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 Open Access
Jessen’s type inequality and exponential convexity for positive \(C_{0}\)semigroups
 Gul I Hina Aslam^{1}Email author and
 Matloob Anwar^{1}
https://doi.org/10.1186/s136600150877z
© Hina Aslam and Anwar 2015
 Received: 1 April 2015
 Accepted: 29 October 2015
 Published: 9 November 2015
Abstract
In this paper a Jessen’s type inequality for normalized positive \(C_{0}\)semigroups is obtained. An adjoint of Jessen’s type inequality has also been derived for the corresponding adjoint semigroup, which does not give the analogous results but the behavior is still interesting. Moreover, it is followed by some results regarding positive definiteness and exponential convexity of complex structures involving operators from a semigroup.
Keywords
 positive semigroups
 Jessen’s inequality
 Banach lattice algebra
 exponential convexity
MSC
 47D03
 46B42
 43A35
 43A17
1 Introduction and preliminaries
A significant theory regarding inequalities and exponential convexity for realvalued functions has been developed [1, 2]. The intention to generalize such concepts for the \(C_{0}\)semigroup of operators is motivated from [3].
In the present article, we shall derive a Jessen type inequality and the corresponding adjoint inequality, for some \(C_{0}\)semigroup and the adjoint semigroup, respectively.
The notion of Banach lattice was introduced to get a common abstract setting, within which one could talk about the ordering of elements. Therefore, the phenomena related to positivity can be generalized. It had mostly been studied in various types of spaces of realvalued functions, e.g. the space \(C(K)\) of continuous functions over a compact topological space K, the Lebesque space \(L^{1}(\mu)\) or even more generally the space \(L^{p}(\mu)\) constructed over measure space \((X,\Sigma,\mu)\) for \(1\leq p\leq\infty\). We shall use without further explanation the terms: order relation (ordering), ordered set, supremum, infimum.
First, we shall go through the definition of a vector lattice.
Definition 1
[4]
 \(O_{1}\)::

\(f\leq g\) implies \(f+h\leq g+h\) for all \(f,g,h\in V\),
 \(O_{2}\)::

\(f\geq0\) implies \(\lambda f\geq0\) for all \(f\in V\) and \(\lambda\geq0\),

An ordered vector space V is called a vector lattice, if any two elements \(f,g\in V\) have a supremum, which is denoted by \(\sup(f,g)\) and an infimum denoted by \(\inf(f,g)\).
It is trivially understood that the existence of supremum of any two elements in an ordered vector space implies the existence of supremum of finite number of elements in V. Furthermore, \(f\geq g\) implies \(f\leqg\), so the existence of finite infima is therefore implied.

Some important quantities are defined as follows:$$\begin{aligned}& \sup(f,f) = \vert f\vert \quad (\mathit{absolute\ value\ of\ }f), \\& \sup(f,0) = f^{+}\quad (\mathit{positive\ part\ of\ }f), \\& \sup(f,0) = f^{} \quad (\mathit{negative\ part\ of\ }f). \end{aligned}$$

Some compatibility axiom is required between norm and order. This is given in the following short way:The norm defined on a vector lattice is called a lattice norm.$$ \vert f\vert \leq \vert g\vert \quad \mbox{implies}\quad \Vert f\Vert \leq \Vert g\Vert . $$(1)
Definition 2
A Banach lattice is a Banach space V endowed with an ordering ≤, such that \((V,\leq)\) is a vector lattice with a lattice norm defined on it.
A Banach lattice transforms to Banach lattice algebra, provided \(u,v\in V_{+}\) implies \(uv\in V_{+}\).
Thus \(\psi:V\rightarrow V\) is positive if and only if \(\vert \psi f\vert \leq\psi \vert f\vert \) holds for any \(f\in V\).
Lemma 1
([4], p.249)
A bounded linear operator ψ on a Banach lattice V is a positive contraction if and only if \(\Vert (\psi f)^{+}\Vert \leq \Vert f^{+}\Vert \) for all \(f\in V\).
Definition 3
 (i)
\(Z(s)Z(t)=Z(s+t)\) for all \(s,t\in\mathbb{R}^{+}\).
 (ii)
\(Z(0)=I\), the identity operator on V.
 (iii)
For each fixed \(f\in V\), \(Z(t)f\rightarrow f\) (with respect to the norm on V) as \(t\rightarrow0^{+}\).
Here \(B(V)\) denotes the space of all bounded linear operators defined on a Banach space V.
Definition 4
2 Jessen’s type inequality
In 1931, Jessen [5] gave the generalization of the Jensen’s inequality for a convex function and positive linear functionals. See [6], p.47. We shall prove this inequality for a normalized positive \(C_{0}\)semigroup and a convex operator defined on a Banach lattice.
Throughout the present section, V will always denote a unital Banach lattice algebra, endowed with an ordering ≤.
Definition 5
Let \(\mathfrak{D}_{c}(V)\) denotes the set of all differentiable convex functions \(\phi:V\rightarrow V\).
Theorem 1
(Jessen’s type inequality)
Proof
The existence of an identity element and condition (3), imposed in the hypothesis of the above theorem, is necessary. We shall elaborate all this by the following examples.
Example 1
Example 2
Let \(\Gamma:=\{z\in\mathbb{C}:\vert z\vert =1\}\), and \(X=C(\Gamma)\). The rotation semigroup \(\{Z(t)\}_{t\geq0}\) is defined as \(T(t)f(z)=f(e^{it}\cdot z)\), \(f\in X\). The identity element \(E\in X\), s.t. for all \(z\in\Gamma\), \(E(z)=z\). Then \(Z(t)E(z)=E(e^{it}\cdot z)=e^{it}\cdot z\). Or we can say that any complex number \(z=e^{ix}\) is mapped to \(e^{i(x+t)}\). \(Z(t)\) satisfies (3), only when t is a multiple of 2π. Let \(f(z)=\mathfrak{R}(z)+1 >0\), then \((Z(t)f)(e^{iz})=f(e^{i(t+z)})=\cos(t+z)+1\), hence \(\phi (Z(t)f)(e^{iz})=\cos(tz)+1\). On the other hand, \(\phi(f)=f\), and \(Z(t)(\phi f)(e^{iz})=(Z(t)f)(e^{iz})=\cos(t+z)+1\). Hence, equality holds in (5) when t is a multiple of 2π, but the two sides are not comparable in general. It can easily be verified that \(Z'=\{Z(2\pi t)\}_{t=0}^{\infty}\) is a subgroup of \(Z=\{ Z(t)\}_{t=0}^{\infty}\), as \(Z(2\pi t)Z(2\pi s)=Z(2\pi(t+s))\). Therefore \(Z'\) is a normalized semigroup. See Figure 1(b).
3 Adjoint Jessen’s type inequality
In the previous section, a Jessen type inequality has been derived for a normalized positive \(C_{0}\)semigroup \(\{Z(t)\}_{t\geq0}\). This gives us the motivation toward finding the behavior of its corresponding adjoint semigroup \(\{Z^{\ast}(t)\}_{t\geq0}\) on \(V^{\ast}\). As the theory for dual spaces gets more complicated, we do not expect to have the analogous results. One may ask for a detailed introduction toward a part of the dual space \(V^{\ast}\), for which an adjoint of Jessen’s type inequality makes sense.
Definition 6
If X is an ordered vector space, we say that a functional \(x^{\ast}\) on X is positive if \(x^{\ast}(x)\geq0\), for each \(x\in X\). By the linearity of \(x^{\ast}\), this is equivalent to \(x^{\ast}\) being order preserving; i.e. \(x\leq y\) implies \(x^{\ast}(x)\leq x^{\ast}(y)\). The set P of all positive linear functionals on X is a cone in \(X^{\ast}\).
We are mainly interested in the study of the space \(V^{\ast}\), where in our case V is a Banach lattice algebra. Let us consider the regular ordering among the elements of \(V^{\ast}\), i.e. \(v_{1}^{\ast}\geq v_{2}^{\ast}\), whenever \(v_{1}^{\ast}(v)\geq v_{2}^{\ast}(v)\), for each \(v\in V\).
Consider the convex operator (4). In the case of equality, F is simply a linear operator and the adjoint F can be defined as above. But how can it be defined in the other case? This question has already been answered.
Definition 7
This is, of course, a straightforward generalization of (6); in fact, for linear operators L we have \(L^{\sharp}_{Y^{\ast}} = L^{\ast}\); i.e. the restriction of the pseudoadjoint to the dual space is the classical adjoint.
For the sake of convenience, we shall denote the adjoint of the operator F by \(F^{\ast}\) throughout the present section. Either it is a classical adjoint or the pseudoadjoint (depending upon the operator F).
Similarly, the considered dual space of the vector lattice algebra V will be denoted by \(V^{\ast}\), which can be the intersection of the pseudodual and classical dual spaces in the case of a nonlinear convex operator.
Lemma 2
Let F be the convex operator on a Banach space X, then the adjoint operator \(F^{\ast}\) on the dual space \(X^{\ast}\) is also convex.
Proof
Theorem 2
(Adjoint Jessen’s inequality)
Proof
4 Exponential convexity
In this section we shall define the exponential convexity of an operator. Moreover, some complex structures, involving the operators from a semigroup, will be proved to be exponentially convex.
Definition 8
Proposition 1
 (i)
H is exponentially convex.
 (ii)H is continuous and for all \(n\in\mathbb{N}\)where \(\xi_{i}\in\mathbb{R}\) and \(x_{i}\in I\subseteq\mathbb{R}\), \(1\leq i\leq n\).$$ \sum_{i,j=1}^{n} \xi_{i}\xi_{j} H \biggl(\frac{x_{i}+x_{j}}{2} \biggr)f\geq0,\quad f \in V, $$(10)
Proof
Remark 1
For \(U\subseteq V\), let us assume that \(F:U\rightarrow V\) is continuously differentiable on U, i.e. the mapping \(F': U\rightarrow\mathfrak{L}(V)\), is continuous. Moreover, \(F''(f)\), will be a continuous linear transformation from V to \(\mathfrak{L}(V)\). A bilinear transformation B defined on \(V\times V\) is symmetric if \(B(f,g)=B(g,f)\) for all \(f,g\in V\). Such a transformation is positive definite (nonnegative definite), if for every nonzero \(f\in V\), \(B(f,f)>0\) (\(B(f,f)\geq0\)). Then \(F''(f)\) is symmetric wherever it exists. See [7], p.69.
Theorem 3
([7], p.100)
Let F be continuously differentiable and suppose that the second derivative exists throughout an open convex set \(U\subseteq V\). Then F is convex on U if and only if \(F''(f)\) is nonnegative definite for each \(f\in U\). If \(F''(f)\) is positive definite on U, then F is strictly convex.
Definition 9
[11]
Lemma 3
Theorem 4
 (i)For every \(n\in\mathbb{N}\) and for every \(p_{k}\in I\), \(k=1,2,\ldots,n\),$$ [\Lambda_{\frac{p_{i}+p_{j}}{2}} ]_{i,j=1}^{n}\geq0. $$(13)
 (ii)
If the mapping \(f\rightarrow\Lambda_{t}\) is continuous on I, then it is exponentially convex on I.
Proof
Lemma 4
Theorem 5
For \(\Lambda_{t} := Z(t)(H_{t}(f))H_{t}(Z(t)f)\), (i) and (ii) from Theorem 4 hold.
Proof
Similar to the proof of Theorem 4. □
Declarations
Acknowledgements
The authors of this paper are grateful to Prof. András Bátkai for his generous help in construction of the examples.
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Authors’ Affiliations
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