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

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