Uniform continuous functions on bounded sets are Lipschitz?

Question

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.

Answer 1

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 $\infty.$

Answer 2

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?

Answer 3

Note that $\arccos$, being the very mean function that it is near $ 1 $, does not verify your claim (because of its vertical tangent).