# Uniform continuous functions on bounded sets are Lipschitz?

I'm trying to prove the following:

if $f: E \rightarrow \mathbb{R}$ where $E$ is a bounded subset of $\mathbb{R}$, and $f$ is uniformly continuous then there exists $K$ such that $$|f(x)-f(y)|\leq K|x-y|$$ for all $x,y\in E$

now I have written down this proof which I'm unsure of:

Assume for a contradiction that for all $n \in \mathbb{Z}$ there exists $x_n,y_n \in E$ such that $$|f(x_n)-f(y_n)|\geq n|x_n-y_n|$$ then since $E$ is bounded there exists a convergent sub sequence $\{x_{n_k}\}$ which converges to some number $p$, now since $f$ is uniformly continuous $\{f(x_{n_k})\}$ converges to some number $l$. Thus by the triangle inequality we have that $|f(y_{n_k})-l|\geq ||f(y_{n_k})-f(x_{n_k})|-|f(x_{n_k})-l||\rightarrow \infty$ as $k \rightarrow \infty$ thus $f$ is unbounded which is a contradiction.

• The claim is false, so you cannot prove it. Counterexample: $f: [0,1]\to \Bbb {R}, x \mapsto \sqrt {x}$. Apr 17, 2017 at 13:01
• Note that you didn't require the domain to be closed (important if you take convergent subsequences and want them to converge in the domain) Apr 17, 2017 at 13:03
• Assuming that you take $E$ to be closed (as the counterexample provided by @PhoemueX shows that is ok to do), the first steps of your proof are ok (convergence of the subsequence and its image). The problem is that you take also a subsequence of the $(y_n)$, and for this one, the property you assume to arrive at a contradiction need not hold anymore. In particular, if $|x_{n_k} - y_{n_k}|$ goes to zero faster than $1/n$, the first term in your lower bound won't explode. Apr 17, 2017 at 13:17

Your main error is in the interpretation of the last formula:$$|f(y_{n_k})-l|\geq |f(y_{n_k})-f(x_{n_k})|-|f(x_{n_k})-l|.$$ We have $|f(y_{n_k})-f(x_{n_k})|\geq n_k|y_{n_k}-x_{n_k}|$ but this does not imply that $|f(y_{n_k})-f(x_{n_k})|\to \infty .$ We have $|y_{n_k}-x_{n_k}|\to 0$ so $n_k|y_{n_k}-x_{n_k}|$ need not go to $\infty.$
One way to guess this isn't true: Suppose $|f(y)-f(x)| \le C|y-x|^{1/2}$ on $E.$ Then $f$ is uniformly continuous on $E.$ If your result were true, we would have this result: If $|f(y)-f(x)| \le C_1|y-x|^{1/2}$ on $E,$ then $|f(y)-f(x)| \le C_2|y-x|$ on $E.$ Does that sound right?
Note that $\arccos$, being the very mean function that it is near $1$, does not verify your claim (because of its vertical tangent).