Question

Any continuous function f defined on a closed interval I=[a,b] is uniformly continuous.Prove

Answer 1

If $f$ is not continuous then there exist $\epsilon >0$ such that for any $\delta >0,$ must exist points $x_k, y_k$ such that $|x_k-y_k|<\frac{\delta}{2^k}$ but $|f(x_k)-f(y_k)|>\epsilon$ for each $k>0.$ Now the sequence is bounded by the interval, hence by Bolzano-Weirstrass a subsequence of ${x_k}$ which converges. For simplicity let us call this sequence also as ${x_k}.$ Now let $x_k\rightarrow x.$ Then we claim $y_k\rightarrow x.$ Let $\epsilon'>0$ be given. Then there exists $k_1, N>0,$ such that $\frac{\delta}{2^{k_1}}<\frac{\epsilon^{'}}{2}$ and $|x_{k}-x|<\epsilon' \ \forall \ k\geq N.$ We take $k=max\{k_{1},N\}.$

By triangle inequality $|y_k-x|\leq |y_k-x_k|+|x_k-x|$$<\frac{\delta}{2^{k_1}}+\frac{\epsilon^{'}}{2} \leq \frac{\delta}{2^k}+\frac{\epsilon'}{2}< \frac{\epsilon'}{2}+\frac{\epsilon'}{2}= \epsilon'.$

Now $f$ being continuous, $f(y_k)\rightarrow f(x).$ Also, $f(x_k)\rightarrow f(x).$ So given $\epsilon>0$ there exist $N_1,N_2$ such that for all $k>N_1,N_2$ $,|f(x_k)-f(x)|<\frac{\epsilon}{2}, |f(y_k)-f(x)|<\frac{\epsilon}{2}.$ Hence $|f(x_k)-f(y_k)|\leq |f(x_k)-f(x)|+|f(y_k)-f(x)|$ $\leq \epsilon$ contradicting $|f(x_k)-f(y_k)|>\epsilon .$