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

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One Response to Convexity, Continuity, and Jensen’s Inequality Revision

  1. Pingback: Discontinuous Midpoint Convex Functions | Math by Matt

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