$|f(a+h)-2f(a)+f(a-h')|<\varepsilon \Rightarrow f$ is continuous at a

Asked Dec 19 '18 at 16:22

Active Dec 19 '18 at 16:30

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I need to prove or disprove that if for every $\varepsilon>0$ there exists $\delta>0$ such that for every $h,h'\in(-\delta,\delta): |f(a+h)-2f(a)+f(a-h')|<\varepsilon$, then f is continuous at a (f is defined at a).

If the statement is true then I need to prove it, otherwise a counterexample would be enough. I could not find one (thus I believe the statement is true), yet I cannot prove it.

Thank you very much and have a good day!!

edited Dec 19 '18 at 16:30

asked Dec 19 '18 at 16:22

Amit Zach

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Your condition holds in particular for $h'=0$, therefore ... – Martin R Dec 19 '18 at 16:29
@MartinR Oh! Can't believe I didn't see it, thank you very much! – Amit Zach Dec 19 '18 at 16:35
Marginally related: Prove the set of points where $f$ is differentiable is dense – Dave L. Renfro Dec 19 '18 at 16:35
Note that the conclusion is wrong if you the estimate holds only for $h=h'$. – Martin R Dec 19 '18 at 16:37
@MartinR Correct. I thought of $f(x)=\frac{\cos(x)}{x}$ when $f(0)$ is defined to be $0$. – Amit Zach Dec 19 '18 at 17:30

$|f(a+h)-2f(a)+f(a-h')|<\varepsilon \Rightarrow f$ is continuous at a

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