Further generalizations, refinements, and reverses of the Young and Heinz inequalities

TL;DR

This paper introduces a new inequality for convex functions and applies it to generalize, refine, and reverse classical Young and Heinz inequalities, with further applications to matrix norm inequalities.

Contribution

It presents a novel convex function inequality and extends classical inequalities with new bounds and reverses, including matrix norm applications.

Findings

New convex function inequality established

Generalizations and reverses of Young and Heinz inequalities derived

Applications to matrix norm inequalities demonstrated

Abstract

In this paper, we give a new inequality for convex functions of real variables, and we apply this inequality to obtain considerable generalizations, refinements, and reverses of the Young and Heinz inequalities for positive scalars. Applications to unitarily invariant norm inequalities involving positive semidefinite matrices are also given.

Equations

$$a^{\nu }{b}^{1-\nu }\le \nu a+(1-\nu )b,$$

$$ab\le \frac{a^p}{p}+\frac{b^q}{q},$$

$$a^{\nu }{b}^{1-\nu }+r_0{\left(\sqrt{a}-\sqrt{b}\right)}^2\le \nu a+(1-\nu )b$$

$$\nu a+(1-\nu )b\le a^{\nu }{b}^{1-\nu }+R_0{\left(\sqrt{a}-\sqrt{b}\right)}^2,$$

$$r_0\left((a+b)-2\sqrt{ab}\right)\le \nu a+(1-\nu )b-a^{\nu }{b}^{1-\nu }\le R_0\left((a+b)-2\sqrt{ab}\right).$$

$${\left(a^{\nu }{b}^{1-\nu }\right)}^2+r_0^2{\left(a-b\right)}^2\le (\nu a+(1-\nu )b{)}^2,$$

$${\left(\nu a+(1-\nu )b\right)}^2\le {\left(a^{\nu }{b}^{1-\nu }\right)}^2+R_0^2{\left(a-b\right)}^2,$$

$$r_0^2\left((a+b{)}^2-4ab\right)\le {\left(\nu a+(1-\nu )b\right)}^2-{\left(a^{\nu }{b}^{1-\nu }\right)}^2\le R_0^2\left((a+b{)}^2-4ab\right).$$

$${\left(a^{\nu }{b}^{1-\nu }\right)}^m+r_0^m{\left(a^{\frac{m}{2}}-b^{\frac{m}{2}}\right)}^2\le {\left(\nu a+(1-\nu )b\right)}^m,$$

$$H_{\nu }(a,b)=\frac{a^{\nu }{b}^{1-\nu }+a^{1-\nu }{b}^{\nu }}{2}$$

$$\sqrt{ab}\le H_{\nu }(a,b)\le \frac{a+b}{2}.$$

$$2r_0\left(a+b-2\sqrt{ab}\right)\le a+b-\left(a^{\nu }{b}^{1-\nu }+a^{1-\nu }{b}^{\nu }\right)\le 2R_0\left(a+b-2\sqrt{ab}\right).$$

$$2r_0\left(\frac{a+b}{2}-\sqrt{ab}\right)\le \frac{a+b}{2}-H_{\nu }(a,b)\le 2R_0\left(\frac{a+b}{2}-\sqrt{ab}\right),$$

$$\frac{f(z)-f(w)}{z-w}\le \frac{f(y)-f(w)}{y-w}\le \frac{f(y)-f(z)}{y-z}\le \frac{f(x)-f(y)}{x-y}$$

$$z-w\le x-y,$$

$$(0\le )\varphi (z)-\varphi (w)\le \varphi (x)-\varphi (y).$$

$$\varphi (z)-\varphi (w)\le \varphi (x)-\varphi (y).$$

$$\varphi \left(r_0\left(a+b\right)\right)-\varphi \left(2r_0\sqrt{ab}\right)$$

$$z-w\le x-y\le z^{\prime }-w^{\prime }.$$

$$r_0^p\left((a+b{)}^p-{\left(2\sqrt{ab}\right)}^p\right)$$

$$r_0^m{\left(a^{\frac{m}{2}}-b^{\frac{m}{2}}\right)}^2$$

$$\le$$

$$(a+b{)}^m$$

$${\left(a+b\right)}^m-{\left(2\sqrt{ab}\right)}^m$$

$$r_0^m{\left(a^{\frac{m}{2}}-b^{\frac{m}{2}}\right)}^2\le r_0^m\left({\left(a+b\right)}^m-{\left(2\sqrt{ab}\right)}^m\right)\le {\left(\nu a+(1-\nu )b\right)}^m-{\left(a^{\nu }{b}^{1-\nu }\right)}^m$$

$$\varphi \left(2r_0\left(a+b\right)\right)-\varphi \left(4r_0\sqrt{ab}\right)$$

$$\varphi \left(2r_0\left(\frac{a+b}{2}\right)\right)-\varphi \left(2r_0\sqrt{ab}\right)$$

$$(2r_0{)}^p\left({\left(a+b\right)}^p-{\left(2\sqrt{ab}\right)}^p\right)$$

$${\Vert A\Vert }_p={\left(\sum _{i=1}^n{s}_i^p(A)\right)}^{\frac{1}{p}},$$

$${\Vert A\Vert }_2={\left(\sum _{i,j=1}^n{\left|a_{ij}\right|}^2\right)}^{\frac{1}{2}}.$$

Full text

 semidefinite matrix, unitarily invariant norm.

1. Introduction

The numerical Young’s inequality for positive real numbers says that

[TABLE]

where $a,b>0$ and $\ 0\leq\nu\leq 1$. Equivalently,

[TABLE]

where $p,q>0$ and $\frac{1}{p}+\frac{1}{q}=1$.

Kittaneh and Manasrah in [6] and [7], respectively, refined the inequality (1.1) and gave a reverse of it in the following forms:

[TABLE]

and

[TABLE]

where $r_{0}=\min\left\{\nu,1-\nu\right\}$* and $R_{0}=\max\left\{\nu,1-\nu\right\}$*.

For our purpose in this paper, the inequalities (1.2) and (1.3) are combined and expressed so that

[TABLE]

In [4], Hirzallah and Kittaneh proved that if $a,b>0$ and $\ 0\leq\nu\leq 1$, then

[TABLE]

where $r_{0}=\min\left\{\nu,1-\nu\right\}.$

In [7], Kittaneh and Manasrah gave a reverse of the inequality (1.5) in the following form:

[TABLE]

where $R_{0}=\max\left\{\nu,1-\nu\right\}$.

Also, for our purpose in this paper, the inequalities (1.5) and (1.6) are combined and expressed so that

[TABLE]

Recently, the authors [1] proved the following theorem.

Theorem 1**.**

If $a,b>0$ and $\ 0\leq\nu\leq 1$, then for $m=1,2,3,...$, we have

[TABLE]

where $r_{0}=\min\left\{\nu,1-\nu\right\}.$

In fact, this is a generalization of the inequalities (1.2) and (1.5), which correspond to the cases $m=1$ and $m=2$, respectively.

The Heinz means are defined as

[TABLE]

for $a,b>0$ and $0\leq\nu\leq 1$. These interesting means interpolate between the geometric and arithmetic means. In fact, the Heinz inequalities assert that

[TABLE]

Interchaning a and b in the inequalities (1.4), and adding the resulting inequalities to the inequalities (1.4), we have

[TABLE]

Equivalently,

[TABLE]

where $r_{0}=\min\left\{\nu,1-\nu\right\}$* and $R_{0}=\max\left\{\nu,1-\nu\right\}.$*

In Section 2, we present a new inequality for convex functions of real variables. We apply this inequality to obtain considerable generalizations, refinements, and reverses of the Young and Heinz inequalities (1.2)-(1.10). Applications to unitarily invariant norm inequalities involving positive semidefinite matrices are given in Section 3.

2. Main results

Let $f$ be a convex function defined on an interval $I$. If $x$, $y,z$ and $w$ are points in $I$ such that $w<z<y<x$, then it follows by a slope argument that

[TABLE]

(see, e.g., [8, p. 21]).

Our first result is a consequence of the inequalities (2.1) for convex functions of real variables.

Theorem 2**.**

Let $\phi$ be a stricly increasing convex function defined on an interval $I$. If $x,y,z,$ and $w$ are points in $I$ such that

[TABLE]

where $w\leq z\leq x$ and $y\leq x$, then

[TABLE]

Proof.

First of all, observe that if $x=y$, then $z=w$, and so the inequality (2.3) becomes an equality. If $y=w$ or $z=w$, then the inequality (2.3) holds. Also, if $x=z$, then according to the inequality (2.2), we have $w\geq y$, so $\phi(w)\geq\phi(y)$, and hence $\phi(z)-\phi(w)\leq\phi(x)-\phi(y)$.

Assume that $x\neq y$, $y\neq w$, $z\neq w$, and $x\neq z$. Then we have three cases for ordering the points $x,y,z,w$  as follows:

Case 1: $w<y\leq z<x$

Case 2: $w<z<y<x$

Case 3: $y<w<z<x$.

Now, if $y=z$ , then the case 1 becomes $w<y=z<x$, and so by the inequalities (2.1), we have $\frac{\phi(z)-\phi(w)}{z-w}\leq\frac{\phi(x)-\phi(y)}{x-y}$, which implies that $\phi(z)-\phi(w)\leq\phi(x)-\phi(y)$. Suppose $y\neq z$. Then apply the inequalities (2.1) to the cases 1 and 2 to get the inequality (2.3).

To discuss the third case, we apply the strictly increasing property of the function $\phi$ to the sequence of points $y<w<z<x$, so we have $\phi(y)<\phi(w)<\phi(z)<\phi(x)$ and this implies that $\phi(y)-\phi(w)<0<\phi(x)-\phi(z)$, and hence $\phi(z)-\phi(w)<\phi(x)-\phi(y)$.

Thus, from the discussion above we have

[TABLE]

This completes the proof.

As a direct consequence of Theorem 2, we have the following generalizations of the inequalities (1.4).

Corollary 1**.**

Let $\phi:[0,\infty)\rightarrow\mathbb{R}$ be a strictly increasing convex function. If $a,b>0$ and $\ 0\leq\nu\leq 1$, then we have

[TABLE]

where $r_{0}=\min\left\{\nu,1-\nu\right\}$ and $R_{0}=\max\left\{\nu,1-\nu\right\}$.

Proof.

Let $x=\nu a+(1-\nu)b$,   $y=a^{\nu}b^{1-\nu},$   $z=r_{0}\left(a+b\right)$, $w=2r_{0}\sqrt{ab}$ , $z^{\prime}=R_{0}\left(a+b\right)$, and $w^{\prime}=2R_{0}\sqrt{ab}$. Then based on the inequalities (1.4) and the arithmetic-geometric mean inequality, we have

[TABLE]

The first and the second inequalities in (2.4) follow directly by applying Theorem 2 to the inequalities $z-w\leq x-y$, with $w\leq z\leq x$, $y\leq x$ and $x-y\leq z^{\prime}-w^{\prime}$ with $y\leq x\leq z^{\prime}$, $w^{\prime}\leq z^{\prime}$, respectively. This completes the proof.

A particular case of Corollary 1, which is obtained by taking $\phi(x)=x^{p}$ $(p\in\mathbb{R}$, $p\geq 1)$ can be stated as follows.

Corollary 2**.**

If $a,b>0$ and $\ 0\leq\nu\leq 1$, then for $p\in\mathbb{R}$, $p\geq 1$, we have

[TABLE]

where $r_{0}=\min\left\{\nu,1-\nu\right\}$ and $R_{0}=\max\left\{\nu,1-\nu\right\}$.

One can observe that the inequalities (2.5) are reduced to the inequalities (1.4) and (1.7) when $p=1$ and $p=2$, respectively.

The next theorem demonstrates the relationship between the inequalities (2.5) and (1.8). In fact, it confirms that the first inequality in (2.5) is uniformly better than the inequality (1.8), and the second inequality in (2.5) provides a reverse of the same inequality. It should be noted here that the variable $p$  in the inequalities (2.5) is continuous $(p\in\mathbb{R}$, $p\geq 1)$, while the variable $m$ in the inequalitiy (1.8) is discrete $(m=1,2,3,...)$.

Theorem 3**.**

If $a,b>0$ and $\ 0\leq\nu\leq 1$, then we have

[TABLE]

for $m=1,2,3,...$, where $r_{0}=\min\left\{\nu,1-\nu\right\}$.

Proof.

It is clear that the second and third inequalities in (2.6) are special cases of the inequalities (2.5). So, it is enough to prove the first inequality in (2.6).

To do this, observe that the cases $m=1$ and $m=2$ degenerate to an equality. For $m=3,4,5,...$, we discuss two cases.

Case 1: If $m$ is even, we have

[TABLE]

Case 2: If $m$ is odd, we have

[TABLE]

Now,

for $m=3,4,5,...,$we have

[TABLE]

Thus,

[TABLE]

for $m=1,2,3,...$ This completes the proof.

Applying Theorem 2 again, we have the following generalizations of the inequalities (1.9) and (1.10), respectively:

[TABLE]

and

[TABLE]

In particular, if $\phi(x)=x^{p}$ $(p\in\mathbb{R},$ $p\geq 1),$we have

[TABLE]

3.

Some inequalities for unutarily invariant norms

In this section, we obtain matrix versions of the scalar inequalities presented in Sections 1 and 2.

Let $M_{n}(\mathbb{C})$ be the space of $n\times n$ complex matrices. A norm $\left|\left|\left|.\right|\right|\right|$ on $M_{n}(\mathbb{C})$ is called unitarily invariant if $\left|\left|\left|UAV\right|\right|\right|=\left|\left|\left|A\right|\right|\right|$ for all $A\in M_{n}(\mathbb{C})$ and for all unitary matrices $U,V$ $\in M_{n}(\mathbb{C})$. An example of unitarly invariant norms is the Schatten $p$-norm, denoted by $\left\|.\right\|_{p}$ and defined, for $1\leq p<\infty$, by

[TABLE]

where $s_{1}(A)\geq s_{2}(A)\geq...\geq s_{n}(A)$ are the singular values of $A\in M_{n}(\mathbb{C})$. The Schatten $1$-norm of $A$ is the trace norm, which can be expressed as $\left\|A\right\|_{1}=$ tr $\left|A\right|$. The Schatten $2-$norm of $A=\left[a_{ij}\right]$ is known as the Hilbert-Schmidt (or the Frobenius) norm, which can be expressed as

[TABLE]

Another important example of unitarily invariant norms on $M_{n}(\mathbb{C})$ is the spectral (or the usual operator) norm $\left\|.\right\|$, given by

[TABLE]

Based on the refined and reversed Young inequalities (1.5) and (1.6), Hirzallah and Kittaneh [4], and Kittaneh and Manasrah [7], respectively, proved that if $A,B,X\in M_{n}(\mathbb{C})$ such that $A$ and $B$ are positive semidefinite, then

[TABLE]

and

[TABLE]

where $0\leq\nu\leq 1$, $r_{0}=\min\left\{\nu,1-\nu\right\}$, and $R_{0}=\max\left\{\nu,1-\nu\right\}.$

It can be easily shown that

[TABLE]

Thus, the inequalities (3.1) and (3.2) can be combined and expressed so that

[TABLE]

Applying Theorem 2 to the inequalities (LABEL:ineq_21), the following general result holds.

Theorem 4**.**

Let $A,B,X\in M_{n}(\mathbb{C})$ such that $A$ and $B$ are positive semidefinite. If $\ \phi:[0,\infty)\rightarrow\mathbb{R}$ is a strictly increasing convex function, then we have

[TABLE]

where $r_{0}=\min\left\{\nu,1-\nu\right\}$ and $R_{0}=\max\left\{\nu,1-\nu\right\}$.

In particular, when $\phi(x)=x^{\frac{p}{2}}$ $(p\in\mathbb{R}$ and $p\geq 2)$, we have

[TABLE]

If $A,B,X\in M_{n}(\mathbb{C})$ such that $A$ and $B$ are positive semidefinite, then it is known that for any unitarily invariant norm, the function $f(\nu)=\left|\left|\left|A^{\nu}XB^{1-\nu}+A^{1-\nu}XB^{\nu}\right|\right|\right|$ is convex on $[0,1]$ and attains its minimum at $\nu=\frac{1}{2}$ (see, e.g., [2, p. 265]).

Bhatia and Davis [3] proved that if $A,B,X\in M_{n}(\mathbb{C})$ such that $A$ and $B$ are positive semidefinite, then

[TABLE]

where $0\leq\nu\leq 1$. These inequalities are known as Heinz norm inequalities.

Kittaneh [5] proved that if $A,B,X\in M_{n}(\mathbb{C})$ such that $A$ and $B$ are positive semidefinite, then

[TABLE]

where $0\leq\nu\leq 1$ and $r_{0}=\min\{\nu,1-\nu\}.$

In the next theorem, we obtain a reverse of the inequalitiy (3.4) as follows.

Theorem 5**.**

If $A,B,X\in M_{n}(\mathbb{C})$ such that $A$ and $B$ are positive semidefinite, then

[TABLE]

where $\ 0\leq\nu\leq 1$ and $R_{0}=\max\{\nu,1-\nu\}$.

Proof.

If $\nu=0$, $\nu=\frac{1}{2}$, or $\nu=1$, then the inequality (3.5) is obviously true. Suppose that $0<\nu<1$, $\nu\neq\frac{1}{2}$.

If $0<\nu<\frac{1}{2}<1$, then based on the convexity of the function $f(\nu)=\left|\left|\left|A^{\nu}XB^{1-\nu}+A^{1-\nu}XB^{\nu}\right|\right|\right|$, we have

[TABLE]

and so

[TABLE]

.

Adding $f(1)$ to both sides, gives

[TABLE]

i.e.,

[TABLE]

If $0<\frac{1}{2}<\nu<1$, then

[TABLE]

and so

[TABLE]

i.e.,

[TABLE]

Thus, from the above two norm inequalities, we get the inequality (3.5).

The inequalities (3.4) and (3.5) can be combined so that

[TABLE]

Applying Theorem 2 to the inequalities (3.6), we have the following general result.

Corollary 3**.**

Let $A,B,X\in M_{n}(\mathbb{C})$ such that $A$ and $B$ are positive semidefinite. If $\phi:[0,\infty)\rightarrow\mathbb{R}$ is a stricly increasing convex function,  then we have

[TABLE]

where $0\leq\nu\leq 1$, $r_{0}=\min\left\{\nu,1-\nu\right\}$, and  $R_{0}=\max\left\{\nu,1-\nu\right\}$.

In particular, if $\phi(x)=x^{q}$ $\left(q\in\mathbb{R},q\geq 1\right)$ and $\left|\left|\left|.\right|\right|\right|=\left\|.\right\|_{p}$ (the Schatten p-norm $p\in\mathbb{R},$ $p\geq 1$), we have

[TABLE]