Suppose we have a function defined as follows:

$\alpha(f,x_o)=\limsup\{|f(a)-f(b)|:a,b\in (x_o-\frac{1}{n},x_o+\frac{1}{n})\}$

with $f:R\rightarrow R$ and $x_o \in R$. I need to prove that $\alpha=0$ iff $f$ is continuous at $x_o$.

It seems trivial to show the $\alpha=0 \Rightarrow$ direction, since $|f(a)-f(b)|\geq0$, so if the $lim\,sup$ is $0$ we must have $f(a)=f(b)$ everywhere on the interval.

However, I'm really drawing a blank on the other direction. How can I prove that continuity of $f$ implies that $\alpha=0$? It seems like I should use the $\epsilon-\delta$ definition of continuity, since I'm working with $|f(a)-f(b)|$, but I think the $lim\,sup$ stuff is throwing me off the scent. Any help would be much appreciated!

  • $\begingroup$ Write out the definition of lim sup. It will become very obvious. $\endgroup$ – Don Thousand Nov 21 '18 at 2:59
  • $\begingroup$ Is it simply that as $n\rightarrow \infty$, $(a, b)\rightarrow (x_o, x_o)$? But where do I use the fact that $f$ is continuous? $\endgroup$ – notadoctor Nov 21 '18 at 3:07
  • $\begingroup$ That's not the definition of lim sup... $\endgroup$ – Don Thousand Nov 21 '18 at 3:08
  • 1
    $\begingroup$ @notadoctor think of the $\limsup$ as being two parts: the limit as $n\rightarrow \infty$, and the $\sup$ over all pairs $a,b$ in that interval. $\endgroup$ – user25959 Nov 21 '18 at 3:10
  • $\begingroup$ I'm still not able to convince myself adequately, I'm afraid. I think I'll have to spend some more time getting familiar with continuity. $\endgroup$ – notadoctor Nov 21 '18 at 4:02

Let $\epsilon >0$. If $n$ is sufficiently large then (by continuity) $|f(a)-f(b)| \leq |f(a)-f(x_0)|+|f(x_0)-f(b)|<2\epsilon$ whenever $a,b\in (x_0-\frac 1 n, x_0+\frac 1 n)$. Take sup over all such $a,b$ and then let $n \to \infty$. You will get $\alpha (f,x_0) \leq 2\epsilon$. Since this is true for all $\epsilon >0$ we must have $\alpha (f,x_0)=0$.

  • $\begingroup$ Thank you! This helps a lot. However, I am unsure where your inequality $\alpha (f,x_0) \leq 2\epsilon$ comes from. If the sup is $\frac{2}{n}$, doesn't $n \rightarrow \infty$ just give me $0$ right away from that? $\endgroup$ – notadoctor Nov 21 '18 at 6:44
  • $\begingroup$ $|f(x_0)-f(a)| <\epsilon$ for $|a-x_0| <\delta$. If $\frac 1 n <\delta$ we get $|f(x_0)-f(a)| <\epsilon$. Similarly, $|f(x_0)-f(b)| <\epsilon$. $\endgroup$ – Kabo Murphy Nov 21 '18 at 7:42

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