## Midpoint Convexity and Lower Semicontinuity

I’ve written a couple posts now on continuity as a sufficient condition for midpoint convexity to imply that a function $f: I \rightarrow\mathbb{R}$, where $I\subset\mathbb{R}$ is an interval, is convex. I now want to weaken the hypothesis of continuity to that of lower semicontinuity.

Recall that an extended real-valued function $f$ defined on a topological space $E$ is said to be lower semicontinuous (l.s.c.) if for every $\lambda\in\mathbb{R}$, the level set $\left\{x\in E: f(x)\leq\lambda\right\}$ is closed in $E$. If $E$ is a second-countable topological space, then a function $f$ is lower semicontinuous if and only if

$x_{n}\rightarrow x\Longrightarrow f(x)\leq\liminf_{n\rightarrow\infty}f(x_{n})$

Lemma. If $[a,b]\subset\mathbb{R}$ is a compact interval and $f: [a,b]\rightarrow\mathbb{R}$ is a l.s.c. function, then $f$ attains its infimum at a point $c\in[a,b]$.

Proof.  Set $\alpha:=\inf_{x\in[a,b]}f(x)$, and let $(x_{n})_{n=1}^{\infty}$ be a sequence in $[a,b]$ such that $f(x_{n})\rightarrow\alpha$. Since $[a,b]$ is compact, the sequence $(x_{n})$ has a convergent subsequence $(x_{n_{k}})_{k=1}^{\infty}$ with limit $c\in[a,b]$. By l.s.c.,

$\displaystyle f(c)\leq\liminf_{k\rightarrow\infty}f(x_{n_{k}})=\lim_{n\rightarrow\infty}f(x_{n})=\alpha$,

which implies that $f(c)=\alpha$. $\Box$

Proposition. A lower-semicontinuous function $f: I \rightarrow\mathbb{R}$ is convex if and only if $f$ is midpoint convex.

Proof. Necessity is obvious, so we concern ourselves for sufficiency. Let $[a,b]\subset I$ be a compact interval. By the preceding lemma, $f$ attains its infimum at a point $c\in [a,b]$. We will show that $f$ is continuous on $[c,b]$ and on $[a,c]$.

We have that $f(c)\leq f(b)$ and therefore by midpoint convexity, $f(\frac{b+c}{2})\leq f(b)$. By definition of $c$, we have the inequality

$\displaystyle f(c)\leq f\left(\frac{b+c}{2}\right)\leq f(b)$

Suppose that we have shown that

$\displaystyle f(c)\leq f\left(c+\frac{b-c}{2^{n}}\right)\leq\cdots\leq f\left(c+(2^{n}-1)\frac{b-c}{2^{n}}\right)\leq f(b)$

for some $n\in\mathbb{Z}^{\geq 1}$. Then by midpoint convexity, we have that

$\begin{array}{lcl}\displaystyle f\left(c+\frac{b-c}{2^{n+1}}\right)=\displaystyle f\left(\dfrac{c+c+\frac{b-c}{2^{n}}}{2}\right)&\leq&\displaystyle\dfrac{f\left(c\right)}{2}+\dfrac{f\left(c+\frac{b-c}{2^{n}}\right)}{2}\end{array}$

By definition of $c$, we have that

$\displaystyle f(c)\leq f\left(c+\frac{b-c}{2^{n+1}}\right)\leq f\left(c+2\frac{b-c}{2^{n+1}}\right)$

Suppose we have shown that, for some even $j\in\left\{0,\cdots,2^{n}\right\}$, we have the inequality

$\displaystyle f\left(c+j\frac{b-c}{2^{n+1}}\right)\leq f\left(c+(j+1)\frac{b-c}{2^{n+1}}\right)\leq f\left(c+(j+2)\frac{b-c}{2^{n+1}}\right)$

Then by midpoint convexity,

$\begin{array}{lcl}\displaystyle f\left(c+(j+2)\frac{b-c}{2^{n+1}}\right)&\leq&\displaystyle\dfrac{f\left(c+(j+1)\frac{b-c}{2^{n+1}}\right)}{2}+\dfrac{f\left(c+(j+3)\frac{b-c}{2^{n+1}}\right)}{2}\\&\leq&\dfrac{f\left(c+(j+2)\frac{b-c}{2^{n+1}}\right)}{2}+\dfrac{f\left(c+(j+3)\frac{b-c}{2^{n+1}}\right)}{2}\end{array},$

which implies that $f\left(c+(j+2)\frac{b-c}{2^{n+1}}\right)\leq f\left(c+(j+3)\frac{b-c}{2^{n+1}}\right)$; and

$\displaystyle f\left(c+(j+3)\frac{b-c}{2^{n+1}}\right)\leq\dfrac{f\left(c+(j+2)\frac{b-c}{2^{n+1}}\right)}{2}+\dfrac{f\left(c+(j+4)\frac{b-c}{2^{n+1}}\right)}{2}\leq f\left(c+(j+4)\frac{b-c}{2^{n+1}}\right)$

Define the set $\mathbb{D}$ of “dyadic points” in $[c,b]$ by $\mathbb{D}:=\left\{\frac{j(b-c)}{2^{n}}: n\in\mathbb{Z}^{\geq 0},1\leq j\leq2^{n}\right\}$. By induction, we see that $x,y\in\mathbb{D}$ and $x\leq y$ imply that $f(x)\leq f(y)$. Hence,

$\displaystyle\tilde{f}(y):=\lim_{{x\in\mathbb{D}}\atop{x\rightarrow y^{-}}}f(x)=\sup_{{x\in\mathbb{D}}\atop{x\rightarrow y^{-}}}f(x)$

exists by the monotone convergence theorem. Observe that $\tilde{f}$ defines a monotone function on $[c,b]$ and $\lim_{x\rightarrow y^{-}}\tilde{f}(x)$. If we can show that $\tilde{f}=f$ and is continuous, then it follows from previous results (see my earlier posts on midpoint convexity and continuity) that $\tilde{f}$ is convex. Choose sequences $(x_{n})_{n=1}^{\infty}$ and $(z_{n})_{n=1}^{\infty}$ in $\mathbb{D}$ such that

$x_{n}\rightarrow y^{-}$, $z_{n}\rightarrow y^{+}$, and $\dfrac{x_{n}+z_{n}}{2}\geq y$

Set $\tilde{f}(y^{+}):=\lim_{x\rightarrow y^{+}}\tilde{f}(x)=\lim_{{x\rightarrow y^{+}}\atop{x\in\mathbb{D}}}f(x)$. Since $\frac{x_{n}+z_{n}}{2}\in\mathbb{D}$, we have that

$\begin{array}{lcl}\displaystyle\tilde{f}(y^{+})=\lim_{n\rightarrow\infty}\tilde{f}\left(\dfrac{x_{n}+z_{n}}{2}\right)=\lim_{n\rightarrow\infty}f\left(\dfrac{x_{n}+z_{n}}{2}\right)&\leq&\displaystyle\lim_{n\rightarrow\infty}\dfrac{f(x_{n})+f(z_{n})}{2}\\[.7 em]&=&\displaystyle\lim_{n\rightarrow\infty}\dfrac{\tilde{f}(x_{n})+\tilde{f}(z_{n})}{2}\\[.7 em]&=&\dfrac{\tilde{f}(y)+\tilde{f}(y^{+})}{2}\end{array}$,

which implies that $\tilde{f}(y^{+})\leq\tilde{f}(y)$ and therefore $\tilde{f}$ is continuous, being monotone. To see that $\tilde{f}=f$ on $[c,b]$, observe that by lower semicontinuity, $f(y)\leq\tilde{f}(y)$. For the reverse inequality, let $(x_{n})_{n=1}^{\infty}$ be a sequence in $\mathbb{D}$ such that $x_{n}\uparrow y$, and define $z_{n}$ so that $x_{n}=\frac{y+z_{n}}{2}$. Then by midpoint convexity,

$\displaystyle f(x_{n})\leq\dfrac{f(z_{n})+f(y)}{2}\leq\dfrac{\tilde{f}(z_{n})+f(y)}{2}$

Letting $n\rightarrow\infty$ and applying the continuity of $\tilde{f}$, we conclude that

$\displaystyle\tilde{f}(y)\leq\dfrac{\tilde{f}(y)+f(y)}{2}\Longleftrightarrow\tilde{f}(y)\leq f(y)$

Applying the same argument to $f$ on the interval $[a,c]$, we arrive at the desired conclusion. $\Box$