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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.

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    $\begingroup$ The claim is false, so you cannot prove it. Counterexample: $f: [0,1]\to \Bbb {R}, x \mapsto \sqrt {x} $. $\endgroup$
    – PhoemueX
    Apr 17, 2017 at 13:01
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    $\begingroup$ 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) $\endgroup$
    – Miguel
    Apr 17, 2017 at 13:03
  • $\begingroup$ 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. $\endgroup$
    – Miguel
    Apr 17, 2017 at 13:17

3 Answers 3

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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 $\infty.$

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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?

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

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