1 Introduction

Let \(f: J\rightarrow {\mathbb {R}}\) be a real-valued function defined on the interval J. We say that f is convex on J if it satisfies the simple inequality

$$\begin{aligned} f((1-t)a+tb)\le (1-t)f(a)+tf(b), \end{aligned}$$

for every \(a,b\in J\) and \(0\le t\le 1\). Though simple, this inequality has been used to obtain numerous useful inequalities, such as the arithmetic–geometric mean inequality, the Cauchy–Schwarz inequality, the Bellman inequality, and many others.

Refining and reversing this inequality, Dragomir [10] showed that if \(a,b\in J\) and \(f: J\rightarrow {\mathbb {R}}\) is convex, then

$$\begin{aligned} \begin{aligned} 2\gamma \left( \frac{f\left( a \right) +f\left( b \right) }{2}-f\left( \frac{a+b}{2} \right) \right)&\le \left( 1-t \right) f\left( a \right) +tf\left( b \right) -f\left( \left( 1-t \right) a+tb \right) \\&\le 2\Gamma \left( \frac{f\left( a \right) +f\left( b \right) }{2}-f\left( \frac{a+b}{2} \right) \right) , \end{aligned} \end{aligned}$$
(1.1)

where \(\gamma =\min \left\{ t,1-t \right\} \), \(\Gamma =\max \left\{ t,1-t \right\} \) and \(0\le t\le 1\). Although this result is known for functions f defined on real numbers, it is valid for convex functions defined on normed spaces.

This inequality has received the attention of numerous researchers due to its usage in sharpening and reversing some celebrated inequalities in the literature. We refer the reader to [6, 21, 22, 27] as a sample of possible applications and related results. A stronger form of a convex function is known as a log-convex function. We recall that a function \(f:J\rightarrow (0,\infty )\) is said to be log-convex if \(\log f\) is convex. Equivalently, if \(f((1-t)a+tb)\le f(a)^{1-t}f(b)^t,\) when \(a,b\in J\) and \(0\le t\le 1.\)

The main goal of this paper is to present some interesting results on the so-called norm-angular distance, the skew angular distance, and matrix norms. Our approach will be based on delicate treatments of convex and log-convex results.

The organization of this paper is as follows. In the next section, we employ convexity as a new approach to obtain newly refined bounds for the angular distance between vectors in normed spaces and inner product spaces. We then study properties of log-convex functions with applications to matrix norm inequalities. For the sake of convenience, we introduce each subsequent section there, with an emphasis on the recent progress in that direction.

2 The Angular Distance

This section further explores the norm and skew angular distances, with related results in inner product spaces. Let \({\mathscr {X}}=({\mathscr {X}},\Vert \cdot \Vert )\) be a normed space (real or complex). Among the most important inequalities in \({\mathscr {X}}, \) is the triangle inequality that states

$$\begin{aligned} \Vert a+b \Vert \le \Vert a \Vert +\Vert b\Vert , a,b\in {\mathscr {X}}. \end{aligned}$$
(2.1)

A significant application of this inequality is its role in obtaining convergence results, for example. We refer the reader to [2, 5, 20, 24] for a relatively new discussion of triangle inequality.

In mathematical inequalities, sharpening and reversing known inequalities is a demanding topic, where researchers try to find better bounds than those known ones and to find reversed versions. This helps understand such inequalities and obtain further applications.

In [16], Maligranda showed the following interesting double inequality, which provides an improvement and a reverse of (2.1):

$$\begin{aligned} A[a,b]\cdot \min \{\Vert a\Vert ,\Vert b\Vert \}\le \Vert a\Vert +\Vert b\Vert - \Vert a + b\Vert \le A[a,b]\cdot \max \{\Vert a\Vert ,\Vert b\Vert \}, \end{aligned}$$
(2.2)

where \(A[a,b]=\Big ( 2 - \Big \Vert \dfrac{a}{\Vert a\Vert }+\dfrac{b}{\Vert b\Vert }\Big \Vert \Big )\ge 0\), and a and b are nonzero vectors in a normed space \({\mathscr {X}} = ({\mathscr {X}},\Vert \cdot \Vert )\).

Using this inequality Maligranda [17] obtained a lower bound and an upper bound for the norm angular distance or Clarkson distance (see, e.g., [4]) between nonzero vectors a and b in a normed space \(({\mathscr {X}},\Vert \cdot \Vert ).\) More precisely, given non-zero elements \(a,b\in {\mathscr {X}}\), the norm angular distance between a, b was defined in [4] as

$$\begin{aligned} \alpha [a,b] = \Big \Vert \dfrac{a}{\Vert a\Vert }-\dfrac{b}{\Vert b\Vert }\Big \Vert . \end{aligned}$$

In [17], the following double inequality was shown

$$\begin{aligned} \frac{\Vert a-b\Vert -|\Vert a\Vert -\Vert b\Vert |}{\min \{ \Vert a\Vert ,\Vert b\Vert \}} \le \alpha [a,b] \le \frac{\Vert a-b\Vert +|\Vert a\Vert -\Vert b\Vert |}{\max \{ \Vert a\Vert ,\Vert b\Vert \}}, \end{aligned}$$
(2.3)

as an improvement and a reverse of the Massera–Schäffer inequality, which states [18]

$$\begin{aligned} \alpha [a,b] \le \frac{2\Vert a-b\Vert }{\max \{ \Vert a\Vert ,\Vert b\Vert \}}, \end{aligned}$$
(2.4)

for every nonzero vectors a and b in \({\mathscr {X}}\). The latter inequality is stronger than the Dunkl–Williams inequality, given in [12] as follows:

$$\begin{aligned} \alpha [a,b] \le \frac{4\Vert a-b\Vert }{|\Vert a\Vert +\Vert b\Vert |}. \end{aligned}$$

Other results related to the angular distance, named Dunkl–Williams type theorems (see [12]), are given by Moslehian et al. [23].

Dehghan in [7] presented a new refinement of the triangle inequality and defined the skew angular distance between nonzero vectors x and y by \(\beta [a,b] = \Big \Vert \dfrac{a}{\Vert b\Vert }-\dfrac{b}{\Vert a\Vert }\Big \Vert \). In [7], the following double inequality is shown

$$\begin{aligned} \frac{\Vert a-b\Vert }{\min \{ \Vert a\Vert ,\Vert b\Vert \}}-\frac{|\Vert a\Vert -\Vert b\Vert |}{\max \{ \Vert a\Vert ,\Vert b\Vert \}} \le \beta [a,b] \le \frac{\Vert a-b\Vert }{\max \{ \Vert a\Vert ,\Vert b\Vert \}}+\frac{|\Vert a\Vert -\Vert b\Vert |}{\min \{ \Vert a\Vert ,\Vert b\Vert \}}, \end{aligned}$$
(2.5)

for any nonzero elements a and b in a real normed linear space \({\mathscr {X}} = ({\mathscr {X}},\Vert \cdot \Vert )\).

Recently, numerous improvements and generalizations of bounds for the angular distance and the skew angular distance have been established in [11, 14, 15, 25]. In this direction, we establish several estimates for the angular distance and the skew angular distance between two non-zero vectors in a normed space. We also obtain some refinements of Maligranda’s inequality and some refinements of Dehghan’s inequality. Our approach is based on delicate treatments of convex functions and their inequalities.

2.1 Normed Spaces

In this subsection, we use convex functions to study norm and skew angular distances. We state several corollaries and remarks that explain the relationship with the existing results. In particular, improvements of (2.3), (2.4) and (2.5) will be presented. The significance of these results is not only the results themselves but also the convexity approach that allows obtaining such results.

Theorem 2.1

Let \(\left( {\mathscr {X}},\left\| \cdot \right\| \right) \) be a real or complex normed space, and let \(p,q>0\) be such that \(\frac{1}{p}+\frac{1}{q}=1\). If \(a,b\in {\mathscr {X}}\) and \(r\ge 1,\) then

$$\begin{aligned} \begin{aligned} 2\lambda \left( \frac{{{p}^{r}}{{\left\| a \right\| }^{r}}+{{q}^{r}}{{\left\| b \right\| }^{r}}}{2}-{{\left\| \frac{pa+qb}{2} \right\| }^{r}} \right)&\le {{p}^{r-1}}{{\left\| a \right\| }^{r}}+{{q}^{r-1}}{{\left\| b \right\| }^{r}}-{{\left\| a+b \right\| }^{r}} \\&\le 2\mu \left( \frac{{{p}^{r}}{{\left\| a \right\| }^{r}}+{{q}^{r}}{{\left\| b \right\| }^{r}}}{2}-{{\left\| \frac{pa+qb}{2} \right\| }^{r}} \right) , \end{aligned} \end{aligned}$$
(2.6)

where \(\lambda =\min \left\{ {1}/{p},{1}/{q}\right\} \) and \(\mu =\max \left\{ {1}/{p},{1}/{q} \right\} \).

Proof

Let \(f:{\mathscr {X}}\rightarrow {\mathbb {R}}\) be a convex function, \(a,b\in {\mathscr {X}}\) and let \(p,q>0\) with \({1}/{p}\;+{1}/{q}\;=1\). If we replace \(1-t\) and t by \({1}/{p}\;\) and \({1}/{q}\;\) in (1.1), we deduce

$$\begin{aligned} 2\lambda \left( \frac{f\left( a \right) +f\left( b \right) }{2}-f\left( \frac{a+b}{2} \right) \right){} & {} \le \frac{1}{p}f\left( a \right) +\frac{1}{q}f\left( b \right) -f\left( \frac{1}{p}a+\frac{1}{q}b \right) \nonumber \\{} & {} \le 2\mu \left( \frac{f\left( a \right) +f\left( b \right) }{2}-f\left( \frac{a+b}{2} \right) \right) ,\nonumber \\ \end{aligned}$$
(2.7)

where \(\lambda =\min \left\{ {1}/{p}\;,{1}/{q} \right\} \) and \(\mu =\max \left\{ {1}/{p}\;,{1}/{q} \right\} \). Since this is valid for all \(a,b\in {\mathscr {X}}\), we may replace a by pa and b by qb in (2.7), to get

$$\begin{aligned} \begin{aligned} 2\lambda \left( \frac{f\left( pa \right) +f\left( qb \right) }{2}-f\left( \frac{pa+qb}{2} \right) \right)&\le \frac{1}{p}f\left( pa \right) +\frac{1}{q}f\left( qb \right) -f\left( a+b \right) \\&\le 2\mu \left( \frac{f\left( pa \right) +f\left( qb \right) }{2}-f\left( \frac{pa+qb}{2} \right) \right) . \end{aligned} \end{aligned}$$
(2.8)

Noting that the function \(f:{\mathscr {X}}\rightarrow {\mathbb {R}}\) defined by \(f\left( x \right) ={\Vert x\Vert ^{r}}\) (\(x\in {\mathscr {X}}\) and \(1\le r<\infty \)) is convex, applying (2.8) implies the desired result. \(\square \)

In the following corollary, we have an easier form that easily follows from (2.6).

Corollary 2.1

Let a and b two nonzero vectors in a normed linear space \(\left( {\mathscr {X}},\left\| \cdot \right\| \right) \) and let \(p,q>0\) be such that \(\frac{1}{p}+\frac{1}{q}=1.\) If \(r\ge 1\), then

$$\begin{aligned} \lambda \left( p^{r}+q^{r}-2^{1-r}\Big \Vert p\frac{a}{\Vert a\Vert }+q\frac{b}{\Vert b\Vert }\Big \Vert ^r\right){} & {} \le p^{r-1}+q^{r-1}-\Big \Vert \frac{a}{\Vert a\Vert }+\frac{b}{\Vert b\Vert }\Big \Vert ^r \nonumber \\{} & {} \le \mu \left( p^{r}+q^{r}-2^{1-r}\Big \Vert p\frac{a}{\Vert a\Vert }+q\frac{b}{\Vert b\Vert }\Big \Vert ^r\right) ,\nonumber \\ \end{aligned}$$
(2.9)

where \(\lambda =\min \left\{ {1}/{p},{1}/{q}\right\} \) and \(\mu =\max \left\{ {1}/{p},{1}/{q} \right\} \).

Proof

In (2.6), replace a and b by \(\dfrac{a}{\left\| a \right\| }\) and \(\dfrac{b}{\left\| b \right\| },\) respectively, to get the desired inequality. \(\square \)

Now we explain the significance (2.9).

Remark 2.1

Taking \(r=1\) in (2.9), we get

$$\begin{aligned} 0\le \lambda \left( pq-\Big \Vert p\frac{a}{\Vert a\Vert }+q\frac{b}{\Vert b\Vert }\Big \Vert \right){} & {} \le 2-\Big \Vert \frac{a}{\Vert a\Vert }+\frac{b}{\Vert b\Vert }\Big \Vert \\ {}{} & {} \le \mu \left( pq-\Big \Vert p\frac{a}{\Vert a\Vert }+q\frac{b}{\Vert b\Vert }\Big \Vert \right) . \end{aligned}$$

Using inequalities (2.2) and (2.9), we deduce

$$\begin{aligned} \Big ( 2 - \Big \Vert \dfrac{a}{\Vert a\Vert }+\dfrac{b}{\Vert b\Vert }\Big \Vert \Big )\ge \lambda \left( pq-\Big \Vert p\dfrac{a}{\Vert a\Vert }+q\dfrac{b}{\Vert b\Vert }\Big \Vert \right) \ge 0. \end{aligned}$$

Therefore, we obtain the following refinement of the triangle inequality

$$\begin{aligned} 0\le \lambda \left( pq-\Big \Vert p\dfrac{a}{\Vert a\Vert }+q\dfrac{b}{\Vert b\Vert }\Big \Vert \right) \cdot \min \{\Vert a\Vert ,\Vert b\Vert \}\le \Vert a\Vert +\Vert b\Vert - \Vert a + b\Vert , \end{aligned}$$

for all nonzero vectors a and b in a normed space \(({\mathscr {X}},\Vert \cdot \Vert )\).

Another refinement of the triangle inequality can be stated as follows, with the same parameters as before.

Corollary 2.2

Let \(\left( {\mathscr {X}},\left\| \cdot \right\| \right) \) be a real or complex normed linear space. The following inequality holds:

$$\begin{aligned} \begin{aligned} \lambda \left( p\left\| a \right\| +q\left\| b \right\| -\left\| pa+qb \right\| \right)&\le \left\| a \right\| +\left\| b \right\| -\left\| a+b \right\| \\&\le \mu \left( p\left\| a \right\| +q\left\| b \right\| -\left\| pa+qb \right\| \right) , \end{aligned} \end{aligned}$$
(2.10)

for all vectors a and b in \({\mathscr {X}}\).

Proof

Letting \(r=1\) in (2.6), we obtain the desired inequality. \(\square \)

Remark 2.2

Inequality (2.10) can be obtained as a consequence of [19, Theorem 1]. It is easy to see that inequality (2.10) improves and reverses the triangle inequality.

In the next result, we present a more elaborated form that improves both (2.3) and (2.4).

Theorem 2.2

Let \(\left( {\mathscr {X}},\left\| \cdot \right\| \right) \) be a real or complex normed space, and let \(p,q>0\) be such that \(\frac{1}{p}+\frac{1}{q}=1\). If \(a,b\in {\mathscr {X}}\backslash \left\{ 0 \right\} \), then

$$\begin{aligned} \left\| \frac{a}{\left\| a \right\| }-\frac{b}{\left\| b \right\| } \right\| +\lambda \min \left\{ \frac{\gamma _1}{\Vert a\Vert },\frac{\gamma _2}{\Vert b\Vert }\right\} \le \frac{\left\| a-b \right\| +\left| \;\left\| a \right\| -\left\| b \right\| \; \right| }{\max \left\{ \left\| a \right\| ,\left\| b \right\| \right\} } \end{aligned}$$
(2.11)

and

$$\begin{aligned} \frac{\left\| a-b \right\| +\left| \;\left\| a \right\| -\left\| b \right\| \; \right| }{\min \left\{ \left\| a \right\| ,\left\| b \right\| \right\} }\le \left\| \frac{a}{\left\| a \right\| }-\frac{b}{\left\| b \right\| } \right\| +\mu \max \left\{ \frac{\gamma _1}{\Vert a\Vert },\frac{\gamma _2}{\Vert b\Vert }\right\} , \end{aligned}$$
(2.12)

where

$$\begin{aligned} {{\gamma }_{1}}=p\left\| a-b \right\| +q\left| \;\left\| a \right\| -\left\| b \right\| \; \right| -\left\| pa+\left( q-p \right) b-\frac{\left\| a \right\| }{\left\| b \right\| }qb \right\| \ge 0 , \\ {{\gamma }_{2}}=q\left\| a-b \right\| +p\left| \;\left\| a \right\| -\left\| b \right\| \;\right| -\left\| \left( p-q\right) a+qb-\frac{\left\| b \right\| }{\left\| a \right\| }pa \right\| \ge 0, \end{aligned}$$

and \(\lambda =\min \left\{ {1}/{p},{1}/{q}\right\} \) and \(\mu =\max \left\{ {1}/{p},{1}/{q} \right\} \).

Proof

By employing the triangle inequality, it is easy to see that \(\gamma _1,\gamma _2\ge 0.\) Let \(a,b\in {\mathscr {X}}\backslash \left\{ 0 \right\} \). Replacing a and b by \({\left( a-b \right) }/{\left\| a \right\| }\;\) and \(b\left( {1}/{\left\| a \right\| }\;-{1}/{\left\| b \right\| }\; \right) \), in (2.10), we obtain

$$\begin{aligned} \begin{aligned}&\frac{\lambda }{\left\| a \right\| }\left( p\left\| a-b \right\| +q\left| \;\left\| a \right\| -\left\| b \right\| \;\right| -\left\| pa+\left( q-p \right) b-\frac{\left\| a \right\| }{\left\| b \right\| }qb \right\| \right) \\&\le \frac{\left\| a-b \right\| +\left| \;\left\| a \right\| -\left\| b \right\| \; \right| }{\left\| a \right\| }-\left\| \frac{a}{\left\| a \right\| }-\frac{b}{\left\| b \right\| } \right\| \\&\le \frac{\mu }{\left\| a \right\| }\left( p\left\| a-b \right\| +q\left| \;\left\| a \right\| -\left\| b \right\| \; \right| -\left\| pa+\left( q-p \right) b-\frac{\left\| a \right\| }{\left\| b \right\| }qb \right\| \right) . \end{aligned}\nonumber \\ \end{aligned}$$
(2.13)

If we replace a and b by \(a\left( {1}/{\left\| a \right\| }\;-{1}/{\left\| b \right\| }\; \right) \) and \({\left( a-b \right) }/{\left\| b \right\| },\) in (2.10), then we have

$$\begin{aligned} \begin{aligned}&\frac{\lambda }{\left\| b \right\| }\left( q\left\| a-b \right\| +p\left| \;\left\| a \right\| -\left\| b \right\| \;\right| -\left\| \left( p-q\right) a+qb-\frac{\left\| b \right\| }{\left\| a \right\| }pa \right\| \right) \\&\le \frac{\left\| a-b \right\| +\left| \;\left\| a \right\| -\left\| b \right\| \; \right| }{\left\| b \right\| }-\left\| \frac{a}{\left\| a \right\| }-\frac{b}{\left\| b \right\| } \right\| \\&\le \frac{\mu }{\left\| b \right\| }\left( q\left\| a-b \right\| +p\left| \;\left\| a \right\| -\left\| b \right\| \;\right| -\left\| \left( p-q\right) a+qb-\frac{\left\| b \right\| }{\left\| a \right\| }pa \right\| \right) . \end{aligned}\nonumber \\ \end{aligned}$$
(2.14)

We deduce the desired result using inequalities in (2.13) and (2.14). \(\square \)

The significance of the above theorem and its relation with (2.3) and (2.4) is explained next.

Remark 2.3

In terms of the angular distance, inequalities (2.11) and (2.12) become

$$\begin{aligned} \begin{aligned}&\frac{\left\| a-b \right\| +\left| \;\left\| a \right\| -\left\| b \right\| \; \right| }{\min \left\{ \left\| a \right\| ,\left\| b \right\| \right\} }-\mu \max \left\{ \frac{\gamma _1}{\Vert a\Vert },\frac{\gamma _2}{\Vert b\Vert }\right\} \le \alpha [a,b]\\&\le \frac{\left\| a-b \right\| +\left| \;\left\| a \right\| -\left\| b \right\| \; \right| }{\max \left\{ \left\| a \right\| ,\left\| b \right\| \right\} }-\lambda \min \left\{ \frac{\gamma _1}{\Vert a\Vert },\frac{\gamma _2}{\Vert b\Vert }\right\} , \end{aligned}\nonumber \\ \end{aligned}$$
(2.15)

for all vectors a and b in \({\mathscr {X}}\backslash \left\{ 0 \right\} \), where \(\gamma _1\) and \(\gamma _2\) are as given above. Therefore, (2.15) improves (2.3) proved by Maligranda.

We also have the inequality \(|\Vert \alpha \Vert -\Vert \beta \Vert |\le \Vert \alpha -\beta \Vert \). So

$$\begin{aligned}{} & {} \frac{2\left| \;\left\| a \right\| -\left\| b \right\| \; \right| }{\min \left\{ \left\| a \right\| ,\left\| b \right\| \right\} }-\mu \max \left\{ \frac{\gamma _1}{\Vert a\Vert },\frac{\gamma _2}{\Vert b\Vert }\right\} \le \alpha [a,b]\nonumber \\{} & {} \qquad \le \frac{2\left\| a-b \right\| }{\max \left\{ \left\| a \right\| ,\left\| b \right\| \right\} }-\lambda \min \left\{ \frac{\gamma _1}{\Vert a\Vert },\frac{\gamma _2}{\Vert b\Vert }\right\} , \end{aligned}$$
(2.16)

for all vectors a and b in \({\mathscr {X}}\backslash \left\{ 0 \right\} \), where \(\gamma _1\) and \(\gamma _2\) are as given above. The right part of inequality (2.16) represents another improvement of the Massera–Schäffer inequality (2.4).

So far, we have studied the norm angular distance. Now we investigate the skew angular distance.

Theorem 2.3

Let \(\left( {\mathscr {X}},\left\| \cdot \right\| \right) \) be a real or complex normed space, and let \(p,q>0\) be such that \(\frac{1}{p}+\frac{1}{q}=1\). If \(a,b\in {\mathscr {X}}\backslash \left\{ 0 \right\} \), then

$$\begin{aligned} \left\| \frac{a}{\left\| b \right\| }-\frac{b}{\left\| a \right\| } \right\| +\lambda \min \{\rho _1,\rho _2\}\le \frac{\left\| a-b \right\| }{\max \left\{ \left\| a \right\| ,\left\| b \right\| \right\} } +\frac{\left| \;\left\| a \right\| -\left\| b \right\| \; \right| }{\min \left\{ \left\| a \right\| ,\left\| b \right\| \right\} } \end{aligned}$$
(2.17)

and

$$\begin{aligned} \frac{\left\| a-b \right\| }{\min \left\{ \left\| a \right\| ,\left\| b \right\| \right\} } +\frac{\left| \;\left\| a \right\| -\left\| b \right\| \; \right| }{\max \left\{ \left\| a \right\| ,\left\| b \right\| \right\} }\le \left\| \frac{a}{\left\| b \right\| }-\frac{b}{\left\| a \right\| } \right\| +\mu \max \{\rho _1,\rho _2\}, \end{aligned}$$
(2.18)

where

$$\begin{aligned} {\rho }_{1}=p\frac{\left\| a-b \right\| }{\Vert b\Vert }+q\frac{|\Vert a\Vert -\Vert b\Vert |}{\Vert a\Vert }-\Big \Vert \frac{p}{\Vert b\Vert }a +\left( \frac{q-p}{\Vert b\Vert }-\frac{q}{\Vert a\Vert }\right) b\Big \Vert \ge 0, \\ {\rho }_{2}=q\frac{\left\| a-b \right\| }{\Vert a\Vert }+p\frac{|\Vert a\Vert -\Vert b\Vert |}{\Vert b\Vert }-\Big \Vert \left( \frac{p-q}{\Vert a\Vert } -\frac{p}{\Vert b\Vert }\right) a+\frac{q}{\Vert a\Vert }b\Big \Vert \ge 0, \end{aligned}$$

and \(\lambda =\min \left\{ {1}/{p},{1}/{q}\right\} \) and \(\mu =\max \left\{ {1}/{p},{1}/{q} \right\} \).

Proof

Using the triangle inequality, it is straightforward to see that \(\rho _1,\rho _2\ge 0.\) Let \(a,b\in {\mathscr {X}}\backslash \left\{ 0 \right\} \). Replacing the vectors a and b by the vectors \({\left( a-b \right) }/{\left\| b \right\| }\;\) and \(b\left( {1}/{\left\| b \right\| }\;-{1}/{\left\| a \right\| }\; \right) \), in (2.10), we get

$$\begin{aligned} \begin{aligned}&\lambda \rho _1=\lambda \left( p\frac{\left\| a-b \right\| }{\Vert b\Vert }+q\frac{|\Vert a\Vert -\Vert b\Vert |}{\Vert a\Vert }-\Big \Vert \frac{p}{\Vert b\Vert }a +\left( \frac{q-p}{\Vert b\Vert }-\frac{q}{\Vert a\Vert }\right) b\Big \Vert \right) \\&\le \frac{\left\| a-b \right\| }{\Vert b\Vert } +\frac{\left| \;\left\| a \right\| -\left\| b \right\| \; \right| }{\Vert a\Vert }-\Big \Vert \frac{a}{\Vert b\Vert }-\frac{b}{\Vert a\Vert }\Big \Vert \\&\le \mu \left( p\frac{\left\| a-b \right\| }{\Vert b\Vert }+q\frac{|\Vert a\Vert -\Vert b\Vert |}{\Vert a\Vert }-\Big \Vert \frac{p}{\Vert b\Vert }a +\left( \frac{q-p}{\Vert b\Vert }-\frac{q}{\Vert a\Vert }\right) b\Big \Vert \right) =\mu \rho _1. \end{aligned} \end{aligned}$$
(2.19)

If we replace a and b by \(a\left( {1}/{\left\| b \right\| }\;-{1}/{\left\| a \right\| }\; \right) \) and \({\left( a-b \right) }/{\left\| a \right\| },\) in (2.10), then we have

$$\begin{aligned} \begin{aligned}&\lambda \rho _2=\lambda \left( q\frac{\left\| a-b \right\| }{\Vert a\Vert }+p\frac{|\Vert a\Vert -\Vert b\Vert |}{\Vert b\Vert }-\Big \Vert \left( \frac{p-q}{\Vert a\Vert }-\frac{p}{\Vert b\Vert }\right) a+\frac{q}{\Vert a\Vert }b\Big \Vert \right) \\&\le \frac{\left\| a-b \right\| }{\Vert a\Vert } +\frac{\left| \;\left\| a \right\| -\left\| b \right\| \; \right| }{\Vert b\Vert }-\Big \Vert \frac{a}{\Vert b\Vert }-\frac{b}{\Vert a\Vert }\Big \Vert \\&\le \mu \left( q\frac{\left\| a-b \right\| }{\Vert a\Vert }+p\frac{|\Vert a\Vert -\Vert b\Vert |}{\Vert b\Vert }-\Big \Vert \left( \frac{p-q}{\Vert a\Vert }-\frac{p}{\Vert b\Vert }\right) a+\frac{q}{\Vert a\Vert }b\Big \Vert \right) =\mu \rho _2. \end{aligned} \end{aligned}$$
(2.20)

We deduce the desired inequalities using (2.19) and (2.20). \(\square \)

In the following remark, we explain how the above theorem improves (2.5).

Remark 2.4

In terms of the skew angular distance, inequalities (2.17) and (2.18) become

$$\begin{aligned} \begin{aligned}&\frac{\left\| a-b \right\| }{\min \left\{ \left\| a \right\| ,\left\| b \right\| \right\} }+\frac{\left| \;\left\| a \right\| -\left\| b \right\| \; \right| }{\max \left\{ \left\| a \right\| ,\left\| b \right\| \right\} }-\mu \max \{\rho _1,\rho _2\}\le \beta [a,b]\\&\quad \le \frac{\left\| a-b \right\| }{\max \left\{ \left\| a \right\| ,\left\| b \right\| \right\} }+\frac{\left| \;\left\| a \right\| -\left\| b \right\| \; \right| }{\min \left\{ \left\| a \right\| ,\left\| b \right\| \right\} }-\lambda \min \{\rho _1,\rho _2\}, \end{aligned} \end{aligned}$$
(2.21)

for all vectors a and b in \({\mathscr {X}}\backslash \left\{ 0 \right\} \), where \(\rho _1\) and \(\rho _2\) are given above. Therefore, (2.21) improves (2.5) proved by Dehghan. We also have the inequality \(|\Vert \alpha \Vert -\Vert \beta \Vert |\le \Vert \alpha -\beta \Vert \). Thus, we obtain

$$\begin{aligned}{} & {} \left| \;\left\| a \right\| -\left\| b \right\| \; \right| \frac{\Vert a\Vert +\Vert b\Vert }{\Vert a\Vert \Vert b\Vert }-\mu \max \{\rho _1,\rho _2\}\le \beta [a,b]\nonumber \\{} & {} \quad \le \left\| a-b \right\| \frac{\Vert a\Vert +\Vert b\Vert }{\Vert a\Vert \Vert b\Vert }-\lambda \min \{\rho _1,\rho _2\}, \end{aligned}$$
(2.22)

for all vectors a and b in \({\mathscr {X}}\backslash \left\{ 0 \right\} \), where \(\rho _1\) and \(\rho _2\) are as given above.

In the following result, we present a more elaborated form of the angular distance, where arbitrary powers of \(\Vert a\Vert \) and \(\Vert b\Vert \) are studied.

Proposition 2.1

Let \({\mathscr {X}}\) be a normed space, \(a,b\in {\mathscr {X}}\backslash \left\{ 0 \right\} \) and \(m\ge 0\). If \(p,q>0\) with \({1}/{p}\;+{1}/{q}\;=1\), then

$$\begin{aligned} \Big \Vert \Vert a\Vert ^{m-1} a-\Vert b\Vert ^{m-1}b \Big \Vert +\lambda \max \left\{ {{\alpha }_{1}},{{\alpha }_{2}} \right\} \le \max \left\{ {{\beta }_{1}},{{\beta }_{2}}\right\} , \end{aligned}$$

where

$$\begin{aligned} {{\alpha }_{1}}= & {} \left( p\Vert a\Vert ^{m-1}\left\| a-b \right\| +q\left\| b \right\| \Big |\left\| a \right\| ^{m-1}-\left\| b \right\| ^{m-1}\Big |-\Big \Vert p\Vert a\Vert ^{m-1}a\right. \\{} & {} \quad \left. +\left( q-p \right) \Vert a\Vert ^{m-1}b-q\left\| b \right\| ^{m-1}b \Big \Vert \right) , \\ {{\alpha }_{2}}= & {} \left( p\Vert b\Vert ^{m-1}\left\| a-b \right\| +q\left\| a \right\| \Big | \left\| a \right\| ^{m-1}-\left\| b \right\| ^{m-1}\Big |-\Big \Vert p\Vert b\Vert ^{m-1}b\right. \\{} & {} \quad \left. +\left( q-p \right) \Vert b\Vert ^{m-1}a-q\left\| a \right\| ^{m-1}a \Big \Vert \right) , \end{aligned}$$
$$\begin{aligned} {{\beta }_{1}}= \Vert a\Vert ^{m-1}\Vert a-b\Vert +\Vert b\Vert \Big |\Vert a\Vert ^{m-1}-\Vert b\Vert ^{m-1}\Big |, \\ {{\beta }_{2}}= \Vert b\Vert ^{m-1}\Vert a-b\Vert +\Vert a\Vert \Big |\Vert a\Vert ^{m-1}-\Vert b\Vert ^{m-1}\Big |, \end{aligned}$$

and \(\lambda =\min \{1/p,1/q\}.\)

Proof

Replacing a and b by \({\Vert a\Vert ^{m-1}\left( a-b \right) }\;\) and \(b\left( \Vert a \Vert ^{m-1}\;-\left\| b \right\| ^{m-1}\; \right) \), in (2.10), we get

$$\begin{aligned} \begin{aligned}&\lambda \left( p\Vert a\Vert ^{m-1}\left\| a-b \right\| +q\left\| b \right\| \Big |\left\| a \right\| ^{m-1}-\left\| b \right\| ^{m-1}\Big |-\Big \Vert p\Vert a\Vert ^{m-1}a\right. \\&\quad \left. +\left( q-p \right) \Vert a\Vert ^{m-1}b-\left\| b \right\| ^{m-1}qb \Big \Vert \right) \\&\le \Vert a\Vert ^{m-1}\Vert a-b\Vert +\Vert b\Vert \Big |\Vert a\Vert ^{m-1}-\Vert b\Vert ^{m-1}\Big |-\Big \Vert \Vert a\Vert ^{m-1} a-\Vert b\Vert ^{m-1}b \Big \Vert \\&\le \mu \left( p\Vert a\Vert ^{m-1}\left\| a-b \right\| +q\left\| b \right\| \Big |\left\| a \right\| ^{m-1}-\left\| b \right\| ^{m-1}\Big |-\Big \Vert p\Vert a\Vert ^{m-1}a\right. \\&\quad \left. +\left( q-p \right) \Vert a\Vert ^{m-1}b-\left\| b \right\| ^{m-1}qb \Big \Vert \right) . \end{aligned} \end{aligned}$$
(2.23)

We get the desired result if we interchange the roles of a and b in (2.23). \(\square \)

In particular, we may state the following refinement of some results in the proof of [16, Theorem 2].

Corollary 2.3

Let \(a,b\in {\mathscr {X}}\backslash \left\{ 0 \right\} , m\ge 0\) and \(p,q>0\) be such that \({1}/{p}\;+{1}/{q}\;=1\).

  1. (1)

    If \(0\le m<1,\) then

    $$\begin{aligned} \Big \Vert \Vert a\Vert ^{m-1} a-\Vert b\Vert ^{m-1}b \Big \Vert +\lambda \max \left\{ {{\alpha }_{1}},{{\alpha }_{2}} \right\} \le (2-m)\frac{\Vert a-b\Vert }{\max \{\Vert a\Vert , \Vert b\Vert \}^{1-m}}. \end{aligned}$$
  2. (2)

    If \(m\ge 1\), then

    $$\begin{aligned} \Big \Vert \Vert a\Vert ^{m-1} a-\Vert b\Vert ^{m-1}b \Big \Vert +\lambda \max \left\{ {{\alpha }_{1}},{{\alpha }_{2}} \right\} \le m\max \{\Vert a\Vert , \Vert b\Vert \}^{m-1}\Vert a-b\Vert , \end{aligned}$$

where

$$\begin{aligned} {{\alpha }_{1}}= & {} \left( p\Vert a\Vert ^{m-1}\left\| a-b \right\| +q\left\| b \right\| \Big |\left\| a \right\| ^{m-1}-\left\| b \right\| ^{m-1}\Big |-\Big \Vert p\Vert a\Vert ^{m-1}a\right. \\{} & {} \quad \left. +\left( q-p \right) \Vert a\Vert ^{m-1}b-\left\| b \right\| ^{m-1}qb \Big \Vert \right) , \\ {{\alpha }_{2}}= & {} \left( p\Vert b\Vert ^{m-1}\left\| a-b \right\| +q\left\| a \right\| \Big |\left\| a \right\| ^{m-1}-\left\| b \right\| ^{m-1}\Big |-\Big \Vert p\Vert b\Vert ^{m-1}b\right. \\{} & {} \quad \left. +\left( q-p \right) \Vert b\Vert ^{m-1}a-\left\| a \right\| ^{m-1}qa \Big \Vert \right) , \end{aligned}$$

and \(\lambda =\min \{1/p,1/q\}.\)

In particular, if we consider the case \(m=0\) in Proposition 2.1, we obtain the following improvement of (2.3).

Corollary 2.4

Let \(a,b\in {\mathscr {X}}\backslash \left\{ 0 \right\} \). If \(p,q>0\) are such that \({1}/{p}\;+{1}/{q}\;=1\), then

$$\begin{aligned} \left\| \frac{a}{\left\| a \right\| }-\frac{b}{\left\| b \right\| } \right\| +\lambda \max \left\{ {{\alpha }_{1}},{{\alpha }_{2}} \right\} \le \frac{\left\| a-b \right\| +\left| \;\left\| a \right\| -\left\| b \right\| \; \right| }{\min \left\{ \left\| a \right\| ,\left\| b \right\| \right\} }, \end{aligned}$$

where

$$\begin{aligned} {{\alpha }_{1}}=\frac{1}{\left\| a \right\| }\left( p\left\| a-b \right\| +\frac{q}{\left\| a \right\| }\left| \;\left\| a \right\| -\left\| b \right\| \; \right| -\left\| pa+\left( q-p \right) b-\frac{\left\| a \right\| }{\left\| b \right\| }qb \right\| \right) , \\ {{\alpha }_{2}}=\frac{1}{\left\| b \right\| }\left( p\left\| a-b \right\| +\frac{q}{\left\| b \right\| }\left| \;\left\| a \right\| -\left\| b \right\| \; \right| -\left\| pb+\left( q-p \right) a-\frac{\left\| b \right\| }{\left\| a \right\| }qa \right\| \right) , \end{aligned}$$

and \(\lambda =\min \{1/p,1/q\}.\)

Remark 2.5

In the reverse direction, we also have

$$\begin{aligned} \frac{\left\| a-b \right\| +\left| \;\left\| a \right\| -\left\| b \right\| \; \right| }{\max \left\{ \left\| a \right\| ,\left\| b \right\| \right\} }\le \mu \min \left\{ {{\beta }_{1}},{{\beta }_{2}} \right\} +\left\| \frac{a}{\left\| a \right\| }-\frac{b}{\left\| b \right\| } \right\| , \end{aligned}$$

where

$$\begin{aligned} {{\beta }_{1}}=\frac{1}{\left\| a \right\| }\left( p\left\| a-b \right\| +\frac{q}{\left\| a \right\| }\left| \;\left\| a \right\| -\left\| b \right\| \; \right| -\left\| pa+\left( q-p \right) b-\frac{\left\| a \right\| }{\left\| b \right\| }qb \right\| \right) , \\ {{\beta }_{2}}=\frac{1}{\left\| b \right\| }\left( p\left\| a-b \right\| +\frac{q}{\left\| b \right\| }\left| \;\left\| a \right\| -\left\| b \right\| \; \right| -\left\| pb+\left( q-p \right) a-\frac{\left\| b \right\| }{\left\| a \right\| }qa \right\| \right) , \end{aligned}$$

and \(\mu =\max \{1/p,1/q\}.\)

2.2 Inner Product Spaces

Let \({\mathscr {X}}=\left( {\mathscr {X}},\left<\cdot ,\cdot \right>\right) \) be a real or complex inner product space, with induced norm \(\Vert \cdot \Vert \). A significant inequality in this space is the Cauchy–Schwarz inequality that states

$$\begin{aligned} |\langle a,b \rangle | \le \Vert a\Vert \Vert b\Vert , \end{aligned}$$
(2.24)

for all \(a,b \in {\mathscr {X}}\). In addition, equality in (2.24) holds if and only if a and b are linearly dependent. We refer the reader to [1, 9, 29] for further discussion of this inequality.

This short section presents some similar results to the previously discussed angular distances. As a consequence, we obtain an identity for \({\mathfrak {R}}\left<a.b\right>\) for the nonzero vectors \(a,b\in {\mathscr {X}}\), where \({\mathfrak {R}}\) is the real part. This identity is related to the angle between two vectors in an inner product space.

Theorem 2.4

Let \({\mathscr {X}}\) be an inner product space, and let \(a,b\in {\mathscr {X}}\backslash \{0\}\). If \(p,q>0\) are such that \(\frac{1}{p}+\frac{1}{q}=1,\) then

$$\begin{aligned} {\mathfrak {R}}\left\langle a,b \right\rangle \le \left( \frac{p+q}{2}-\left( 1+\lambda \left( \frac{{{p}^{2}}+{{q}^{2}}}{2}-{{\left\| \frac{pa\left\| b \right\| +qb\left\| a \right\| }{2\left\| a \right\| \left\| b \right\| } \right\| }^{2}} \right) \right) \right) \left\| a \right\| \left\| b \right\| , \end{aligned}$$

and

$$\begin{aligned} \left( \frac{p+q}{2}-\left( 1+\mu \left( \frac{{{p}^{2}}+{{q}^{2}}}{2}-{{\left\| \frac{pa\left\| b \right\| +qb\left\| a \right\| }{2\left\| a \right\| \left\| b \right\| } \right\| }^{2}} \right) \right) \right) \left\| a \right\| \left\| b \right\| \le {\mathfrak {R}}\left\langle a,b \right\rangle ,\end{aligned}$$

where \(\lambda =\min \{1/p,1/q\}\) and \(\mu =\max \{1/p,1/q\}.\)

Proof

Assume that \(\left\| a \right\| =\left\| b \right\| =1\) and \({\mathscr {X}}\) is an inner product space. Letting \(r=2\) in the first inequality in (2.6) implies

$$\begin{aligned}\begin{aligned} 2\lambda \left( \frac{{{p}^{2}}+{{q}^{2}}}{2}-{{\left\| \frac{pa+qb}{2} \right\| }^{2}} \right)&\le p+q-{{\left\| a+b \right\| }^{2}} \\&=p+q-\left( 2+\left\langle a,b \right\rangle +\left\langle b,a \right\rangle \right) \\&=p+q-\left( 2+2{\mathfrak {R}}\left\langle a,b \right\rangle \right) . \end{aligned} \end{aligned}$$

Thus,

$$\begin{aligned} {\mathfrak {R}}\left\langle a,b \right\rangle \le \frac{p+q}{2}-\left( 1+\lambda \left( \frac{{{p}^{2}}+{{q}^{2}}}{2}-{{\left\| \frac{pa+qb}{2} \right\| }^{2}} \right) \right) . \end{aligned}$$

Now, replacing a and b by \({a}/{\left\| a \right\| }\;\) and \({b}/{\left\| b \right\| }\;\), respectively, gives the first inequality. If we apply the same method for the second inequality in (2.6), we infer the second inequality. This completes the proof. \(\square \)

Remark 2.6

If we let \(a, b \in {\mathscr {X}}\backslash \{0\}\) and \(p=q=2\) in Theorem 2.4, then

$$\begin{aligned} \frac{{\mathfrak {R}}\left\langle a,b \right\rangle }{\left\| a \right\| \left\| b \right\| }= -1+\frac{1}{2} {{\left\| \frac{a\left\| b \right\| +b\left\| a \right\| }{\left\| a \right\| \left\| b \right\| } \right\| }^{2}} . \end{aligned}$$

Using the admissible definition for the angle between the vectors a and b, given by [28]

$$\begin{aligned} \cos \phi _{a, b}=\frac{{\mathfrak {R}}\left\langle a,b \right\rangle }{\left\| a \right\| \left\| b \right\| }, 0\le \phi _{a, b}\le \pi , \end{aligned}$$

we have

$$\begin{aligned} \cos \phi _{a, -b} =-1+\frac{1}{2} {\left\| \frac{a}{\left\| a \right\| }-\frac{b}{\left\| b \right\| } \right\| }^{2}. \end{aligned}$$

Note that the last identity is related to the notion of the angle between two non-zero vectors a and b in a real normed space introduced by Dimmine et al. in [8],

$$\begin{aligned} \cos \psi _{a, b}=1-\frac{1}{2} {\left\| \frac{a}{\left\| a \right\| }-\frac{b}{\left\| b \right\| } \right\| }^{2}. \end{aligned}$$

3 Log-Convex Functions

In the previous section, we have seen how Theorem 2.1 plays a major role in obtaining our results about the angular distance between vectors in normed spaces. We notice that (1.1) is the key tool behind Theorem 2.1. Thus, convexity has been used to obtain those results concerning the angular distance.

As mentioned in the introduction, a more potent form of convexity is the so-called log-convexity. In this section, we present some inequalities for log-convex functions and then employ them to obtain new inequalities for norms of matrices.

In this context, we use the notation \({\mathcal {M}}_n\) to denote the algebra of all \(n\times n\) complex matrices. If \(\left<Ax,x\right>\ge 0\) for all \(x\in {\mathbb {C}}^n\), we say that A is positive semi-definite. On the other hand, if \(\left<Ax,x\right>>0\) for all non-zero vectors \(x\in {\mathbb {C}}^n\), we say that A is positive definite. A matrix norm \(\Vert \cdot \Vert \) defined on \({\mathcal {M}}_n\) is said to be unitarily invariant if \(\Vert UAV\Vert =\Vert A\Vert \) for all \(A\in {\mathcal {M}}_n\) and all unitary matrices \(U,V\in {\mathcal {M}}_n\). In the literature, researchers showed interest in studying possible bounds of \(\Vert A^tXB^{1-t}\Vert \) where \(X\in {\mathcal {M}}_n,\) \(A,B\in {\mathcal {M}}_n\) are positive (definite or semi-definite) and \(0\le t\le 1.\)

The following two related lemmas will be needed in our analysis. The first one has been shown in the proof of [13, Theorem 2.1], while the second one is shown in [26, Proposition 2.21].

Lemma 3.1

Let \(A\in {\mathcal {M}}_n\) be positive definite and \(x\in {\mathbb {C}}^n\). Then

$$\begin{aligned} f\left( t \right) =\left\langle {{A}^{t}}x,x \right\rangle \end{aligned}$$

is log-convex on \(\left( -\infty ,\infty \right) \).

Some applications of Lemma 3.1 can be found in [22].

Lemma 3.2

Let \(A,B,X\in {{{\mathcal {M}}}_{n}}\) be such that A, B are positive. Then the function

$$\begin{aligned} f\left( t \right) ={{\left\| {{A}^{t}}X{{B}^{1-t}} \right\| }} \end{aligned}$$

is log-convex on \(\left[ 0,1 \right] \), for every unitarily invariant norm \(\Vert \cdot \Vert .\)

Before proceeding, we remind the reader that every convex function \(f:[0,\infty )\rightarrow [0,\infty )\) with \(f\left( 0 \right) = 0\) satisfies the following two inequalities

$$\begin{aligned} f\left( tx \right) \le tf\left( x \right) ;\text { }0\le t\le 1 \end{aligned}$$

and

$$\begin{aligned} f\left( a \right) +f\left( b \right) \le f\left( a+b \right) . \end{aligned}$$

Now we prove some inequalities for log-convex functions; then we employ them to obtain matrix norm results. After that, we show further inequalities among real numbers’ powers based on a more delicate treatment of log-convex functions. The first result is a multiplicative-additive type inequality for log-convex functions.

Theorem 3.1

Let \(f:[0,\infty )\rightarrow \left( 0,\infty \right) \) be a log-convex function. Then for any \(a,b \ge 0\),

$$\begin{aligned} f\left( a \right) f\left( b \right) \le {{\left( \frac{f\left( \frac{a+b}{2} \right) }{\sqrt{f\left( 0 \right) f\left( a+b \right) }} \right) }^{\frac{4}{a+b}r'}}f\left( 0 \right) f\left( a+b \right) , \end{aligned}$$

and

$$\begin{aligned} {{\left( \frac{f\left( \frac{a+b}{2} \right) }{\sqrt{f\left( 0 \right) f\left( a+b \right) }} \right) }^{\frac{4}{a+b}R'}}f\left( 0 \right) f\left( a+b \right) \le f\left( a \right) f\left( b \right) , \end{aligned}$$

where \(r'=\min \left\{ a,b \right\} \) and \(R'=\max \left\{ a,b \right\} \).

Proof

Since f is log-convex, it is convex. So, by (1.1), we have

$$\begin{aligned} f\left( \left( 1-t \right) a+tb \right) \le \left( 1-t \right) f\left( a \right) +tf\left( b \right) -2r\left( \frac{f\left( a \right) +f\left( b \right) }{2}-f\left( \frac{a+b}{2} \right) \right) , \end{aligned}$$

where \(r=\min \left\{ t,1-t \right\} \) and \(0\le t\le 1\) for \(a,b\ge 0.\) Since f is a log-convex function, \(\log f\) is convex. Accordingly,

$$\begin{aligned}\begin{aligned}&\log f\left( \left( 1-t \right) a+tb \right) \\&\le \left( 1-t \right) \log f\left( a \right) +t\log f\left( b \right) -2r\left( \frac{\log f\left( a \right) +\log f\left( b \right) }{2}-\log f\left( \frac{a+b}{2} \right) \right) \\&=\log \left( {{\left( \frac{f\left( \frac{a+b}{2} \right) }{\sqrt{f\left( a \right) f\left( b \right) }} \right) }^{2r}}{{f}^{1-t}}\left( a \right) {{f}^{t}}\left( b \right) \right) . \\ \end{aligned} \end{aligned}$$

Consequently (see also [6, (5)]),

$$\begin{aligned} f\left( \left( 1-t \right) a+tb \right) \le {{\left( \frac{f\left( \frac{a+b}{2} \right) }{\sqrt{f\left( a \right) f\left( b \right) }} \right) }^{2r}}{{f}^{1-t}}\left( a \right) {{f}^{t}}\left( b \right) . \end{aligned}$$
(3.1)

Now, from the inequality (3.1), we can write

$$\begin{aligned} f\left( tx \right) \le {{\left( \frac{f\left( \frac{x}{2} \right) }{\sqrt{f\left( 0 \right) f\left( x \right) }} \right) }^{2r}}{{f}^{1-t}}\left( 0 \right) {{f}^{t}}\left( x \right) , \end{aligned}$$
(3.2)

for any \(x\ge 0\) and \(0\le t\le 1\). Employing (3.2), we obtain

$$\begin{aligned}\begin{aligned}&f\left( a \right) f\left( b \right) \\&=f\left( \frac{a}{a+b}\left( a+b \right) \right) f\left( \frac{b}{a+b}\left( a+b \right) \right) \\&\le {{\left( \frac{f\left( \frac{a+b}{2} \right) }{\sqrt{f\left( 0 \right) f\left( a+b \right) }} \right) }^{2r_1}}{{f}^{\frac{b}{a+b}}}\left( 0 \right) {{f}^{\frac{a}{a+b}}}\left( a+b \right) \\&\quad {{\left( \frac{f\left( \frac{a+b}{2} \right) }{\sqrt{f\left( 0 \right) f\left( a+b \right) }} \right) }^{2r_1}}{{f}^{\frac{a}{a+b}}}\left( 0 \right) {{f}^{\frac{b}{a+b}}}\left( a+b \right) \\&={{\left( \frac{f\left( \frac{a+b}{2} \right) }{\sqrt{f\left( 0 \right) f\left( a+b \right) }} \right) }^{\frac{4}{a+b}r'}}f\left( 0 \right) f\left( a+b \right) , \end{aligned}\end{aligned}$$

where \(r'=\min \left\{ a,b \right\} \) and \(r_1=\min \{\frac{a}{a+b},\frac{b}{a+b}\}=\frac{r'}{a+b}\). So,

$$\begin{aligned} f\left( a \right) f\left( b \right) \le {{\left( \frac{f\left( \frac{a+b}{2} \right) }{\sqrt{f\left( 0 \right) f\left( a+b \right) }} \right) }^{\frac{4}{a+b}r'}}f\left( 0 \right) f\left( a+b \right) . \end{aligned}$$

For the reverse side, the second inequality in (1.1) assures that

$$\begin{aligned} \left( 1-t \right) f\left( a \right) +tf\left( b \right) \le f\left( \left( 1-t \right) a+tb \right) +2R\left( \frac{f\left( a \right) +f\left( b \right) }{2}-f\left( \frac{a+b}{2} \right) \right) \end{aligned}$$

holds, where \(R=\max \left\{ t,1-t \right\} \) and \(0\le t\le 1\). Therefore, for the log-convex function f, we have (see also [6, (8)])

$$\begin{aligned} {{f}^{1-t}}\left( a \right) {{f}^{t}}\left( b \right) \le {{\left( \frac{\sqrt{f\left( a \right) f\left( b \right) }}{f\left( \frac{a+b}{2} \right) } \right) }^{2R}}f\left( \left( 1-t \right) a+tb \right) . \end{aligned}$$
(3.3)

Hence,

$$\begin{aligned} {{f}^{1-t}}\left( 0 \right) {{f}^{t}}\left( x \right) \le {{\left( \frac{\sqrt{f\left( 0 \right) f\left( x \right) }}{f\left( \frac{x}{2} \right) } \right) }^{2R}}f\left( tx \right) , \end{aligned}$$
(3.4)

for any \(x\ge 0\) and \(0\le t\le 1\). By (3.4), we conclude that

$$\begin{aligned} {{\left( \frac{f\left( \frac{a+b}{2} \right) }{\sqrt{f\left( 0 \right) f\left( a+b \right) }} \right) }^{\frac{4}{a+b}R'}}f\left( 0 \right) f\left( a+b \right) \le f\left( a \right) f\left( b \right) , \end{aligned}$$

where \(R'=\max \left\{ a,b \right\} \). This completes the proof. \(\square \)

Remark 3.1

Since, for log-convex functions,

$$\begin{aligned} {{\left( \frac{f\left( \frac{a+b}{2} \right) }{\sqrt{f\left( 0 \right) f\left( a+b \right) }} \right) }^{\frac{4}{a+b}r'}}\le 1 \end{aligned}$$

and \(f\left( \frac{a+b}{2}\right) \le \sqrt{f(0)f(a+b)}\), we have

$$\begin{aligned} f\left( a \right) f\left( b \right) \le {{\left( \frac{f\left( \frac{a+b}{2} \right) }{\sqrt{f\left( 0 \right) f\left( a+b \right) }} \right) }^{\frac{4}{a+b}r'}}f\left( 0 \right) f\left( a+b \right) \le f\left( 0 \right) f\left( a+b \right) . \end{aligned}$$

If f enjoy the inequality \(f\left( a \right) f\left( b \right) \le f\left( a+b \right) \), and if 0 is in its domain, then \(f\left( 0 \right) \le 1\). To see this, \(f\left( 0 \right) \le \frac{f\left( a+0 \right) }{f\left( a \right) }=1\). The following corollary provides an interesting additive-multiplicative inequality for log-convex functions.

Corollary 3.1

Let the assumptions of Theorem 3.1 hold. If \(f\left( 0 \right) =1\), then

$$\begin{aligned} {{\left( \frac{f\left( \frac{a+b}{2} \right) }{\sqrt{f\left( a+b \right) }} \right) }^{\frac{4}{a+b}R'}}f\left( a+b \right) \le f\left( a \right) f\left( b \right) \le {{\left( \frac{f\left( \frac{a+b}{2} \right) }{\sqrt{f\left( a+b \right) }} \right) }^{\frac{4}{a+b}r'}}f\left( a+b \right) . \end{aligned}$$

On the other hand, the following theorem presents super and sub-additive inequalities for certain inner products.

Theorem 3.2

Let \(A\in {\mathcal {M}}_n\) be positive definite and let \(x\in {\mathbb {C}}^n\). Then for any \(s,t\ge 0\),

$$\begin{aligned} \left\langle {{A}^{s}}x,x \right\rangle \left\langle {{A}^{t}}x,x \right\rangle \le {{\left( \frac{\left\langle {{A}^{\frac{s+t}{2}}}x,x \right\rangle }{\sqrt{\left\langle {{A}^{s+t}}x,x \right\rangle }} \right) }^{\frac{4}{s+t}\lambda }}\left\langle {{A}^{s+t}}x,x \right\rangle \end{aligned}$$

and

$$\begin{aligned} {{\left( \frac{\left\langle {{A}^{\frac{s+t}{2}}}x,x \right\rangle }{\sqrt{\left\langle {{A}^{s+t}}x,x \right\rangle }} \right) }^{\frac{4}{s+t}\nu }}\left\langle {{A}^{s+t}}x,x \right\rangle \le \left\langle {{A}^{s}}x,x \right\rangle \left\langle {{A}^{t}}x,x \right\rangle , \end{aligned}$$

where \(\lambda =\min \left\{ s,t \right\} \) and \(\nu =\max \left\{ s,t \right\} \).

Proof

The result follows from Lemma 3.1 and Corollary 3.1 by taking into account that \({{A}^{0}}=I\). \(\square \)

Now we are ready to present the following matrix norm inequalities, related to the quantity \(\Vert A^tXB^{1-t}\Vert ,\) where we use Lemma 3.2 and Corollary 3.1.

Theorem 3.3

Let \(A,B,X\in {{{\mathcal {M}}}_{n}}\) be such that A, B are positive definite. Then for any \(0\le s,t \le 1\) and any unitarily invariant norm \(\Vert \cdot \Vert \),

$$\begin{aligned} {{\left( \frac{{{\left\| {{A}^{\frac{t+s}{2}}}X{{B}^{1-\left( \frac{t+s}{2} \right) }} \right\| }}}{\sqrt{{{\left\| {{A}^{t+s}}X{{B}^{1-\left( t+s \right) }} \right\| }}}} \right) }^{\frac{4}{t+s}\nu }}{{\left\| {{A}^{t+s}}X{{B}^{1-\left( t+s \right) }} \right\| }}\le {{\left\| {{A}^{t}}X{{B}^{1-t}} \right\| }}{{\left\| {{A}^{s}}X{{B}^{1-s}} \right\| }} \end{aligned}$$

and

$$\begin{aligned} {{\left\| {{A}^{t}}X{{B}^{1-t}} \right\| }}{{\left\| {{A}^{s}}X{{B}^{1-s}} \right\| }}\le {{\left( \frac{{{\left\| {{A}^{\frac{t+s}{2}}}X{{B}^{1-\left( \frac{t+s}{2} \right) }} \right\| }}}{\sqrt{{{\left\| {{A}^{t+s}}X{{B}^{1-\left( t+s \right) }} \right\| }}}} \right) }^{\frac{4}{t+s}\lambda }}{{\left\| {{A}^{t+s}}X{{B}^{1-\left( t+s \right) }} \right\| }}, \end{aligned}$$

where \(\lambda =\min \left\{ s,t \right\} \) and \(\nu =\max \left\{ s,t \right\} \).

Remark 3.2

At this point, we remark that Theorems 3.2 and 3.3 have been shown for positive definite matrices \(A,B\in {\mathcal {M}}_n.\) It is natural to ask about the validity of these results in infinite dimensional Hilbert spaces. In this context, let \({\mathcal {B}}({\mathcal {H}})\) be the algebra of all bounded linear operators on a Hilbert space \({\mathcal {H}}.\) If \(A\in {\mathcal {B}}({\mathcal {H}})\) is such that \(\left<Ax,x\right>>0\) for all non-zero \(x\in {\mathcal {H}}\), then A is said to be a strictly positive operator.

The proof of Theorem 3.2 was based on Lemma 3.1, which asserts log-convexity of the function \(f(t)=\left<A^tx,x\right>\) when \(A\in {\mathcal {M}}_n\) is positive definite.

Referring to [13, Theorem 2.1], it is shown that when \(A\in {\mathcal {B}}({\mathcal {H}})\) is strictly positive, then \(f(t)=\left<A^tx,x\right>\) is log-convex, for each \(x\in {\mathcal {H}}\). Thus, Theorem 3.2 is valid for \(A\in {\mathcal {B}}({\mathcal {H}})\) and \(x\in {\mathcal {H}}.\)

On the other hand, Theorem 3.3 was based on Lemma 3.2, which asserts log-convexity of the function \(f(t)= \Vert A^{t}XB^{1-t}\Vert \), for positive definite \(A,B\in {\mathcal {M}}_n\) and arbitrary \(X\in {\mathcal {M}}_n\), where \(\Vert \cdot \Vert \) is an arbitrary unitarily invariant norm on \({\mathcal {M}}_n\). Recall that unitarily invariant norms on \({\mathcal {B}}({\mathcal {H}})\) are defined on norm ideals associated with these norms. Let \(\Vert \;\;\Vert \) be a unitarily invariant norm defined on a norm ideal \({\mathcal {N}}_{\Vert \;\;\Vert }\) of \({\mathcal {B}}({\mathcal {H}})\). When we write \(\Vert A\Vert \), we implicitly mean that \(A\in {\mathcal {N}}_{\Vert \;\;\Vert }\). In [3, Theorem 2], it is shown that when \(\Vert \cdot \Vert \) is a unitarily invariant norm on \({\mathcal {B}}({\mathcal {H}})\), with associated norm ideal \({\mathcal {N}}_{\Vert \;\Vert }\), then the function \(f(t)= \Vert A^{t}XB^{1-t}\Vert \) is log-convex, where \(A,B,X\in {\mathcal {N}}_{\Vert \;\Vert }\). Therefore, we deduce that Theorem 3.3 is true for Hilbert space operators.

So, the conclusion of this remark is that both Theorems 3.2 and 3.3 are valid for Hilbert space operators, not only for the algebra \({\mathcal {M}}_n\).

As a conclusion of this paper, we present a delicate treatment of log-convex functions. This treatment is related to the fact that if f is a convex function on the interval J and \(x,y,z\in J\), with \(x< y< z\), then

$$\begin{aligned} \frac{f\left( y \right) -f\left( x \right) }{y-x}\le \frac{f\left( z \right) -f\left( y \right) }{z-y}. \end{aligned}$$

This inequality can be translated to the log-convex setting as

$$\begin{aligned} {{\left( \frac{f\left( y \right) }{f\left( x \right) } \right) }^{\frac{1}{y-x}}}\le {{\left( \frac{f\left( z \right) }{f\left( y \right) } \right) }^{\frac{1}{z-y}}}. \end{aligned}$$

The following theorem provides an improvement and a reverse of this inequality.

Theorem 3.4

Let \(f:J\rightarrow \left( 0,\infty \right) \) be a log-convex function and let \(x,y,z\in J\). Then for any \(x< y< z\),

$$\begin{aligned} {{\left( \frac{f\left( y \right) }{f\left( x \right) } \right) }^{\frac{1}{y-x}}}\le {{\left( \frac{f\left( \frac{z+x}{2} \right) }{\sqrt{f\left( z \right) f\left( x \right) }} \right) }^{\frac{z-x-\left| 2y-\left( x+z \right) \right| }{\left( y-x \right) \left( z-y \right) }}}{{\left( \frac{f\left( z \right) }{f\left( y \right) } \right) }^{\frac{1}{z-y}}}, \end{aligned}$$
(3.5)

and

$$\begin{aligned} {{\left( \frac{f\left( \frac{z+x}{2} \right) }{\sqrt{f\left( z \right) f\left( x \right) }} \right) }^{\frac{z-x+\left| 2y-\left( x+z \right) \right| }{\left( y-x \right) \left( z-y \right) }}}{{\left( \frac{f\left( z \right) }{f\left( y \right) } \right) }^{\frac{1}{z-y}}}\le {{\left( \frac{f\left( y \right) }{f\left( x \right) } \right) }^{\frac{1}{y-x}}}. \end{aligned}$$
(3.6)

Proof

If we let \(1-t=\frac{y-x}{z-x}\), \(t=\frac{z-y}{z-x}\), \(a=z\), and \(b=x\), we get from (3.1) that

$$\begin{aligned} f\left( y \right) \le {{\left( \frac{f\left( \frac{z+x}{2} \right) }{\sqrt{f\left( z \right) f\left( x \right) }} \right) }^{1-\frac{\left| 2y-\left( x+z \right) \right| }{z-x}}}{{f}^{\frac{y-x}{z-x}}}\left( z \right) {{f}^{\frac{z-y}{z-x}}}\left( x \right) . \end{aligned}$$
(3.7)

Raise the two sides of (3.7) to the power of \(z-x\), and we obtain

$$\begin{aligned} {{f}^{z-x}}\left( y \right) \le {{\left( \frac{f\left( \frac{z+x}{2} \right) }{\sqrt{f\left( z \right) f\left( x \right) }} \right) }^{z-x-\left| 2y-\left( x+z \right) \right| }}{{f}^{y-x}}\left( z \right) {{f}^{z-y}}\left( x \right) . \end{aligned}$$

Multiply both sides of the above inequality by \(\frac{{{f}^{x-y}}\left( y \right) }{{{f}^{z-y}}\left( x \right) }\) to get

$$\begin{aligned} {{\left( \frac{f\left( y \right) }{f\left( x \right) } \right) }^{z-y}}\le {{\left( \frac{f\left( \frac{z+x}{2} \right) }{\sqrt{f\left( z \right) f\left( x \right) }} \right) }^{z-x-\left| 2y-\left( x+z \right) \right| }}{{\left( \frac{f\left( z \right) }{f\left( y \right) } \right) }^{y-x}}. \end{aligned}$$

From this, we can write

$$\begin{aligned} {{\left( \frac{f\left( y \right) }{f\left( x \right) } \right) }^{\frac{1}{y-x}}}\le {{\left( \frac{f\left( \frac{z+x}{2} \right) }{\sqrt{f\left( z \right) f\left( x \right) }} \right) }^{\frac{z-x-\left| 2y-\left( x+z \right) \right| }{\left( y-x \right) \left( z-y \right) }}}{{\left( \frac{f\left( z \right) }{f\left( y \right) } \right) }^{\frac{1}{z-y}}}, \end{aligned}$$

as desired. We can deduce the second inequality by applying the same technique to the inequality (3.3). \(\square \)

It is worth mentioning here that inequalities (3.1) and (3.5) (resp. (3.3) and (3.6)) are equivalent. (3.1) (resp. (3.3)) \(\Rightarrow \) (3.5) (resp. (3.6)) has been presented in the proof of Theorem 3.4, so we need to show (3.5) (resp. (3.6)) \(\Rightarrow \) (3.1) (resp. (3.3)). If we put \(x=0\), \(y=t\), \(z=1\), in (3.5), we get

$$\begin{aligned} f\left( t \right) \le {{\left( \frac{f\left( \frac{1}{2} \right) }{\sqrt{f\left( 1 \right) f\left( 0 \right) }} \right) }^{2r}}{{f}^{t}}\left( 1 \right) {{f}^{1-t}}\left( 0 \right) \end{aligned}$$
(3.8)

for any \(0\le t\le 1\). Define

$$\begin{aligned} g\left( t \right) \equiv f\left( \left( 1-t \right) a+tb \right) . \end{aligned}$$

It is not hard to check that if f is log-convex, then g is log-convex too. Therefore, g satisfies the inequality (3.8). We reach (3.1) since \(g\left( \frac{1}{2} \right) =f\left( \frac{a+b}{2} \right) \), \(g\left( 0 \right) =f\left( a \right) \), and \(g\left( 1 \right) =f\left( b \right) \). (3.6) \(\Rightarrow \) (3.3) can be obtained in the same way.

Our last result in this direction is the following Clarkson-type inequality, where we explore the relation between \(\frac{|a|^p+|b|^p}{2}\) and \(\left| \frac{a^p+b^p}{2}\right| ,\) for the real numbers a, b and \(p\ge 1.\) Such inequalities are important when studying the \(p-\)norm in \(L^p\) spaces.

Theorem 3.5

Let \(a,b \in {\mathbb {R}}\). Then for any \(p\ge 1\),

$$\begin{aligned} \frac{{{\left| a \right| }^{p}}+{{\left| b \right| }^{p}}}{2}\le \frac{1}{2r}\log \left( \frac{{{\exp }^{1-t}}\left( {{\left| a \right| }^{p}} \right) {{\exp }^{t}}\left( {{\left| b \right| }^{p}} \right) }{\exp \left( {{\left| \left( 1-t \right) a+tb \right| }^{p}} \right) } \right) +{{\left| \frac{a+b}{2} \right| }^{p}}, \end{aligned}$$

and

$$\begin{aligned} {{\left| \frac{a+b}{2} \right| }^{p}}\le \frac{1}{2R}\log \left( \frac{\exp \left( {{\left| \left( 1-t \right) a+tb \right| }^{p}} \right) }{{{\exp }^{1-t}}\left( {{\left| a \right| }^{p}} \right) {{\exp }^{t}}\left( {{\left| b \right| }^{p}} \right) } \right) +\frac{{{\left| a \right| }^{p}}+{{\left| b \right| }^{p}}}{2}. \end{aligned}$$

In particular,

$$\begin{aligned} \left| a \right| +\left| b \right| \le \frac{1}{r}\log \left( \frac{{{\exp }^{1-t}}\left( \left| a \right| \right) {{\exp }^{t}}\left( \left| b \right| \right) }{\exp \left( \left| \left( 1-t \right) a+tb \right| \right) } \right) +\left| a+b \right| , \end{aligned}$$

and

$$\begin{aligned} \left| a+b \right| \le \left| a \right| +\left| b \right| +\frac{1}{R}\log \left( \frac{\exp \left( \left| \left( 1-t \right) a+tb \right| \right) }{{{\exp }^{1-t}}\left( \left| a \right| \right) {{\exp }^{t}}\left( \left| b \right| \right) } \right) , \end{aligned}$$

where \(r=min\{t,1-t\}\), \(R=max\{t,1-t\}\), and \(0 \le t \le 1\).

Proof

We can write the inequality (3.1) in the following form:

$$\begin{aligned} \sqrt{f\left( a \right) f\left( b \right) }\le {{\left( \frac{{{f}^{1-t}}\left( a \right) {{f}^{t}}\left( b \right) }{f\left( \left( 1-t \right) a+tb \right) } \right) }^{\frac{1}{2r}}}f\left( \frac{a+b}{2} \right) . \end{aligned}$$

The function \(f\left( x \right) =\exp \left( {{\left| x \right| }^{p}} \right) \left( p\ge 1 \right) \) is a log-convex function. Thus, for any \(p\ge 1\), we have

$$\begin{aligned} \begin{aligned} \sqrt{\exp \left( {{\left| a \right| }^{p}}+{{\left| b \right| }^{p}} \right) }&=\sqrt{\exp \left( {{\left| a \right| }^{p}} \right) \exp \left( {{\left| b \right| }^{p}} \right) } \\&\le {{\left( \frac{{{\exp }^{1-t}}\left( {{\left| a \right| }^{p}} \right) {{\exp }^{t}}\left( {{\left| b \right| }^{p}} \right) }{\exp \left( {{\left| \left( 1-t \right) a+tb \right| }^{p}} \right) } \right) }^{\frac{1}{2r}}}\exp \left( {{\left| \frac{a+b}{2} \right| }^{p}} \right) . \end{aligned}\nonumber \\ \end{aligned}$$
(3.9)

Take the logarithm on both sides of (3.9); we get

$$\begin{aligned} \begin{aligned} \frac{{{\left| a \right| }^{p}}+{{\left| b \right| }^{p}}}{2}&=\log \sqrt{\exp \left( {{\left| a \right| }^{p}}+{{\left| b \right| }^{p}} \right) } \\&\le \log \left( {{\left( \frac{{{\exp }^{1-t}}\left( {{\left| a \right| }^{p}} \right) {{\exp }^{t}}\left( {{\left| b \right| }^{p}} \right) }{\exp \left( {{\left| \left( 1-t \right) a+tb \right| }^{p}} \right) } \right) }^{\frac{1}{2r}}}\exp \left( {{\left| \frac{a+b}{2} \right| }^{p}} \right) \right) \\&=\log \left( {{\left( \frac{{{\exp }^{1-t}}\left( {{\left| a \right| }^{p}} \right) {{\exp }^{t}}\left( {{\left| b \right| }^{p}} \right) }{\exp \left( {{\left| \left( 1-t \right) a+tb \right| }^{p}} \right) } \right) }^{\frac{1}{2r}}} \right) +{{\left| \frac{a+b}{2} \right| }^{p}} \\&=\frac{1}{2r}\log \left( \frac{{{\exp }^{1-t}}\left( {{\left| a \right| }^{p}} \right) {{\exp }^{t}}\left( {{\left| b \right| }^{p}} \right) }{\exp \left( {{\left| \left( 1-t \right) a+tb \right| }^{p}} \right) } \right) +{{\left| \frac{a+b}{2} \right| }^{p}}. \end{aligned}\end{aligned}$$

This completes the first inequality. The third inequality can be obtained using this inequality by setting \(p=1\). The inequality (3.3) also can be written as

$$\begin{aligned} f\left( \frac{a+b}{2} \right) \le {{\left( \frac{f\left( \left( 1-t \right) a+tb \right) }{{{f}^{1-t}}\left( a \right) {{f}^{t}}\left( b \right) } \right) }^{\frac{1}{2R}}}\sqrt{f\left( a \right) f\left( b \right) }. \end{aligned}$$

Using the same arguments as above, we infer the other inequalities. \(\square \)