## Convexity, Continuity, and Jensen’s Inequality Revision

A comment to my post Convexity, Continuity, and Jensen’s Inequality asks how we know that $x_{n}\rightarrow x_{0}$, as $n\rightarrow\infty$, in the proof of Lemma 3. It’s actually false that the sequence converges to $x_{0}$. What I believe to be a correct proof follows below. I will replace the false proof in the original post soon.

Lemma 3. If $f$ is midpoint convex and has a point of discontinuity $x_{0}$, then $\limsup_{x\rightarrow x_{0}}f(x)=+\infty$.

Proof. $f$ is discontinuous at $x_{0}$, so there exists a $\varepsilon>0$ such that

$\displaystyle\forall\delta>0,\ \exists x\in B_{\delta}(x_{0})\text{ s.t. }\left|f(x)-f(x_{0})\right|>\varepsilon$

Let $\delta>0$ be given. For such an $x$, define $x':=2x_{0}-x$ (i.e. the reflection of $x$ by $x_{0}$). By midpoint convexity,

$\displaystyle\dfrac{f(x)+f(x')}{2}\geq f\left(\dfrac{x+x'}{2}\right)=f(x_{0})$

which implies that $\left[f(x)-f(x_{0})\right]+\left[f(x')-f(x_{0})\right]\geq0$. I claim that there is $y\in\left\{x,x'\right\}$ such that $f(y)-f(x_{0})>\varepsilon$. Observe that if $f(x)-f(x_{0})<-\varepsilon$, then the estimate obtained above implies that $f(x')-f(x_{0})>\varepsilon$. But

$\displaystyle\left|x'-x_{0}\right|=\left|(2x_{0}-x)-x_{0}\right|=\left|x_{0}-x\right|<\delta$

which proves the claim. Hence, there exists a point $x_{1}\in B_{\delta}(x_{0})$ such that $f(x_{1})-f(x_{0})>\varepsilon$.

Suppose that we have constructed $x_{1},\ldots,x_{n}$ as desired. Set $x_{n+1}:=2x_{n}-x_{0}$. Note that $\left|x_{n+1}-x_{0}\right|=2\left|x_{n}-x_{0}\right|$. By midpoint convexity,

$\displaystyle\dfrac{f(x_{n+1})+f(x_{0})}{2}\geq f\left(\dfrac{x_{n+1}+x_{0}}{2}\right)=f(x_{n})$

which implies that

$\begin{array}{lcl}\displaystyle f(x_{n+1})-f(x_{0})&\geq&\left[f(x_{n+1})-f(x_{n})\right]+\left[f(x_{n})-f(x_{0})\right]\\[2 em]\displaystyle&\geq&2\left[f(x_{n})-f(x_{0})\right]\\[2 em]&>&\displaystyle2^{n}\varepsilon\end{array}$

By induction, we obtain a sequence $(x_{n})_{n=1}^{\infty}$ with the property that $f(x_{n})-f(x_{0})>2^{n-1}\varepsilon$ and $\left|x_{n}-x_{0}\right|<2^{n}\delta$.

I claim that every open neighborhood of $x_{0}$ contains a point $z$ such that $f(z)>M$, for any $M>0$. Indeed, let a neighborhood $B_{\delta_{0}}(x_{0})$ and quantity $M>0$ be given. Let $i$ be a sufficiently large integer so that $2^{i-1}\varepsilon>M$. Run the argument above with $\delta=2^{-i-1}\delta_{0}$ to obtain a sequence $(x_{n})_{n=1}^{\infty}$. Observe that

$\displaystyle\left|x_{i}-x_{0}\right|<2^{i}\delta=2^{i}\left(2^{-i-1}\delta_{0}\right)=\dfrac{\delta_{0}}{2}$

and

$\displaystyle f(x_{i})-f(x_{0})>2^{i-1}\varepsilon>M$

$\Box$