I have two separate conditions:

a) $\lim\limits_{h\to 0}|f(x+h)-f(x-h)|=0$ for every $x \in \Bbb R$ and

b) $\lim_\limits{h\to 0}|f(x+h)+f(x-h)-2f(x)|=0$ for every $x \in \Bbb R$.

My question is do the each of them imply $f$ is continuous? $f(x)$ is said to be continuous at $x_0$ if $\lim_\limits{x\to x_0} f(x) = f(x_0) = c$ For a) it seems correct but I don't know how to prove it. For b), it seems wrong but I can't think of a counterexample.

  • 2
    $\begingroup$ I think both conditions combined imply continuity. $\endgroup$
    – M.Herzkamp
    Sep 26, 2017 at 11:40
  • 1
    $\begingroup$ @M.Herzkamp I agree. Because then $\lim|f(x+h)-f(x)|=\frac{1}{2}\lim|f(x+h)+f(x-h)-2f(x)+f(x+h)-f(x-h)|\le \frac{1}{2}\lim(|f(x+h)+f(x-h)-2f(x)|+|f(x+h)-f(x-h)|)=\frac{1}{2}\lim|f(x+h)+f(x-h)-2f(x)|+\frac{1}{2}\lim|f(x+h)-f(x-h)|=0+0=0$. $\endgroup$
    – velut luna
    Sep 26, 2017 at 12:11

1 Answer 1


Both of them do not imply continuity.

For (a), consider $f(x)=a$ for $x \ne x_0$ and $f(x_0) =b \ne a$.

A simple example may be $f(x)=0$ for $x \ne 0$ and $f(0) =1$.

For (b), consider $f(x)=c$ for $x>x_0$ and $f(x)=d \ne c$ for $x<x_0$ and $f(x_0)=\frac{c+d}{2}$.

A simple example may be $f(x)=1$ for $x>0$ and $f(x)=-1$ for $x<0$ and $f(0)=0$.

Both are counterexamples that satisfy your conditions but are not continuous at $x=x_0$.

  • $\begingroup$ Ah yes thanks I understand perfectly, I just posted another qns about continuity of functions as well hope you may answer! $\endgroup$
    – Homaniac
    Sep 26, 2017 at 8:01

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.