0
$\begingroup$

I'm reading a proof of the following statement:

Let $f: (a,b) \to \mathbb{R}$ be a midpoint convex (i.e. $f(1/2(x+y)) \leq 1/2f(x) + 1/2f(y)$) function that is bounded. Then $f$ is continuous.

Here is a proof that I found here: Proving continuity of $f$

To prove that a bounded midpoint convex is continuous, argue by contradiction. Supose $f$ is discontinuous at $x_0\in(a,b)$. Without loss of generality we may assume $x_0=0$, $f(x_0)=0$.

First step. There exists a sequence $\{x_n\}\subset(a,b)$, such that $\lim_{n\to\infty}x_n=0$ and $\lim_{n\to\infty}f(x_n)=m\ne0$. We may assume that $m>0$.

Second step. The sequence $\{2\,x_n\}$ also converges to $0$ and $$ f(x_n)=f\Bigl(\frac{0+2\,x_n}2\Bigr)\le\frac{f(0)+f(2\,x_n)}2\implies f(2\,x_n)\ge2\,f(x_n)\implies\liminf f(2\,x_n)\ge2\,m. $$ Iteration shows that $$ \liminf f(2^k\,x_n)\ge2^k\,m, $$ which is impossible since $f$ is bounded.

Question: Why can we assume $m > 0$?

$\endgroup$
11
  • 1
    $\begingroup$ Consider the function $f+2|m|$ which is also mid point convex. $\endgroup$
    – copper.hat
    Apr 21 '20 at 21:14
  • $\begingroup$ Thanks @copper.hat $\endgroup$
    – user745578
    Apr 21 '20 at 21:16
  • $\begingroup$ but that function wont satisfy $f(x_0)=0$ $\endgroup$
    – Yorch
    Apr 21 '20 at 21:21
  • $\begingroup$ @JorgeFernándezHidalgo Good point. I didn't think of that. $\endgroup$
    – user745578
    Apr 21 '20 at 21:25
  • $\begingroup$ I found a simple proof of the theorem in question if you would like $\endgroup$
    – Yorch
    Apr 21 '20 at 21:28
1
$\begingroup$

We assume $a<0<b$, $f$ is discontinuous at $0$ and $f(0)=0$.

If not we can solve it by shifting the function horizontal and vertically and stretching horizontally.

Notice that since $f$ is discontinuous at $0$ there is an $\epsilon>0$ such that we can find $x$ as small as possible such that $|f(x)|>\epsilon$

Notice that if $f(x) < 0$ we have that $f(-x)>0$ because $f(x)+f(-x)\ge 0$.

So without loss of generality $f(x)>0$ and now we have $f(x)\leq f(2x)/2\leq f(4x)/4\dots$

This tells us $f(2^nx)\geq 2^n\epsilon$.

Of course we need to make $x$ small so that $2^nx$ is inside the interval

$\endgroup$
20
  • 1
    $\begingroup$ Why didn't I think of that! Thanks! $\endgroup$
    – user745578
    Apr 21 '20 at 21:07
  • $\begingroup$ I was wrong, but we can multiply it by $-1$. $\endgroup$
    – Yorch
    Apr 21 '20 at 21:10
  • $\begingroup$ But multiplying by -1 does not preserve midpoint convexity? $\endgroup$
    – user745578
    Apr 21 '20 at 21:10
  • $\begingroup$ I also don't know why they assume that the limit of $f(x_n)$ exists $\endgroup$
    – Yorch
    Apr 21 '20 at 21:10
  • $\begingroup$ They can assume that because $f$ is bounded, so $f(x_n)$ has a convergent subsequence. $\endgroup$
    – user745578
    Apr 21 '20 at 21:11
0
$\begingroup$

Suppose $$\lim_{n\to\infty}f(x_n)=m<0.$$ Let $y_n=-x_n$, then $$0=f(0)=f\left(\frac{x_n+y_n}{2}\right)\leq\frac{f(x_n)+f(y_n)}{2},$$ so $f(y_n)\geq-f(x_n)$, then $$\limsup_{n\to\infty}f(y_n)\geq\limsup_{n\to\infty}(-f(x_n)) =\lim_{n\to\infty}(-f(x_n))=-m>0.$$ So there exists subsequences $\{f(y_{_{n_k}})\}$ such that $$\lim_{k\to\infty}f(y_{_{n_k}})=\limsup_{n\to\infty}f(y_n)\geq-m>0.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy