Another Proof that Midpoint-Convex and Continuous Implies Convex

A few months back, I wrote a post about some elementary results concerning convex functions on the interval. One of these results was that a real-valued continuous function f:I\rightarrow\mathbb{R} defined on an interval I\subset\mathbb{R} is convex if and only if it is midpoint convex. I want to present another proof of this result, one which comes from Constantin Niculescu’s and Lars-Erik Persson’s text Convex Functions and Their Applications.

Proposition 1. Suppose f: I \rightarrow \mathbb{R} is continuous. Then f is convex if and only if f is midpoint convex.

Proof. Necessity is obvious. For sufficiency, suppose that f is not convex; that is, there exist a,b\in I and t\in (0,1) such that

\displaystyle f(ta+(1-t)b)>tf(a)+(1-t)f(b)

Define a function \varphi: I \rightarrow\mathbb{R} by

\displaystyle\varphi(x):=f(x)-\dfrac{f(b)-f(a)}{b-a}(x-a)-f(a),\indent\forall x\in[a,b]

Note that \varphi is the sum of f and an affine function. I claim that \gamma:=\sup_{x\in[a,b]}\varphi(x)>0. Indeed,

\begin{array}{lcl}\displaystyle\varphi(ta+(1-t)b)&=&\displaystyle f(ta+(1-t)b)-\dfrac{f(b)-f(a)}{b-a}\left(ta+(1-t)b-a\right)-f(a)\\&>&\displaystyle tf(a)+(1-t)f(b)-\dfrac{f(b)-f(a)}{b-a}\left(ta+(1-t)b-a\right)-f(a)\\&=&\displaystyle tf(a)+(1-t)f(b)-(1-t)[f(b)-f(a)]-f(a)\\&=&\displaystyle 0\end{array}

Observe that \varphi(a)=\varphi(b)=0 and \varphi is midpoint convex, since

\begin{array}{lcl}\displaystyle\varphi\left(\frac{x+y}{2}\right)&=&\displaystyle f\left(\frac{x+y}{2}\right)-\dfrac{f(b)-f(a)}{b-a}\left(\dfrac{x+y}{2}-a\right)-f(a)\\&\leq&\displaystyle\dfrac{f(x)+f(y)}{2}-\dfrac{f(b)-f(a)}{b-a}\left(\dfrac{x+y}{2}-2\dfrac{a}{2}\right)-2\dfrac{f(a)}{2}\\&=&\displaystyle\dfrac{\varphi(x)+\varphi(y)}{2}\end{array}

Set c:=\inf\left\{x\in[a,b]:\varphi(x)=\gamma\right\}. Since \varphi is continuous, \varphi(c)=\gamma>0, where the positivity implies that c\in(a,b). For any h>0 such that c\pm h\in (a,b), we have that \varphi(c-h)<\varphi(c) and \varphi(c+h)\leq\varphi(c), so by midpoint convexity of \varphi,

\displaystyle\varphi(c)\leq\dfrac{\varphi(c-h)+\varphi(c+h)}{2}<\dfrac{\varphi(c)+\varphi(c)}{2}=\varphi(c),

which is a contradiction. \Box

I claim that any function f: I \rightarrow \mathbb{R} satisfying

\displaystyle f(x+h)-f(x-h)-2f(x)\geq 0,

for all x\in I and h>0 such that x\pm h\in I, is midpoint convex. Indeed, Fix a,b \in I and set h:=\frac{b-a}{2}. Then

\displaystyle a=\dfrac{a+b}{2}-h,\indent b=\dfrac{a+b}{2}+h,

so that

\begin{array}{lcl}\displaystyle f(a)+f(b)-2f\left(\dfrac{a+b}{2}\right)&=&\displaystyle f\left(\dfrac{a+b}{2}-h\right)+f\left(\dfrac{a+b}{2}+h\right)-2f\left(\frac{a+b}{2}\right)\\&\geq&\displaystyle 0\end{array}

Thus, we obtain a corollary.

Corollary 2. A continuous function f: I \rightarrow\mathbb{R} is convex if and only if

\displaystyle f(x+h)+f(x-h)-2f(x)\geq 0,

for all x\in I and h>0 such that x\pm h\in I.

In fact, for Proposition 1, we can relax the sufficiency condition to

\displaystyle f(\alpha x+(1-\alpha y))\leq\alpha x+(1-\alpha)y,\indent\forall x,y\in I

for some fixed parameter \alpha\in (0,1). Indeed, it is easy to verify that \varphi satisfies the same condition as f. I claim that there exist x,y \in [a,b]\setminus\left\{c\right\}, where without loss of generality, x < c, such that

\displaystyle c=\alpha x+(1-\alpha)y

Indeed,

\begin{array}{lcl}\displaystyle\dfrac{c-(1-\alpha)x}{\alpha}-a=\dfrac{c-(1-\alpha)x-\alpha a}{\alpha}=\dfrac{c-x+\alpha(x-a)}{\alpha}&=&\displaystyle\alpha^{-1}(c-x)+(x-a)>0\\&\Longleftrightarrow&\displaystyle\dfrac{\alpha^{-1}c-a}{\alpha^{-1}-1}>x\end{array}

and

\begin{array}{lcl}\displaystyle b-\dfrac{c-(1-\alpha)x}{\alpha}=\dfrac{\alpha b-c+(1-\alpha)x}{\alpha}&=&\displaystyle(b-x)+\alpha^{-1}(x-c)>0\\&\Longleftrightarrow&\displaystyle -\dfrac{b-\alpha^{-1}c}{\alpha^{-1}-1}<x\end{array}

So we can make an appropriate choice of x \in (-\frac{b-\alpha^{-1}c}{\alpha^{-1}-1},\frac{\alpha^{-1}c-a}{\alpha^{-1}-1}) satisfying x<c, which uniquely determines y. Then \varphi(x)<\varphi(c) and \varphi(y)\leq\varphi(c) so that

\displaystyle\varphi(c)\leq\alpha\varphi(x)+(1-\alpha)\varphi(y)<\alpha\varphi(c)+(1-\alpha)\varphi(c)=\varphi(c),

which is a contradiction.

We also can relax the hypothesis that f is continuous to f is bounded from above on compact subintervals of I.

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One Response to Another Proof that Midpoint-Convex and Continuous Implies Convex

  1. Pingback: Weighted Power Means | Math by Matt

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