associated with \(|||\cdot |||\) and for all unitary operators U and V in B(H). For \(1\le p<\infty \), the Schatten p-norm of an operator \(A\in B_{1}(H)\) defined by \(\Vert A\Vert _{p}=({{\mathrm{Tr}}}|A|^{p})^{1/p}\). These Schatten p-norms are unitarily invariant.
In [2], Bhatia and Davis proved the following inequality
for all operators A, B, X and \(r\ge 0\).
As we know, \(\Vert A\Vert _{1}={{\mathrm{Tr}}}|A|\). From (21) for \(p=1\), we have
So, by inequality (22), we can write
for all operators \(A, B\in B_{1}(H)\), \(X\in B(H)\) and \(r\ge 0\).
Let, \(X=I\) in above inequality, we have
Moreover, let \(r=1\) in inequality (23), we have
Put \(X^{*}\) instead of X and applying the property of trace we have
for all \(A, B\in B_{1}(H)\) and \(X\in B(H)\).
Let \(X=I\) in (24).
Corollary 1
Let \(A, B\in B_{1}(H)\). Then
In [5, Theorem 5], Dragomir proved the following inequality for \(X\in B(H)\), \(A,B\in B_{1}(H)\) and \(\alpha \in [0,1]\)
Here, we give a generalization for above inequality when \(\alpha \in {\mathbb {R}}\).
Theorem 6
Let \(X\in B_{1}(H)\), \(A, B\in B(H)\) and \(\alpha \in {\mathbb {R}}\). Then
Proof
Let replace A and B in Corollary 1 with \(|X|^{\alpha }A\) and \(|X|^{(1-\alpha )}B\), where \(\alpha \in {\mathbb {R}}\). It follows that
So, we have
\(\square \)
Let \(A=B=I\) in Theorem 6, we have
for \(X\in B_{1}(H)\) and \(\alpha \in {\mathbb {R}}\). Above inequality is a refinement for [5, Inequality(3.1)].
Also, let \(X\in B_{1}(H)\) and normal operators \(A, B\in B(H)\). For \(\alpha \in {\mathbb {R}}\), we have
In [3, Theorem 2.3], Dannan proved that if \(S_{i}\) and \(T_{i}\) (\(i=1,2,\ldots ,n\)) are positive definite matrices, then we have
Moreover, if \(S_{i}T_{i}\ge 0\), (\(i=1,2,\ldots ,n\)). Then
So,
Here, we prove inequality (27) for arbitrary operators.
Theorem 7
Let \(S_{i}\) and \(T_{i}\) \((i=1,2,\ldots ,n)\) be arbitrary operators in \(B_{1}(H)\). Then,
Proof
Let \(A=\left( \begin{array}{ccccc} S_{1} &{} S_{2}&{}\ldots &{}S_{n} \\ 0&{} 0 &{} \ldots &{}0 \\ \vdots &{}\vdots &{} \ddots &{}\vdots \\ 0 &{} 0 &{} 0 &{} 0\\ \end{array}\right) \) and \(B=\left( \begin{array}{ccccc} T_{1} &{} T_{2}&{}\ldots &{}T_{n} \\ 0&{} 0 &{} \ldots &{}0 \\ \vdots &{}\vdots &{} \ddots &{}\vdots \\ 0 &{} 0 &{} 0 &{} 0\\ \end{array}\right) \). So, we have
Put A and B in inequality (25), by property of trace, we obtain the desired result. \(\square \)
Corollary 2
Let \(S_{i}\) and \(T_{i}\) \((i=1,2,\ldots ,n)\) be positive operators in \(B_{1}(H)\). Then, we have
Proof
By Theorem 7 for positive operators \(S_{i}\) and \(T_{i}\), we obtain
Since \(S_{i}\) and \(T_{i}\) are positive operators, we have \({{\mathrm{Tr}}}(S_{i}T_{i})\ge 0\). It follows that \({{\mathrm{Tr}}}(\sum _{i=1}^{n}S_{i}T_{i})\ge 0\) because \({{\mathrm{Tr}}}\left( \sum _{i=1}^{n}S_{i}T_{i}\right) =\sum _{i=1}^{n}{{\mathrm{Tr}}}(S_{i}T_{i})\). So,
\(\square \)
References
Beckenbach, E.F.: Convex functions. Bull. Am. Math. Soc. 54, 439–460 (1948)
Bhatia, R., Davis, C.: A Cauchy–Schwarz inequality for operators with applications. Linear Algebra Appl. 223(224), 119–129 (1995)
Dannan, F.M.: Matrix and operator inequalities. J. Inequal. Pure Appl. Math. 2 (2001) (article 34)
Dragomir, S.S.: Hermite–Hadamards type inequalities for operator convex functions. Appl. Math. Comput. 218, 766–772 (2011)
Dragomir, S.S.: Some inequalities for trace class operators via a Kato’s result. RGMIA Res. Rep. Coll. 17 (2014) (preprint, article 105)
Horn, R.A., Johnson, C.R.: Matrix Analysis, 2nd edn. Cambridge University Press, Cambridge (2012)
İscan, İ.: Some new Hermite–Hadamard type inequalities for geometrically convex functions. Math. Stat. 1, 86–91 (2013)
İscan, İ.: On some new Hermite–Hadamard type inequalities for \(s\)-geometrically convex functions. Int. J. Math. Math. Sci. 2014 (2014) (article ID 163901, 8 pages)
Mitrinović, D.S., Lacković, I.B.: Hermite and convexity. Aequ. Math. 28, 229–232 (1985)
Nagisa, M., Ueda, M., Wada, S.: Commutativity of operators. Nihonkai Math. J. 17, 1–8 (2006)
Niculescu, C.P.: Convexity according to the geometric mean. Math. Inequal. Appl. 3, 155–167 (2000)
Niculescu, C.P., Persson, L.E.: Convex Functions and their Applications: A Contemporary Approach. Springer, New York (2006)
Pečarić, J.E., Proschan, F., Tong, Y.L.: Convex Functions, Partial Orderings, and Statistical Applications. Academic Press Inc, San Diego (1992)
Simon, B.: Trace Ideals and Their Applications. Cambridge University Press, Cambridge (1979)
Taghavi, A., Darvish, V., Nazari, H.M., Dragomir, S.S.: Some inequalities associated with the Hermite–Hadamard inequalities for operator \(h\)-convex functions. RGMIA Research Report Collection, vol. 18 (2015) (article 22)
Zhan, X.: Matrix Inequalities. Springer, Berlin (2002)
Zhu, K.: An Introduction to Operator Algebras. CRC Press, Boca Raton (1993)
Acknowledgments
This work was written whilst the second author was visiting Victoria University during his short sabbatical leave provided by the Ministry of Science, Research and Technology. He thanks them for the support and hospitality.
Author information
Authors and Affiliations
Corresponding author
Additional information
Communicated by A. Constantin.
Rights and permissions
About this article
Cite this article
Taghavi, A., Darvish, V., Nazari, H.M. et al. Hermite–Hadamard type inequalities for operator geometrically convex functions. Monatsh Math 181, 187–203 (2016). https://doi.org/10.1007/s00605-015-0816-6
Received:
Accepted:
Published:
Issue Date:
DOI: https://doi.org/10.1007/s00605-015-0816-6
Keywords
- Hermite–Hadamard inequality
- Operator geometrically convex function
- Trace inequality
- Unitarily invariant norm