# Proving that continuity and the lim sup of a given set being 0 are equivalent

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!

• Write out the definition of lim sup. It will become very obvious. – Don Thousand Nov 21 '18 at 2:59
• 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? – notadoctor Nov 21 '18 at 3:07
• That's not the definition of lim sup... – Don Thousand Nov 21 '18 at 3:08
• @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. – user25959 Nov 21 '18 at 3:10
• 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. – 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 $$\alpha (f,x_0)=0$$.
• 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? – notadoctor Nov 21 '18 at 6:44
• $|f(x_0)-f(a)| <\epsilon$ for $|a-x_0| <\delta$. If $\frac 1 n <\delta$ we get $|f(x_0)-f(a)| <\epsilon$. Similarly, $|f(x_0)-f(b)| <\epsilon$. – Kabo Murphy Nov 21 '18 at 7:42