If $f$ is symmetrically continuous, then $f$ is not necessarily continuous

Question

I've already proved that continuous functions are symmetrically continuous.

Now I want to show that the converse does not hold.

Define $f:X\rightarrow \mathbb{R}, f$ is symmetrically continuous at $x$

So, $\lim_{h\to 0} [f(x+h)-f(x-h)]=0$

I can't come up with a function that's symmetrically continuous but not continuous. Any ideas or hints? The Cantor set comes to mind, but is the Cantor set symmetrically continuous?

@Tobsn Mark's answer is the Wiki example, and there is a little bit of controversy around it. — Shubham Johri, Nov 10 '20 at 14:53
Basically, symmetric continuity doesn't put constraints on what the function value is at that point at which you are checking symmetric continuity. So even if $\lim_{h \to 0} [f(x+h) - f(x-h)] = 0$, this doesn't say anything about $f(x)$ itself, while continuity certainly imposes constraints on $f(x)$ as well. That's basically what leads to the easy counterexample above and below. — Sarvesh Ravichandran Iyer, Nov 10 '20 at 14:56
Various counterexamples here: Does $\lim_{h\rightarrow 0}\ [f(x+h)-f(x-h)]=0$ imply that $f$ is continuous? — Martin R, Nov 10 '20 at 14:58

score 1 · Answer 1 · answered Nov 10 '20 at 14:50

1

Choose your favourite continuous function $g:\Bbb R\to\Bbb R$ and call $f(t)=\begin{cases}g(x)+1&\text{if }t=x\\ g(t)&\text{if }t\ne x\end{cases}$

answered Nov 10 '20 at 14:50

If $f$ is symmetrically continuous, then $f$ is not necessarily continuous

1 Answers1