# On the Jensen Functional and Strong Convexity

• Flavia-Corina Mitroi-Symeonidis
• Nicuşor Minculete
Article

## Abstract

In this note we describe some results concerning upper and lower bounds for the Jensen functional. We use several known and new results to shed light on the concept of a strongly convex function.

### Keywords

Jensen functional Strongly convex function

### Mathematics Subject Classification

Primary 26B25 Secondary 26D15

## 1 Introduction

The aim of this paper is to discuss new results concerning the Jensen functional in the framework of strong convexity.

For the reader’s convenience, before going into details, we here quote some relevant results.

### Definition 1

We consider a real valued function f defined on an interval I , $$x_{1},x_{2},...,x_{n}\in I$$ and $$p_{1},p_{2},...,p_{n}\in \left( 0,1\right)$$ with $$\sum _{i=1}^{n}p_{i}=1$$. The Jensen functional is defined by
\begin{aligned} {\mathcal {J}}\left( f,\mathbf {p},\mathbf {x}\right) \equiv \sum _{i=1}^{n}p_{i}f\left( x_{i}\right) -f\left( \sum _{i=1}^{n}p_{i}x_{i}\right) \end{aligned}
(see [1]).

### Definition 2

A function f defined on an interval I is strongly convex with modulus $$c>0$$[ or c-strongly convex] if
\begin{aligned} f\left( \left( 1-\lambda \right) x+\lambda y\right) \le \left( 1-\lambda \right) f\left( x\right) +\lambda f\left( y\right) -c\lambda \left( 1-\lambda \right) \left( y-x\right) ^{2}, \end{aligned}
(1.1)
for all$$\ x,y\in I$$ and all $$\lambda \in [0,1]$$. We call f strongly convex if there exists a $$c>0$$ such that is strongly convex with modulus c.

Strongly convex functions were introduced by B. T. Polyak [7]. A function f is called strongly concave with modulus c[or approximately convex of order 2 [6]] if $$-f$$ is strongly convex with modulus c.

### Example 1

Obviously every strongly convex function is convex. Affine functions are not strongly convex. The function $$f\left( x\right) =cx^{2}+bx+a$$ is strongly convex with modulus c and the inequality (1.1) holds with equality sign.

According to Hiriart–Urruty and Lemaréchal [2] we have:

### Proposition 1

The function f is strongly convex with modulus c if and only if the function $$g\left( x\right) =f\left( x\right) -cx^{2}$$ is convex.

In [3, Theorem 4] the following result is proved:

### Proposition 2

Considering $$p_{i}\ge 0,i=1,...,n,$$ with $$\sum _{i=1}^{n}p_{i}=1$$, $$\bar{x}=\sum _{i=1}^{n}p_{i}x_{i}$$ and the function f strongly convex with modulus c,  we have
\begin{aligned} {\mathcal {J}}\left( f,\mathbf {p},\mathbf {x}\right) \ge c\sum _{i=1}^{n}p_{i}(x_{i}-\bar{x})^{2}. \end{aligned}
(1.2)

This is re-proved using the probabilistic approach in a paper of T. Rajba and Sz. Wąsowicz [8, Corollary2.3]. Notice that the set of strongly convex functions is closed under addition and positive scalar multiplication.

In what follows we shall be also interested in a more general Jensen functional and its behaviour in the context of strong convexity.

## 2 Main Results

### Theorem 1

Let f be a strongly convex function with modulus c defined on an interval $$I, x_{1},x_{2},...,x_{n}\in I$$ and $$p_{1},p_{2},...,p_{n}\in \left( 0,1\right)$$ with $$\sum _{i=1}^{n}p_{i}=1$$. Then
\begin{aligned} \sum _{i=1}^{n}p_{i}f\left( \left( 1-\lambda \mu \right) \bar{x}+\lambda \mu x_{i}\right)\le & {} \left( 1-\lambda \right) f\left( \bar{x}\right) +\lambda \sum _{i=1}^{n}p_{i}f\left( \left( 1-\mu \right) \bar{x}+\mu x_{i}\right) \\&-\, c\lambda \left( 1-\lambda \right) \mu ^{2}\sum _{i=1}^{n}p_{i}(x_{i}-\bar{x} )^{2} \end{aligned}
for all $$\lambda ,\mu \in \left[ 0,1\right]$$.

### Proof

In (1.1) we replace y by $$\left( 1-\mu \right) \bar{x}+\mu x_{i}$$ and x by $$\bar{x}.$$ Hence,
\begin{aligned}&f\left( \left( 1-\lambda \mu \right) \bar{x}+\lambda \mu x_{i}\right) \\&\quad \le \left( 1-\lambda \right) f\left( \bar{x}\right) +\lambda f\left( \left( 1-\mu \right) \bar{x}+\mu x_{i}\right) -c\lambda \left( 1-\lambda \right) \mu ^{2}(x_{i}-\bar{x})^{2} \end{aligned}
for all $$i=1,...,n.$$ We multiply by $$p_{i}$$ and then sum up from 1 to n, which leads us to the conclusion.$$\square$$
The particular case $$\mu =1, \lambda =\frac{1}{2}$$ gives
\begin{aligned} \sum _{i=1}^{n}p_{i}f\left( \frac{\bar{x}+x_{i}}{2}\right) \le \frac{1}{2} \left[ f\left( \bar{x}\right) +\sum _{i=1}^{n}p_{i}f\left( x_{i}\right) \right] -\frac{c}{4}\sum _{i=1}^{n}p_{i}(x_{i}-\bar{x})^{2}, \end{aligned}
which is equivalent to
\begin{aligned} 2\left[ \sum _{i=1}^{n}p_{i}f\left( \frac{\bar{x}+x_{i}}{2}\right) -f\left( \bar{x}\right) \right] +\frac{c}{2}\sum _{i=1}^{n}p_{i}(x_{i}-\bar{x} )^{2}\le {\mathcal {J}}\left( f,\mathbf {p},\mathbf {x}\right) . \end{aligned}
Moreover, from (1.2) for $$x_{i}\rightarrow \frac{\bar{x}+x_{i}}{2}$$ we get a double inequality which refines the Merentes-Nikodem inequality (1.2):

### Corollary 1

Let f be a strongly convex function with modulus c defined on an interval I$$x_{1},x_{2},...,x_{n}\in I$$ and $$p_{1},p_{2},...,p_{n}\in \left( 0,1\right)$$ with $$\sum _{i=1}^{n}p_{i}=1$$. Then
\begin{aligned} c\sum _{i=1}^{n}p_{i}(x_{i}-\bar{x})^{2}\le 2\left[ \sum _{i=1}^{n}p_{i}f \left( \frac{\bar{x}+x_{i}}{2}\right) -f\left( \bar{x}\right) \right] +\frac{ c}{2}\sum _{i=1}^{n}p_{i}(x_{i}-\bar{x})^{2}\le {\mathcal {J}}\left( f,\mathbf {p },\mathbf {x}\right) . \end{aligned}

We introduce in a natural way a more general functional.

### Definition 3

Assume that we have a real valued function f defined on an interval I,  the real numbers $$p_{ij},$$$$i=1,...,k\$$and $$j=1,...,n_{i}$$ such that $$p_{ij}>0,$$$$\sum _{j=1}^{n_{i}}p_{ij}=1$$ for all $$i=1,...,k\$$(we denote $$\mathbf {p}_{i}=$$$$\left( p_{i1},p_{i2},...,p_{in_{i}}\right)$$), $$\mathbf {x}_{i}=$$$$\left( x_{i1},x_{i2},...,x_{in_{i}}\right) \in I^{n_{i}}$$ for all $$i=1,...,k\$$and $$\mathbf {q}=\left( q_{1},q_{2},...,q_{k}\right) ,$$$$q_{i}>0$$ such that $$\sum _{i=1}^{k}q_{i}=1.$$ We define the generalized Jensen functional by
\begin{aligned} {\mathcal {J}}_{k}\left( f,\mathbf {p}_{1},...,\mathbf {p}_{k},\mathbf {q},\mathbf { x}_{1},...,\mathbf {x}_{k}\right):= & {} \sum _{j_{1},...,j_{k}=1}^{n_{1},...,n_{k}}p_{1j_{1}}...p_{kj_{k}}f\left( \sum _{i=1}^{k}q_{i}x_{ij_{i}}\right) \\&-f\left( \sum _{i=1}^{k}q_{i}\sum _{j=1}^{n_{i}}p_{ij}x_{ij}\right) . \end{aligned}

We notice that for $$k=1$$ this definition reduces to Definition 1.

For more results concerning Jensen’s functional the reader is referred to the papers [4, 5].

We state the following lemma about the behaviour of the generalized Jensen functional under the strong convexity condition:

### Lemma 1

Let $$f, \mathbf {p}_{i},\mathbf {x}_{i}\$$and $$\mathbf {q\ }$$be as in Definition 3. If f is strongly convex with modulus c then we have
\begin{aligned} {\mathcal {J}}_{k}\left( f,\mathbf {p}_{1},...,\mathbf {p}_{k},\mathbf {q},\mathbf { x}_{1},...,\mathbf {x}_{k}\right) \ge c\sum _{j_{1},...,j_{k}=1}^{n_{1},...,n_{k}}p_{1j_{1}}...p_{kj_{k}}\left( \sum _{i=1}^{k}q_{i}x_{ij_{i}}-\bar{x}\right) ^{2}, \end{aligned}
where $$\bar{x}=\sum _{i=1}^{k}q_{i}\sum _{j=1}^{n_{i}}p_{ij}x_{ij}.$$

### Proof

One has
\begin{aligned} \sum _{j_{1},...,j_{k}=1}^{n_{1},...,n_{k}}p_{1j_{1}}...p_{kj_{k}}=1 \end{aligned}
and obviously it holds
\begin{aligned} \bar{x}=\sum _{j_{1},...,j_{k}=1}^{n_{1},...,n_{k}}p_{1j_{1}}...p_{kj_{k}} \sum _{i=1}^{k}q_{i}x_{ij_{i}}=\sum _{i=1}^{k}q_{i} \sum _{j=1}^{n_{i}}p_{ij}x_{ij}. \end{aligned}
The conclusion follows from Definition 3 and Proposition 2.$$\square$$

For strongly convex functions we have the following bounds:

### Theorem 2

Let $$f, \mathbf {p}_{i},$$$$\mathbf {x}_{i}\$$and $$\mathbf {q\ }$$be as in Definition 3 and the positive real numbers $$r_{ij},$$$$i=1,...,k\$$and $$j=1,...,n_{i}$$ such that $$\sum _{j=1}^{n_{i}}r_{ij}=1$$    for all $$i=1,...,k.$$ Let
\begin{aligned} \mathbf {r}_{i}= & {} \left( r_{i1},r_{i2},...,r_{in_{i}}\right) \quad \text {\ for all }i=1,...,k, \\ m= & {} \min _{\begin{array}{c} 1\le j_{1}\le n_{1} \\ ... \\ 1\le j_{k}\le n_{k} \end{array}}\left\{ \frac{p_{1j_{1}}...p_{kj_{k}}}{r_{1j_{1}}...r_{kj_{k}}}\right\} ,\\ M= & {} \max _{\begin{array}{c} 1\le j_{1}\le n_{1} \\ ... \\ 1\le j_{k}\le n_{k} \end{array}}\left\{ \frac{p_{1j_{1}}...p_{kj_{k}}}{r_{1j_{1}}...r_{kj_{k}}}\right\} . \end{aligned}
If f is a strongly convex function with modulus c then we have$$\mathrm{:}$$
\begin{aligned}&{\mathcal {J}}_{k}\left( f,\mathbf {p}_{1},...,\mathbf {p}_{k},\mathbf {q}, \mathbf {x}_{1},...,\mathbf {x}_{k}\right) -m{\mathcal {J}}_{k}\left( f,\mathbf {r} _{1},...,\mathbf {r}_{k},\mathbf {q},\mathbf {x}_{1},...,\mathbf {x}_{k}\right) \\&\qquad \ge c\sum _{j_{1},...,j_{k}=1}^{n_{1},...,n_{k}}\left( p_{1j_{1}}...p_{kj_{k}}-mr_{1j_{1}}...r_{kj_{k}}\right) \left( \sum _{i=1}^{k}q_{i}x_{ij_{i}}-\bar{x}\right) ^{2} \\&\qquad \quad +\,mc\left( \sum _{i=1}^{k}q_{i}\sum _{j=1}^{n_{i}}\left( r_{ij}-p_{ij}\right) x_{ij}\right) ^{2} \end{aligned}
and
\begin{aligned}&M{\mathcal {J}}_{k}\left( f,\mathbf {r}_{1},...,\mathbf {r}_{k},\mathbf {q}, \mathbf {x}_{1},...,\mathbf {x}_{k}\right) -{\mathcal {J}}_{k}\left( f,\mathbf {p} _{1},...,\mathbf {p}_{k},\mathbf {q},\mathbf {x}_{1},...,\mathbf {x}_{k}\right) \\&\qquad \ge c\sum _{j_{1},...,j_{k}=1}^{n_{1},...,n_{k}}\left( Mr_{1j_{1}}...r_{kj_{k}}-p_{1j_{1}}...p_{kj_{k}}\right) \left( \sum _{i=1}^{k}q_{i}x_{ij_{i}}-\bar{x}\right) ^{2} \\&\qquad \qquad +\,c\left( \sum _{i=1}^{k}q_{i}\sum _{j=1}^{n_{i}}\left( r_{ij}-p_{ij}\right) x_{ij}\right) ^{2}, \end{aligned}
where $$\bar{x}=\sum _{i=1}^{k}q_{i}\sum _{j=1}^{n_{i}}p_{ij}x_{ij}.$$

### Proof

We prove only the first inequality. Obviously
\begin{aligned}&{\mathcal {J}}_{k}\left( f,\mathbf {p}_{1},...,\mathbf {p}_{k},\mathbf {q}, \mathbf {x}_{1},...,\mathbf {x}_{k}\right) -m{\mathcal {J}}_{k}\left( f,\mathbf {r} _{1},...,\mathbf {r}_{k},\mathbf {q},\mathbf {x}_{1},...,\mathbf {x}_{k}\right) \\&\quad =\sum _{j_{1},...,j_{k}=1}^{n_{1},...,n_{k}}\left( p_{1j_{1}}...p_{kj_{k}}-mr_{1j_{1}}...r_{kj_{k}}\right) f\left( \sum _{i=1}^{k}q_{i}x_{ij_{i}}\right) \\&\qquad +\,mf\left( \sum _{i=1}^{k}q_{i}\sum _{j=1}^{n_{i}}r_{ij}x_{ij}\right) -f\left( \sum _{i=1}^{k}q_{i}\sum _{j=1}^{n_{i}}p_{ij}x_{ij}\right) . \end{aligned}
Since
\begin{aligned} \sum _{i=1}^{k}q_{i}\sum _{j=1}^{n_{i}}p_{ij}x_{ij}= & {} \sum _{j_{1},...,j_{k}=1}^{n_{1},...,n_{k}}\left( p_{1j_{1}}...p_{kj_{k}}-mr_{1j_{1}}...r_{kj_{k}}\right) \sum _{i=1}^{k}q_{i}x_{ij_{i}} \\&+\,m\sum _{i=1}^{k}q_{i}\sum _{j=1}^{n_{i}}r_{ij}x_{ij} \end{aligned}
we obtain the claimed result via Lemma 1.
The proof of the other inequality goes likewise, since
\begin{aligned} \sum _{i=1}^{k}q_{i}\sum _{j=1}^{n_{i}}r_{ij}x_{ij}= & {} \sum _{j_{1},...,j_{k}=1}^{n_{1},...,n_{k}}\frac{ Mr_{1j_{1}}...r_{kj_{k}}-p_{1j_{1}}...p_{kj_{k}}}{M} \sum _{i=1}^{k}q_{i}x_{ij_{i}} \\&+\,\frac{1}{M}\sum _{i=1}^{k}q_{i}\sum _{j=1}^{n_{i}}p_{ij}x_{ij}. \end{aligned}
We omit the details.$$\square$$

The theorem simplifies in some particular cases. Let us give the result for $$\mathbf {p}_{1}=...=\mathbf {p}_{k}=\mathbf {p}$$ and $$\mathbf {x}_{1}=...= \mathbf {x}_{k}=\mathbf {x.}$$

### Corollary 2

Let us consider that we have $$\mathbf {x}=\left( x_{1},x_{2},...,x_{n}\right) \in I^{n}, \mathbf {p}=\left( p_{1},p_{2},...,p_{n}\right) \$$ such that $$p_{i}>0$$, $$\sum _{i=1}^{n}p_{i}=1, \mathbf {q}=\left( q_{1},q_{2},...,q_{k}\right)$$ such that $$q_{i}>0$$, $$\sum _{i=1}^{k}q_{i}=1$$ ($$1\le k\le n$$) and $$\mathbf {r}=\left( r_{1},r_{2},...,r_{n}\right)$$ such that $$r_{i}>0$$, $$\sum _{i=1}^{n}r_{i}=1$$. We denote
\begin{aligned} m=\min _{1\le i_{1},...,i_{k}\le n}\left\{ \frac{p_{i_{1}}...p_{i_{k}}}{ r_{i_{1}}...r_{i_{k}}}\right\} \quad \text { and } \quad M=\max _{1\le i_{1},...,i_{k}\le n}\left\{ \frac{p_{i_{1}}...p_{i_{k}}}{r_{i_{1}}...r_{i_{k}}}\right\} . \end{aligned}
We define
\begin{aligned} {\mathcal {J}}_{k}\left( f,\mathbf {p},\mathbf {q},\mathbf {x}\right) :=\sum _{i_{1},i_{2},...,i_{k}=1}^{n}p_{i_{1}}...p_{i_{k}}f\left( \sum _{j=1}^{k}q_{j}x_{i_{j}}\right) -f\left( \sum _{i=1}^{n}p_{i}x_{i}\right) . \end{aligned}
If f is a strongly convex function with modulus c, then we have:
\begin{aligned}&{\mathcal {J}}_{k}\left( f,\mathbf {p},\mathbf {q},\mathbf {x}\right) -m\mathcal { J}_{k}\left( f,\mathbf {r},\mathbf {q},\mathbf {x}\right) \\&\qquad \ge c\sum _{j_{1},...,j_{k}=1}^{n}\left( p_{j_{1}}...p_{j_{k}}-mr_{j_{1}}...r_{j_{k}}\right) \left( \sum _{i=1}^{k}q_{i}x_{j_{i}}-\bar{x}\right) ^{2} \\&\qquad \quad +\,mc\left( \sum _{j=1}^{n}\left( r_{j}-p_{j}\right) x_{j}\right) ^{2} \end{aligned}
and
\begin{aligned}&M{\mathcal {J}}_{k}\left( f,\mathbf {r},\mathbf {q},\mathbf {x}\right) -\mathcal { J}_{k}\left( f,\mathbf {p},\mathbf {q},\mathbf {x}\right) \\&\quad \ge c\sum _{j_{1},...,j_{k}=1}^{n}\left( Mr_{j_{1}}...r_{j_{k}}-p_{j_{1}}...p_{j_{k}}\right) \left( \sum _{i=1}^{k}q_{i}x_{j_{i}}-\bar{x}\right) ^{2} \\&\qquad +\,c\left( \sum _{j=1}^{n}\left( r_{j}-p_{j}\right) x_{j}\right) ^{2}, \end{aligned}
where $$\bar{x}=\sum _{j=1}^{n}p_{j}x_{j}.$$

For the particular case $$k=1$$ we get$$\mathbf {:}$$

### Corollary 3

For $$i=1,...,n,$$ we consider $$x_{i}\in I, p_{i}>0$$ with $$\sum _{i=1}^{n}p_{i}=1$$ and $$r_{i}>0$$ with $$\sum _{i=1}^{n}r_{i}=1.$$ Let
\begin{aligned} m=\min _{i=1,...,n}\left\{ \frac{p_{i}}{r_{i}}\right\} ,\text { } \quad M=\max _{i=1,...,n}\left\{ \frac{p_{i}}{r_{i}}\right\} . \end{aligned}
If f is a strongly convex function with modulus c, then we have
\begin{aligned}&{\mathcal {J}}\left( f,\mathbf {p},\mathbf {x}\right) -m{\mathcal {J}}\left( f, \mathbf {r},\mathbf {x}\right) \\&\quad \ge c\sum _{j=1}^{n}\left( p_{j}-mr_{j}\right) \left( x_{j}-\bar{x}\right) ^{2}+mc\left( \sum _{j=1}^{n}\left( r_{j}-p_{j}\right) x_{j}\right) ^{2} \end{aligned}
and
\begin{aligned}&M{\mathcal {J}}_{k}\left( f,\mathbf {r},\mathbf {x}\right) -{\mathcal {J}} _{k}\left( f,\mathbf {p},\mathbf {x}\right) \\&\quad \ge c\sum _{j=1}^{n}\left( Mr_{j}-p_{j}\right) \left( x_{j}-\bar{x}\right) ^{2}+c\left( \sum _{j=1}^{n}\left( r_{j}-p_{j}\right) x_{j}\right) ^{2}, \end{aligned}
where $$\bar{x}=\sum _{j=1}^{n}p_{j}x_{j}.$$

### 2.1 Applications to the Gamma Function

The function Gamma is defined via a convergent improper integral as
\begin{aligned} \Gamma \left( t\right) =\int _{0}^{\infty }x^{t-1}e^{-x}\mathrm {d}x,\qquad t>0. \end{aligned}
The following infinite product definition for the Gamma function is due to Weierstrass.
\begin{aligned} \Gamma \left( t\right) =\frac{e^{-\gamma t}}{t}\prod \limits _{n=1}^{\infty }\left( 1+\frac{t}{n}\right) ^{-1}e^{\frac{t}{n}}, \end{aligned}
where $$\gamma =0.577216...$$ is the Euler-Mascheroni constant. This gives
\begin{aligned} \log \Gamma \left( t\right) =-\gamma t-\log t+\sum \limits _{n=1}^{\infty } \left[ \frac{t}{n}-\log \left( 1+\frac{t}{n}\right) \right] . \end{aligned}
(2.1)

### Proposition 3

The function $$f:\left[ 0,\infty \right) \rightarrow \mathbb {R}$$, $$f\left( t\right) =\log \Gamma \left( t^{2}+1\right) +\gamma t^{2}+t\arctan t\$$is strongly convex with modulus 1 on $$\left[ 0,\infty \right) .$$

### Proof

From (2.1) we get
\begin{aligned} \log \Gamma \left( t^{2}+1\right) =-\gamma \left( t^{2}+1\right) -\log \left( t^{2}+1\right) +\sum \limits _{n=1}^{\infty }\left[ \frac{t^{2}+1}{n} -\log \left( 1+\frac{t^{2}+1}{n}\right) \right] . \end{aligned}
(2.2)
We consider the function
\begin{aligned} g\left( t\right) =\log \Gamma \left( t^{2}+1\right) +\gamma t^{2}+t\arctan t-t^{2} \end{aligned}
defined on $$\left[ 0,\infty \right) .$$ One has
\begin{aligned} g^{\prime }\left( t\right) =-\frac{t}{t^{2}+1}+2t\sum \limits _{n=1}^{\infty }\left( \frac{1}{n}-\frac{1}{t^{2}+n+1}\right) +\arctan t-2t \end{aligned}
and
\begin{aligned} g^{\prime \prime }\left( t\right) =\frac{2t^{2}}{\left( t^{2}+1\right) ^{2}} +4t^{2}\sum \limits _{n=1}^{\infty }\frac{1}{\left( t^{2}+n+1\right) ^{2}} +2\sum \limits _{n=1}^{\infty }\left( \frac{1}{n}-\frac{1}{t^{2}+n+1}\right) -2. \end{aligned}
The inequality
\begin{aligned} 2\sum \limits _{n=1}^{\infty }\left( \frac{1}{n}-\frac{1}{t^{2}+n+1}\right) -2\ge 2\sum \limits _{n=1}^{\infty }\left( \frac{1}{n}-\frac{1}{n+1}\right) -2=0 \end{aligned}
yields $$g^{\prime \prime }\left( t\right) \ge 0,$$ therefore g is convex, so f is strongly convex with modulus 1 on $$\left[ 0,\infty \right)$$.

$$\square$$

It is straightforward that:

### Corollary 4

The function $$f:\left[ 0,\infty \right) \rightarrow \mathbb {R}$$, $$f\left( t\right) =\log \Gamma \left( t^{2}+1\right) +t\arctan t\$$is strongly convex with modulus $$1-\gamma$$ on $$\left[ 0,\infty \right)$$.

## Notes

### Acknowledgments

The first author is grateful to Professors S. Abramovich and S.S. Dragomir for the interesting and productive mathematical discussions which have led to some of the questions this work is intended to answer.

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