# Prove or disprove that $f(x) = \sin x/x$ is uniformly continuous over the interval $(0,1)$?

I am thinking of $\lim_{x \to 0} \ {f(x)} = 1$ but I am confused as $x=0$ is not in the domain and also I want to write an $\epsilon$-$\delta$ proof. So any help is much appreciated!

Thanks!

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Here is a non $\epsilon-\delta$ proof: Extend $f$ to a function $g:[0,1]\to \mathbb R$ by setting $g(x)=f(x)$ for all $x\in (0,1)$ and $g(0)=1$, $g(1)=\sin (1)$. Now, $g$ is continuous on its domain which is a closed interval and thus is uniformly continuous. It is very easy to show that the restriction of a uniformly continuous function is itself uniformly continuous and thus $f$ is uniformly continuous.

Giving a direct $\epsilon -\delta$ proof will be quite technically messy. Is there any particular reason you wish to obtain such a proof.

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Because that is how I usually do these proofs and for instance, I don't understand your proof completely because I don't know why the restriction of a uniformly continuous function is uniformly continuous. Also can you explain why g is continuous on the domain? – UH1 Dec 16 '12 at 5:38
g is continuous at every 0<x<1 since f is. g is continuous at 0 since its defines there as 1 which is the limit of sinx/x as x goes to 0. g is continuous at 1 for the analog reason. Now, informally, g uniformly continuous means for a given epsilon a delta can be chosen to work uniformly for all points in the domain of g. Clearly then the same choice will work for all points in a subset of the domain, thus any restriction is also uniformly continuous. You should prove this rigorously. – Ittay Weiss Dec 16 '12 at 5:42
you usually do proofs using epsilon delta arguments only in the beginning and for very simple functions. Very soon this becomes too daunting. The whole point of developing a rich theory is that you learn new, more powerful, tools and techniques to proof things. Of course, any proof in analysis can be reduced to a (potentially horribly complicated) argument with epsilon delta. But it helps no-one to see it. – Ittay Weiss Dec 16 '12 at 5:43
i see, yeah i was really struggling to come up with a neat $\ epsilon-\delta$ proof but this makes some sense now. Can I ask you one more thing? Can I conclude f(x)= |x| is uniformly continuous over [-1,1] because its continuous over all R so it must continuous over [-1,1] and since [-1,1] is compact, its uniformly continuous. THanks for the help, i really appreciate it! – UH1 Dec 16 '12 at 5:49
Yes and you're welcome :) – Ittay Weiss Dec 16 '12 at 5:51

Ittay Weiss described the right way to solve this problem. If you really want an $\varepsilon$-$\delta$ solution, you can do the following.

Given $\varepsilon > 0$, let $\delta = \varepsilon$. Now for $x,y\in (0,1)$ with $|x-y| \leq \delta$, we have $$|f(x) - f(y)| = |(x-y) f'(\xi)| = |x-y| \cdot |f'(\xi)| \leq \delta |f'(\xi)|,$$ where $\xi \in (0,1)$. Note that $$f'(x) = \frac{x\cos x - \sin x}{x^2}.$$ This function takes negative values on $(0,1)$ since $\tan x > x$ (for $x\in (0,1)$). Also $$\frac{x\cos x - \sin x}{x^2} \geq \frac{x\cos x - x}{x^2} = \frac{\cos x - 1}{x} \geq -1.$$ That is, $f'(\xi) \in [-1,0)$. Therefore, $|f(x) - f(y)| \leq \delta = \varepsilon$.

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this is nice! appreciate it! – UH1 Dec 16 '12 at 6:11
Nice solution! Far less messy than I had anticipated. – Ittay Weiss Dec 16 '12 at 6:14
Does this proof hold good if the interval is entire $R$ – Aman Mittal Sep 19 '13 at 13:24
@AmanMittal: Yes,it does. The proof goes through with small modifications. It's no longer true that $f'(\xi) \in (-1,0)$; but it is true that $f'(\xi) \in (-1,1)$. – Yury Sep 19 '13 at 15:38
@yury Thank you !! – Aman Mittal Sep 19 '13 at 18:45