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Order generalised gradient and operator inequalities
Journal of Inequalities and Applications volume 2015, Article number: 49 (2015)
Abstract
We introduce the notion of order generalised gradient, a generalisation of the notion of subgradient, in the context of operator-valued functions. We state some operator inequalities of Hermite-Hadamard and Jensen types. We discuss the connection between the notion of order generalised gradient and the Gâteaux derivative of operator-valued functions. We state a characterisation of operator convexity via an inequality concerning the order generalised gradient.
1 Background
Convex functions play a crucial role in many fields of mathematics, most prominently in optimisation theory. There are two main important inequalities which characterise convex functions, namely Jensen’s and Hermite-Hadamard’s inequalities. In 1905 (1906), Jensen defined convex functions as follows: \(f: I\subset\mathbb{R} \rightarrow\mathbb{R}\) is a convex function if and only if
Inequality (1) is referred to as Jensen’s inequality. Hermite-Hadamard’s inequality provides a refinement for Jensen’s inequality, namely, for a convex function \(f: I\subset\mathbb{R} \rightarrow\mathbb{R}\),
We refer the reader to Section 2 for further details regarding these inequalities.
Similarly to the case of real-valued functions, the operator convexity can be characterised by some operator inequalities. Hansen and Pedersen [1] characterise operator convexity via a non-commutative generalisation of Jensen’s inequality. If f is a real continuous function on an interval I, and \(\mathcal{A}(H)\) is the set of bounded self-adjoint operators on a Hilbert space H with spectra in I, then f is operator convex if and only if
for \(x_{1},\dots, x_{n}\in\mathcal{A}(H)\) and \(a_{1},\dots, a_{n}\in\mathcal{B}(H)\) with \(\sum_{i=1}^{n} a_{i}^{*}a_{i}=\mathbf{1}\). We refer the reader to Section 2 for further details regarding this characterisation.
One of the useful differential properties of convex functions is the fact that their one-sided directional derivatives exist universally [2, p.213]. Just as the ordinary two-sided directional derivatives of a differentiable function can be described in terms of gradient vectors, the one-sided directional derivatives can be described in terms of ‘subgradient’ vectors [2, p.213]. A vector \(x^{*}\) is said to be a subgradient of a convex function \(f:K\subset\mathbb{R}^{n}\rightarrow\mathbb{R}\) at point x if
This condition is referred to as the subgradient inequality [2, p.214]. If (3) holds for every \(x\in K\), then (3) characterises the convexity of f (cf. Eisenberg [3, Theorem 1]).
In this paper, we introduce the notion of order generalised gradient (cf. Section 3) for operator-valued functions, which is a generalisation of (3) (without the assumption of convexity) in the settings of bounded self-adjoint operators on a Hilbert space. Furthermore, we state some inequalities of Hermite-Hadamard and Jensen types for the order generalised gradient in Section 4. Finally, in Section 5, we state the connection between the order generalised gradient and Gâteaux derivative of operator-valued functions. We state a characterisation of convexity analogues to (3) in the context of operator-valued functions.
2 Inequalities for convex functions
This section serves as a point of reference for known results regarding some inequalities related to convex functions (both real-valued and operator-valued functions).
2.1 Jensen’s inequality
Jensen’s inequality for convex functions plays a crucial role in the theory of inequalities due to the fact that other inequalities, such as the arithmetic-geometric mean, Hölder, Minkowski and Ky Fan’s inequalities, can be obtained as particular cases of it.
Let C be a convex subset of the linear space X and f be a convex function on C. If \(\mathbf{p}=(p_{1},\dots, p_{n})\) is a probability sequence and \(\mathbf{x}=(x_{1},\dots,x_{n}) \in C^{n}\), then
This inequality is referred to as Jensen’s inequality. Recently, Dragomir [4] obtained the following refinement of Jensen’s inequality:
where f, \(x_{k}\) and \(p_{k}\) are as defined above. For other refinements of Jensen’s inequality, we refer the reader to Pečarić and Dragomir [5] and Dragomir [6].
The above result provides a different approach to the one that Pečarić and Dragomir [5] obtained in 1989
for \(k\geq1\), and p, x are as defined above.
If \(q_{1},\dots, q_{k}\geq0\) with \(\sum_{j=1}^{k} q_{j} =1\), then the following refinement obtained in 1994 by Dragomir [6] also holds:
where \(1\leq k\leq n\) and p, x are as defined above.
For more refinements and applications related to the generalised triangle inequality, the arithmetic-geometric mean inequality, the f-divergence measures, Ky Fan’s inequality, etc., we refer the readers to [6–10] and [4].
2.2 Hermite-Hadamard’s inequality
The following inequality also holds for any convex function f defined on ℝ:
It was first discovered by Hermite in 1881 in the journal Mathesis [11]. However, this result was nowhere mentioned in the mathematical literature and was not widely known as Hermite’s result [12].
Beckenbach, a leading expert on the history and the theory of convex functions wrote that this inequality was proven by Hadamard in 1893 [13]. In 1974, Mitrinović found Hermite’s note in Mathesis [11]. Since (8) was known as Hadamard’s inequality, the inequality is now commonly referred to as Hermite-Hadamard’s inequality [12].
Hermite-Hadamard’s inequality has been extended in many different directions. One of the extensions of this inequality is in the vector space settings. Firstly, we start with the following definitions and notation: Let X be a vector space and x, y be two distinct vectors in X. We define the segment generated by x and y to be the set
For any real-valued function f defined on the segment \([x,y]\), there exists an associated function \(g_{x,y}: [0,1] \rightarrow\mathbb{R}\) with
We remark that f is convex on \([x,y]\) if and only if g is convex on \([0,1]\). For any convex function defined on a segment \([x,y] \subset X\), we have the Hermite-Hadamard integral inequality (cf. Dragomir [14, p.2] and Dragomir [15, p.2]):
which can be derived by the classical Hermite-Hadamard inequality (8) for the convex function \(g_{x,y}:[0,1]\rightarrow\mathbb {R}\). Consider the function \(f(x)=\Vert x\Vert ^{p}\) (\(x\in X\) and \(1\leq p<\infty \)), which is convex on X, then we have the following norm inequality (derived from (9)) [16, p.106]:
for any \(x,y\in X\).
2.3 Non-commutative generalisation of Jensen’s inequality
Hansen [17] discussed Jensen’s operator inequality for operator monotone functions. Motivated by Aujla’s work [18] on the matrix convexity of functions of two variables, Hansen [19] characterised operator convex functions of two variables in terms of a non-commutative generalisation of Jensen’s inequality (cf. [19, Theorem 3.1]). A simplified proof of this result formulated for matrices is given in Aujla [20]. The case for several variables is given in Hansen [21]. The case for self-adjoint elements in the algebra \(M_{n}\) of n-square matrices is given in Hansen and Pedersen [22]. Finally, Hansen and Pedersen [1] presented a generalisation of the above results for self-adjoint operators defined on a Hilbert space.
Theorem 1
We denote by \(\mathcal{B}(H)\) the Banach algebra of all bounded linear operators on the Hilbert space H. If f is a real continuous function on an interval I, and \(\mathcal {A}(H)\) is the set of bounded self-adjoint operators on a Hilbert space H with spectra in I, then the following conditions are equivalent:
-
(i)
f is operator convex;
-
(ii)
\(f(\sum_{i=1}^{n} a_{i}^{*}x_{i}a_{i})\leq\sum_{i=1}^{n} a_{i}^{*}f(x_{i})a_{i}\) for \(x_{1},\dots, x_{n}\in\mathcal{A}(H)\) and \(a_{1},\dots, a_{n}\in\mathcal{B}(H)\) with \(\sum_{i=1}^{n} a_{i}^{*}a_{i}=\mathbf{1}\);
-
(iii)
\(f(v^{*}xv)\leq v^{*}f(x)v\) for any \(x\in\mathcal{A}(H)\) and any isometry \(v\in\mathcal{B}(H)\);
-
(iv)
\(pf(pxp + s (\mathbf{1}-p))p\leq pf(x)p\) for \(x\in\mathcal{A}(H)\), projection \(p\in\mathcal{B}(H)\), every self-adjoint operator x with spectrum in I and \(s\in I\).
2.4 Subgradient inequality
Recall the following definition of a subgradient [2].
Definition 2
A vector \(x^{*}\) is said to be a subgradient of a convex function \(f:K\subset\mathbb{R}^{n}\rightarrow\mathbb{R}\) at point x if
The following theorem is a useful characterisation of convexity (cf. Eisenberg [3, Theorem 1]).
Theorem 3
If U is a nonempty open subset of \(\mathbb{R}^{n}\), \(f:U \rightarrow \mathbb{R}\) is a differentiable function on U, and K is a convex subset of U, then f is convex on K if and only if
where \(f'(y)\) denotes the gradient of f at y.
This theorem has been generalised and employed in obtaining optimality conditions of a non-differentiable minimax programming problem in complex spaces (cf. Lai and Liu [23]). Note that \((x-y)^{T}f'(y)\) can be written as \(f'(y)\cdot(x-y)\), which can be interpreted as the directional derivative of f at point y in \(x-y\) direction.
3 Order generalised gradient
Throughout the paper, we use the following notation. We denote by \(\mathcal{B}(H)\) the Banach algebra of all bounded linear operators on the Hilbert space \((H,\langle\cdot,\cdot\rangle)\), and by \(\mathcal {A}(H)\) the linear subspace of all self-adjoint operators on H. We denote by \(\mathcal{P}_{+}(H) \subset\mathcal{A}(H)\) the convex cone of all positive definite operators defined on H, that is, \(P \in\mathcal {P}_{+}(H)\) if and only if \(\langle Px,x\rangle\geq0\), and for all \(x\in H\), \(\langle Px,x\rangle=0\) implies \(x=0\). This gives a partial ordering (we refer to it as the operator order) on \(\mathcal{A}(H)\), where two elements \(A,B \in\mathcal{A}(H)\) satisfy \(A \leq B\) if and only if \(B-A\in \mathcal{P}_{+}(H)\).
Definition 4
Let \(\mathcal{C}\) be a convex set in \(\mathcal{A}(H)\). A function \(f:\mathcal{C} \rightarrow\mathcal{A}(H)\) has the function \(\nabla _{f}:\mathcal{C}\times\mathcal{A}(H) \rightarrow\mathcal{A}(H)\) as an order generalised gradient if
in the operator order of \(\mathcal{A}(H)\).
Remark 5
We remark that in (12), if f is a real-valued differentiable function on an open set \(U\subset\mathbb{R}^{n}\), and \(\nabla_{f}\) is the gradient of f, then (12) becomes (11). We also note that there is no assumption of convexity at this point. We discuss the convexity case in Section 5.
Proposition 6
If \(Q \in\mathcal{A}(H)\) and \(f:\mathcal{A}(H) \rightarrow\mathcal{A}(H)\), \(f(A)=QA^{2}Q\), then
is an order generalised gradient for f.
Proof
Observe that \(BX+XB\in\mathcal{A}(H)\) and if \(P \in\mathcal{A}(H)\) then \(P(BX+XB)P \in\mathcal{A}(H)\). We need to prove that
for any \(A,B \in\mathcal{A}(H)\), that is,
Since
hence (14) is equivalent to
which is also equivalent to
which always holds. This completes the proof. □
We denote by \(\mathcal{P}(H) \subset\mathcal{A}(H)\) the convex cone of all nonnegative operators defined on H.
Proposition 7
If \(P\in\mathcal{P}(H)\), then the function \(f: \mathcal{A}(H) \rightarrow\mathcal {A}(H)\), \(f(A)=APA\) has
as an order generalised gradient.
Proof
Observe that \(XPB+BPX \in\mathcal{A}(H)\). We need to prove that
that is,
But \(APA-APB-BPA+BPB = (A-B)P(A-B)\) and since \((A-B)P(A-B)\geq0\), and this completes the proof. □
Recall \(\mathcal{P}_{+}(H) \subset\mathcal{A}(H)\) the convex cone of all positive definite operators defined on H, that is, \(P \in\mathcal{P}_{+}(H)\) if and only if \(\langle Px,x\rangle\geq0\), and for all \(x\in H\), \(\langle Px,x\rangle=0\) implies \(x=0\).
Proposition 8
Let \(f:\mathcal{P}_{+}(H)\rightarrow\mathcal{A}(H)\) defined by
where \(Q\in\mathcal{A}(H)\). The function \(\nabla_{f}:\mathcal{P}_{+}(H) \times \mathcal {P}_{+}(H) \rightarrow\mathcal{A}(H)\) with
is an order generalised gradient for f.
Proof
For \(B\in\mathcal{P}_{+}(H)\), \(B^{-1} \in\mathcal{P}_{+}(H)\) then \(B^{-1}XB^{-1} \in\mathcal{P}_{+}(H)\) for any \(X\in\mathcal{P}_{+}(H)\) and thus \(QB^{-1}XB^{-1}Q \in\mathcal{P}_{+}(H)\) showing that \(\nabla_{f}(B,X) \in\mathcal{A}(H)\). We need to prove that
that is,
or equivalently
or
But
which is true since for \(A\in\mathcal{P}_{+}(H)\) we have that
and \(Q\in\mathcal{A}(H)\). □
4 Inequalities involving order generalised gradients
We start this section by the following definition.
Definition 9
An order generalised gradient \(\nabla_{f}:\mathcal{C} \times\mathcal{A}(H) \rightarrow\mathcal{A}(H)\) is
-
(i)
operator convex if
$$\nabla_{f} (B, \alpha X + \beta Y) \leq\alpha\nabla_{f}(B,X)+ \beta \nabla_{f}(B,Y) $$for any \(B\in\mathcal{C}\), \(X,Y \in\mathcal{A}(H)\) and \(\alpha,\beta\geq0\) with \(\alpha+\beta=1\);
-
(ii)
operator sub-additive if
$$\nabla_{f} (B, X + Y) \leq\nabla_{f}(B,X)+ \nabla_{f}(B,Y) $$for any \(B\in\mathcal{C}\) and \(X,Y \in\mathcal{A}(H)\);
-
(iii)
positive homogeneous if
$$\nabla_{f} (B, \alpha X ) = \alpha\nabla_{f}(B,X) $$for any \(B\in\mathcal{C}\), \(X\in\mathcal{A}(H)\) and \(\alpha\geq0\);
-
(iv)
operator linear if
$$\nabla_{f} (B, \alpha X + \beta Y) = \alpha\nabla_{f}(B,X)+ \beta\nabla_{f}(B,Y) $$for any \(B\in\mathcal{C}\), \(X,Y \in\mathcal{A}(H)\) and \(\alpha,\beta\in \mathbb{R}\).
It can be seen that if \(\nabla_{f}(\cdot,\cdot)\) is operator linear, then it is positive homogeneous and sub-additive. If \(\nabla_{f}(\cdot,\cdot)\) is positive homogeneous and sub-additive, then it is operator convex.
Theorem 10
Let \(f:\mathcal{C}\rightarrow\mathcal{A}(H)\) be operator convex and \(\nabla _{f}:\mathcal {C} \times\mathcal{A}(H) \rightarrow\mathcal{A}(H)\) be an order generalised gradient for f. Then, for any \(A,B, \in\mathcal{C}\) and \(t\in[0,1]\), we have the inequalities
Proof
If we write the definition of \(\nabla_{f}\) for B instead of A, we get
which is equivalent to
Therefore, for any \(A,B\in\mathcal{C}\), we have the gradient inequalities
Since \(\mathcal{C}\) is a convex set, hence by (17) we have
and
for any \(t\in(0,1)\).
If we multiply (18) by t and (19) by \((1-t)\) and add the obtained inequalities, then we get
Since \(\nabla_{f}(\cdot,\cdot)\) is operator convex, we also know that
which completes the proof. □
Corollary 11
Under the assumptions of Theorem 10,
-
(1)
If \(\nabla_{f}(\cdot,\cdot)\) is positive homogeneous, then we have
$$\begin{aligned} & {-}t(1-t)\bigl[\nabla_{f}(B,A-B)+\nabla_{f}(A,B-A) \bigr] \\ &\quad \geq tf(A)+(1-t)f(B) -f\bigl(tA+(1-t)B\bigr)\geq0. \end{aligned}$$(20) -
(2)
If \(\nabla_{f}(\cdot,\cdot)\) is operator linear, then
$$\begin{aligned} & t(1-t)\bigl[\nabla_{f}(B,B-A)-\nabla_{f}(A,B-A) \bigr] \\ &\quad \geq tf(A)+(1-t)f(B) -f\bigl(tA+(1-t)B\bigr)\geq0. \end{aligned}$$(21)
4.1 Hermite-Hadamard type operator inequalities
In this subsection, we will state inequalities of Hermite-Hadamard type for order generalised gradients.
Corollary 12
Under the assumptions of Theorem 10, if \(\nabla_{f}\) is positive homogeneous, then we have the following inequality:
We obtain (22) by integrating (20) over \(t \in[0,1]\).
Example 13
-
1.
We consider the function \(f(A)=QA^{2}Q\) with \(Q\in\mathcal{A}(H)\). We note that the order generalised gradient
$$\nabla_{f}(B,X)=Q(BX+XB)Q $$is operator linear. Then
$$\begin{aligned} \nabla_{f}(B,X) - \nabla_{f}(A,X) =& Q(BX+XB)Q - Q(AX+XA)Q \\ =& Q\bigl[(B-A)X+X(B-A)\bigr]Q. \end{aligned}$$For \(X=B-A\), we then get
$$\nabla_{f}(B,B-A)-\nabla_{f}(A,B-A) =2Q(B-A)^{2}Q. $$Applying inequality (21) we have
$$\begin{aligned} &2t(1-t)Q(B-A)^{2}Q \\ &\quad \geq Q\bigl[tA^{2}+(1-t)B^{2}-\bigl(tA+(1-t)B \bigr)^{2}\bigr]Q \geq0 \end{aligned}$$(23)for any \(A,B \in\mathcal{A}(H)\) and \(Q\in\mathcal{A}(H)\).
-
2.
We consider the function \(f(A)=APA\) with \(P\in\mathcal{P}(H)\). We note that the order generalised gradient
$$\nabla_{f}(B,X)= XPB +BPX $$is operator linear. Then
$$\begin{aligned} \nabla_{f}(B,X) - \nabla_{f}(A,X) =& XPB +BPX-XPA-APX \\ =&XP(B-A)+(B-A)PX. \end{aligned}$$If \(X=B-A\), we then get
$$\nabla_{f}(B,B-A) - \nabla_{f}(A,B-A) = 2(B-A)P(B-A). $$Applying inequality (21) we have
$$\begin{aligned} &2t(1-t) (B-A)P(B-A) \\ &\quad \geq tAPA+(1-t)BPB -\bigl(tA+(1-t)B\bigr)P\bigl(tA+(1-t)B\bigr)\geq0 \end{aligned}$$for any \(A,B \in\mathcal{A}(H)\) and \(P\in\mathcal{P}(H)\).
-
3.
For \(f(A)=QA^{-1}Q\) with \(Q\in\mathcal{A}(H)\) and \(A\in\mathcal{P}_{+}(H)\), we note that the order generalised gradient
$$\nabla_{f}(B,X)= -QB^{-1}XB^{-1}Q $$is operator linear. Then
$$\nabla_{f}(B,X)-\nabla_{f}(A,X) =-QB^{-1}XB^{-1}Q+QA^{-1}XA^{-1}Q. $$For \(X=B-A\), we get
$$\begin{aligned} &\nabla_{f}(B,B-A) - \nabla_{f}(A,B-A) \\ &\quad = -QB^{-1}(B-A)B^{-1}Q + QA^{-1}(B-A)A^{-1}Q \\ &\quad = -Q\bigl(B^{-1}-B^{-1}AB^{-1}\bigr)Q + Q \bigl(A^{-1}BA^{-1}-A^{-1}\bigr)Q \\ &\quad = QA^{-1}BA^{-1}Q + Q B^{-1}AB^{-1}Q -QB^{-1}Q-QA^{-1}Q. \end{aligned}$$By (21) we have the inequality
$$\begin{aligned} & t(1-t)\bigl[QA^{-1}BA^{-1}Q+ QB^{-1}AB^{-1}Q-QB^{-1}Q-QA^{-1}Q \bigr] \\ &\quad \geq tQA^{-1}Q+(1-t)QB^{-1}Q-Q\bigl(tA+(1-t)B \bigr)^{-1}Q\geq0 \end{aligned}$$for any \(A,B\in\mathcal{P}_{+}(H)\) and \(Q\in\mathcal{A}(H)\).
4.2 Jensen type operator inequalities
In this subsection, we will state inequalities of Jensen type for order generalised gradients.
Theorem 14
Let \(f:\mathcal{C}\subset\mathcal{A}(H) \rightarrow\mathcal{A}(H)\) be a function that possesses \(\nabla_{f}: \mathcal{C} \times\mathcal{A}(H) \rightarrow \mathcal{A}(H)\) as an order generalised gradient. Then, for any \(A_{i} \in \mathcal {C}\), \(i\in\{1,\dots, n\}\) and \(p_{i}\geq0\) with \(P_{n}:=\sum_{i=1}^{n} p_{i} >0\), we have the inequalities
Proof
From the definition of an order generalised gradient we have
Now, if we choose \(A=A_{j}\), \(j\in\{1,\dots,n\}\) and \(B=(1/P_{n}) \sum_{i=1}^{n} p_{i} A_{i}\) in (25), then we get
for any \(j\in\{1,\dots, n\}\). We obtain the desired inequalities (24) by multiplying the inequalities in (26) by \(p_{j}\geq0\) and taking the sum over j from 1 to n; and divide the resulted inequalities by \(P_{n}\). □
Corollary 15
Under the assumptions of Theorem 14, we have the following results:
-
(1)
If \(\nabla_{f}:\mathcal{C} \times\mathcal{A}(H)\) is convex, then
$$ \frac{1}{P_{n}} \sum_{j=1}^{n} p_{j}f(A_{j}) - f \Biggl(\frac{1}{P_{n}} \sum _{i=1}^{n} p_{i}A_{i} \Biggr) \geq\nabla_{f} \Biggl(\frac{1}{P_{n}} \sum _{i=1}^{n} p_{i}A_{i},0 \Biggr) . $$(27) -
(2)
If \(\nabla_{f}\) is linear, then \(\nabla_{f}(B,0)=0\) for any B, and we get the Jensen’s inequality
$$ \frac{1}{P_{n}} \sum_{j=1}^{n} p_{j}f(A_{j}) - f \Biggl(\frac{1}{P_{n}} \sum _{i=1}^{n} p_{i}A_{i} \Biggr) \geq0. $$(28) -
(3)
If \(\nabla_{f}\) is linear, we have
$$\begin{aligned} &\frac{1}{P_{n}} \sum_{j=1}^{n}p_{j} \nabla _{f} \Biggl(A_{j},A_{j} - \frac{1}{P_{n}} \sum_{i=1}^{n} p_{i}A_{i} \Biggr) \\ &\quad \geq \frac{1}{P_{n}} \sum_{j=1}^{n} p_{j}f(A_{j}) - f \Biggl(\frac {1}{P_{n}} \sum _{i=1}^{n} p_{i}A_{i} \Biggr) \geq0. \end{aligned}$$(29)
Theorem 16
Under the assumptions of Theorem 14, we have the following results:
Proof
From (25) we also have
If we multiply (30) by \(p_{j}\geq0\) and take the sum over j from 1 to n and divide the resulted inequalities by \(P_{n}\), then
which completes the proof. □
Remark 17
If \(\nabla_{f}\) is linear in Theorem 16, then we get simpler inequalities such as
and
Therefore, if \(A\in\mathcal{A}(H)\) is such that
then we have the Slater type inequality (cf. Slater [24] and Pečarić [25])
5 Connection with Gâteaux derivatives
In this section, we consider the connection between order generalised gradients and Gâteaux derivatives. We refer the reader to Dragomir [26] for some inequalities of Jensen type, involving Gâteaux derivatives of convex functions in linear spaces.
Let \(\mathcal{C} \subset\mathcal{A}(H)\) be a convex set. Then \(f:\mathcal{C} \rightarrow\mathcal{A}(H)\) is said to be operator convex if for all \(t\in[0,1]\) and \(A,B \in\mathcal{C}\), we have
We start with the following lemmas.
Lemma 18
Let \(F:\mathbb{R}\rightarrow\mathcal{B}(H)\) be a function such that \(\lim_{t\rightarrow0^{\pm}} F(t)\) exists. Then \(\lim_{t\rightarrow0^{\pm}} F(t)\) is a bounded linear operator and
for all nonzero \(x,y \in H\).
Proof
We provide the proof for the right-sided limit, as the proof for the left-sided limit follows similarly. Let \(\varepsilon>0\) and for \(x,y \in H\), where \(x,y\neq0\), set \(\varepsilon_{0} = \varepsilon/(\Vert x\Vert _{H} \Vert y\Vert _{H})\). Since \(\lim_{t\rightarrow0^{+}} F(t)=L\), there exists \(\delta_{0}\) such that
when \(0< t<\delta_{0}\). Note that \(L \in\mathcal{B}(H)\) since \(\mathcal{B}(H)\) is a Banach space, hence \(F(t)-L\) is also a bounded linear operator. Now, we have
which completes the proof. □
Lemma 19
Let \(f: \mathcal{A}(H) \rightarrow\mathcal{A}(H)\) be operator convex and \(A \in\mathcal{A}(H)\). Then, for all \(B\in\mathcal{A}(H)\), both limits
and
exist and are bounded self-adjoint operators.
Proof
Fix an arbitrary \(B\in\mathcal{A}(H)\), and let
We want to show that G is nondecreasing. Let \(0 < t_{1}<t_{2}\), then
Thus,
Also,
Note also that
which implies that
and thus
which implies that
By the above expositions, we conclude that G is nondecreasing on \(\mathbb{R}\setminus\{0\}\). This proves that both \((\nabla_{-} f(A))(B)\) and \((\nabla_{+}f(A))(B)\) exist and are bounded linear operators by Lemma 18. Note that for all \(t \in\mathbb{R}\), \(t\neq0\) and \(A,B \in\mathcal{A}(H)\),
is a self-adjoint operator. If \(x,y \in H\), then Lemma 18 gives us
which completes the proof. □
Theorem 20
Let \(\mathcal{C} \subset\mathcal{A}(H)\) be a convex set and \(f:\mathcal{C} \rightarrow\mathcal{A}(H)\) be operator convex. Then \(\nabla_{\pm}f\) defined by
is an order generalised gradient.
Proof
Let \(t\in(0,1)\) and \(A,B \in\mathcal{C}\). Since f is operator convex, we have
This is equivalent to
Note that for all \(x\in H\),
by Lemma 18. Since \(K \in\mathcal{P}_{+}(H)\), \(\langle K x,x \rangle\geq 0\), hence \(\langle [ \lim_{t\rightarrow0^{\pm}} K ] x,x \rangle\geq0\), which implies that
Therefore,
Lemma 19 gives us
which implies that
Thus both \(\nabla_{+} f\) and \(\nabla_{-} f\) are order generalised gradients. □
Proposition 21
Let \(f: \mathcal{A}(H) \rightarrow\mathcal{A}(H)\) be operator convex and \(A \in \mathcal{A}(H)\). The right Gâteaux derivative of f is sub-additive, i.e.
for any \(B,C \in\mathcal{A}(H)\). The left Gâteaux derivative of f is super-additive, i.e.
for any \(B,C \in\mathcal{A}(H)\).
Proof
Since f is operator convex, we have the following for any \(B,C \in \mathcal {A}\) and \(t >0\):
By a similar argument to the proof of Theorem 20, we conclude that
as desired. The proof for the left Gâteaux derivative of f follows similarly. □
Remark 22
We remark that the Gâteaux (lateral) derivative(s) is always positive homogeneous with respect to the second variable, i.e. for any function \(f: \mathcal{A}(H) \rightarrow\mathcal{A}(H)\) and fixed \(A\in\mathcal{A}(H)\),
for all \(\alpha\geq0\) and \(B\in\mathcal{A}(H)\). The Gâteaux derivative, on the other hand, is always homogeneous with respect to the second variable, i.e. for any function \(f: \mathcal{A}(H) \rightarrow \mathcal {A}(H)\) and fixed \(A\in\mathcal{A}(H)\),
for all \(\alpha\in\mathbb{C}\) and \(B\in\mathcal{A}(H)\).
The following result restates Theorem 3 in the setting of operator-valued functions.
Corollary 23
Let \(\mathcal{C} \subset\mathcal{A}(H)\) be a convex set and \(f:\mathcal{C} \rightarrow\mathcal{A}(H)\) be a Gâteaux differentiable function. Then f is operator convex if and only if ∇f defined by
is an order generalised gradient.
Proof
For any \(A,B \in\mathcal{C}\), if f is operator convex, then by Theorem 20
are order generalised gradients. Since f is assumed to be Gâteaux differentiable, both limits are equal, hence
is an order generalised gradient for any \(A,B \in\mathcal{C}\). Conversely, we have the following inequality:
for any \(A,B \in\mathcal{C}\). Let \(C,D \in\mathcal{C}\), \(t\in(0,1)\), and choose \(A=C\) and \(B=tC+(1-t)D\). Then we have
Let \(A=D\) and \(B=tC+(1-t)D\). Then we have
Multiply (33) by t and (34) by \((1-t)\), and add the resulting inequalities to obtain
which completes the proof. □
The following result follows by Corollary 12 and employing the fact that the Gâteaux lateral derivatives are positive homogenous.
Corollary 24
(Hermite-Hadamard type inequality)
Let \(f:\mathcal{C}\subset\mathcal{A}(H) \rightarrow\mathcal{A}(H)\) be operator convex. The following inequality holds:
The above inequality also holds for ∇f when f is Gâteaux differentiable.
Example 25
-
(1)
We note that the function \(f(x)=-\log(x)\) is operator convex. The log function is (operator) Gâteaux differentiable with the following explicit formula for the derivative (cf. Pedersen [27, p.155]):
$$\bigl(\nabla\log(A)\bigr) (B)=\int_{0}^{\infty} (sI+A)^{-1}B(sI+A)^{-1}\,ds $$for \(A,B\in\mathcal{A}(H)\) and I the identity operator. Thus, we have the following inequality:
$$\begin{aligned} &\frac{1}{6} \biggl[\int_{0}^{\infty} (sI+B)^{-1}(A-B) (sI+B)^{-1}\,ds \\ &\qquad {}+\int_{0}^{\infty} (sI+A)^{-1}(B-A) (sI+A)^{-1}\,ds \biggr] \\ &\quad \geq -\frac{\log(A)+\log(B)}{2} + \int_{0}^{1} \log \bigl(tA+(1-t)B\bigr)\,dt \geq0. \end{aligned}$$ -
(2)
We consider the operator convex function \(f(x)=x\log(x)\), and using the following representation (cf. Pedersen [27, p.155])
$$\log(t)=\int_{0}^{\infty} \frac{t-1}{(s+t)(s+1)}\,ds, $$and noting the fact that \(\frac{d}{dt} t\log(t)= \log(t)\), we have
$$\bigl(\nabla f(A)\bigr) (B)=\int_{0}^{\infty} \frac{1}{s+1} (sI+A)^{-1}(A-I)B\, ds. $$Then we have the following inequalities:
$$\begin{aligned} &{-}\frac{1}{6} \biggl[\int_{0}^{\infty} \frac{1}{s+1} (sI+B)^{-1}(B-I) (A-B)\,ds \\ &\qquad {}+\int_{0}^{\infty} \frac{1}{s+1} (sI+A)^{-1}(A-I) (B-A)\,ds \biggr] \\ &\quad \geq\frac{A\log(A)+B\log(B)}{2} - \int_{0}^{1} \bigl[tA+(1-t)B\bigr]\log\bigl(tA+(1-t)B\bigr)\,dt\geq0. \end{aligned}$$
The following results follow by Theorems 14 and 16.
Corollary 26
(Jensen type inequality)
Let \(f:\mathcal{C}\subset\mathcal{A}(H) \rightarrow\mathcal{A}(H)\) be operator convex. Then, for any \(A_{i} \in\mathcal{C}\), \(i\in\{1,\dots, n\}\) and \(p_{i}\geq0\) with \(P_{n}:=\sum_{i=1}^{n} p_{i} >0\), we have the inequalities
We also have
The above inequalities also hold for ∇f when f is Gâteaux differentiable.
Example 27
-
(1)
We have the following inequalities for the operator convex function \(f(x)=-\log(x)\):
$$\begin{aligned} & \frac{1}{P_{n}} \sum _{j=1}^{n} p_{j} \int_{0}^{\infty} (sI+A_{j})^{-1} \Biggl( \frac{1}{P_{n}} \sum _{i=1}^{n} p_{i}A_{i} - A_{j} \Biggr) (sI+A_{j})^{-1}\,ds \\ &\quad \geq{-\frac{1}{P_{n}} \sum_{j=1}^{n}p_{j} \log(A_{j}) + \log \Biggl(\frac {1}{P_{n}} \sum _{i=1}^{n} p_{i}A_{i} \Biggr)} \\ &\quad \geq-\frac{1}{P_{n}} \sum_{j=1}^{n} p_{j} \int_{0}^{\infty} \Biggl(sI+ \frac {1}{P_{n}} \sum_{i=1}^{n}p_{i}A_{i} \Biggr)^{-1}\\ &\qquad {} \times\Biggl( A_{j}-\frac{1}{P_{n}} \sum _{i=1}^{n}p_{i}A_{i} \Biggr) \Biggl(sI+\frac{1}{P_{n}} \sum_{i=1}^{n}p_{i}A_{i} \Biggr)^{-1}\,ds \end{aligned} $$and
$$\begin{aligned} &{-}\frac{1}{P_{n}} \sum_{j=1}^{n} p_{j}\log(A_{j})+ \frac{1}{P_{n}} \sum _{j=1}^{n} p_{j} \int_{0}^{\infty} (sI+A)^{-1}(A_{j}-A) (sI+A)^{-1}\,ds \\ &\quad \geq -\log(A) \\ &\quad \geq- \frac{1}{P_{n}} \sum_{j=1}^{n} p_{j}\log(A_{j})- \frac{1}{P_{n}} \sum _{j=1}^{n} p_{j} \int_{0}^{\infty}(sI+A_{j})^{-1}(A-A_{j}) (sI+A_{j})^{-1}\,ds. \end{aligned}$$ -
(2)
We have the following inequalities for the operator convex function \(f(x)=x\log(x)\):
$$\begin{aligned} & {-}\frac{1}{P_{n}} \sum _{j=1}^{n} p_{j} \int_{0}^{\infty} \frac{1}{s+1} (sI+A_{j})^{-1}(A_{j}-I) \Biggl( \frac{1}{P_{n}} \sum_{i=1}^{n} p_{i}A_{i} - A_{j} \Biggr)\,ds \\ &\quad \geq{\frac{1}{P_{n}} \sum_{j=1}^{n}p_{j}A_{j} \log(A_{j}) - \Biggl(\frac {1}{P_{n}} \sum _{i=1}^{n} p_{i}A_{i} \Biggr) \log \Biggl(\frac{1}{P_{n}} \sum_{i=1}^{n} p_{i}A_{i} \Biggr)} \\ &\quad \geq\frac{1}{P_{n}} \sum_{j=1}^{n} p_{j} \int_{0}^{\infty} \frac{1}{s+1} \Biggl(sI+ \frac{1}{P_{n}} \sum_{i=1}^{n}p_{i}A_{i} \Biggr)^{-1}\\ &\qquad {} \times \Biggl(\frac {1}{P_{n}} \sum _{i=1}^{n}p_{i}A_{i} -I \Biggr) \Biggl( A_{j}-\frac{1}{P_{n}} \sum_{i=1}^{n}p_{i}A_{i} \Biggr)\,ds \end{aligned} $$and
$$\begin{aligned} &\frac{1}{P_{n}} \sum_{j=1}^{n} p_{j}A_{j} \log(A_{j})- \frac{1}{P_{n}} \sum _{j=1}^{n} p_{j} \int _{0}^{\infty} \frac{1}{s+1} (sI+A)^{-1}(A-I) (A_{j}-A)\,ds \\ &\quad \geq A\log(A) \\ &\quad \geq \frac{1}{P_{n}} \sum_{j=1}^{n} p_{j}A_{j}\log(A_{j})+ \frac{1}{P_{n}} \sum _{j=1}^{n} p_{j} \int _{0}^{\infty} \frac{1}{s+1} (sI+A_{j})^{-1}(A_{j}-I) (A-A_{j})\,ds. \end{aligned}$$
6 Conclusions
For a function \(f:\mathcal{C} \rightarrow\mathcal{A}(H)\) defined on a convex set \(\mathcal{C} \subset\mathcal{A}(H)\), the function \(\nabla_{f}:\mathcal{C}\times \mathcal {A}(H) \rightarrow\mathcal{A}(H)\) is an order generalised gradient if
in the operator order of \(\mathcal{A}(H)\). We have the following operator inequalities.
-
(1)
Operator inequalities of Hermite-Hadamard type:
$$\begin{aligned} &{-}\frac{1}{6}\bigl[\nabla_{f}(B,A-B)+\nabla_{f}(A,B-A) \bigr] \\ &\quad \geq\frac{f(A)+f(B)}{2} -\int_{0}^{1} f \bigl(tA+(1-t)B\bigr)\,dt \geq0 \quad \text{for any $A,B\in\mathcal{C}$}. \end{aligned}$$ -
(2)
Operator inequalities of Jensen type:
$$\begin{aligned} & {-}\frac{1}{P_{n}} \sum _{j=1}^{n} p_{j} \nabla_{f} \Biggl(A_{j}, \frac{1}{P_{n}} \sum_{i=1}^{n} p_{i}A_{i} - A_{j} \Biggr) \\ &\quad \geq{\frac{1}{P_{n}} \sum_{j=1}^{n}p_{j}f(A_{j}) - f \Biggl(\frac{1}{P_{n}} \sum_{i=1}^{n} p_{i}A_{i} \Biggr)} \\ &\quad \geq{\frac{1}{P_{n}} \sum_{j=1}^{n} p_{j} \nabla_{f} \Biggl(\frac{1}{P_{n}} \sum _{i=1}^{n}p_{i}A_{i} , A_{j}-\frac{1}{P_{n}} \sum_{i=1}^{n}p_{i}A_{i} \Biggr);} \end{aligned} $$and
$$\begin{aligned} &\frac{1}{P_{n}} \sum_{j=1}^{n} p_{j}f(A_{j})- \frac{1}{P_{n}} \sum _{j=1}^{n} p_{j} \nabla_{f}(A,A_{j}-A) \\ &\quad \geq f(A) \\ &\quad \geq \frac{1}{P_{n}} \sum_{j=1}^{n} p_{j}f(A_{j})+ \frac{1}{P_{n}} \sum _{j=1}^{n} p_{j} \nabla_{f}(A_{j},A-A_{j}) \end{aligned}$$for any \(A\in\mathcal{C}\), \(A_{i} \in\mathcal{C}\), \(i\in\{1,\dots, n\}\) and \(p_{i}\geq0\) with \(P_{n}:=\sum_{i=1}^{n} p_{i} >0\).
-
(3)
Operator inequalities of Slater type: if \(\nabla_{f}\) is linear and \(A\in\mathcal{A}(H)\) is such that
$$\frac{1}{P_{n}} \sum_{j=1}^{n} p_{j} \nabla_{f}(A_{j},A) \geq\frac{1}{P_{n}} \sum_{j=1}^{n} p_{j} \nabla_{f}(A_{j},A_{j}), $$then
$$f(A) \geq\frac{1}{P_{n}} \sum_{j=1}^{n} p_{j}f(A_{j}) $$for any \(A\in\mathcal{C}\), \(A_{i} \in\mathcal{C}\), \(i\in\{1,\dots, n\}\) and \(p_{i}\geq0\) with \(P_{n}:=\sum_{i=1}^{n} p_{i} >0\).
Order generalised gradients extend the notion of subgradients, without the assumption of convexity, for operator-valued functions. This notion is also connected to the Gâteaux (lateral) derivatives. If f is operator convex, then \(\nabla_{\pm}f\) defined by
is an order generalised gradient. Furthermore, if \(f:\mathcal{C} \rightarrow\mathcal{A}(H)\) is a Gâteaux differentiable function, we have the following characterisation: f is operator convex if and only if ∇f defined by
is an order generalised gradient. This characterisation of convexity is a generalised version of Theorem 3 of Section 2 (cf. Eisenberg [3, Theorem 1]).
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Acknowledgements
The research of E Kikianty is supported by the Claude Leon Foundation. The authors would like to thank the anonymous referees for valuable suggestions that have been incorporated in the final version of the manuscript.
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Dragomir, S.S., Kikianty, E. Order generalised gradient and operator inequalities. J Inequal Appl 2015, 49 (2015). https://doi.org/10.1186/s13660-015-0574-y
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DOI: https://doi.org/10.1186/s13660-015-0574-y