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Suppose $\{f_n\}$ is an equicontinuous sequence of functions on a compact set $K$, and $\{f\}$ converges pointwise on $K$. Prove that $\{f_n\}$ converges uniformly on $K$.

Proof: For any $\epsilon>0$ $\exists \delta>0$ such that $\forall x,y\in K$ with $d(x,y)<\delta$ and $\forall n\in \mathbb{N}$ implies that $|f_n(x)-f_n(y)|<\epsilon/3$.

Since $K$ is compact then exists finitely many points $x_1, x_2, \cdots, x_p$ from $K$ such that $K\subset U_{\delta}(x_1)\cup \cdots \cup U_{\delta}(x_p).$

Also $\forall x\in K$ and $\forall \epsilon>0$ $\exists N=N(\epsilon,x)$ such that $\forall n,m\ge N$ we have $|f_n(x)-f_m(x)|<\epsilon/3.$

We have to prove that $\{f_n\}$ converges uniformly.

Let $\epsilon>0$ be given then $\exists N_k=N(\epsilon, x_k)$ such that for any $n,m\ge N_k$ we have $|f_n(x_k)-f_m(x_k)|<\epsilon/3$ for $k\in \{1,2,\dots,p\}.$

For any $x\in K$ $\Rightarrow$ $x\in U_{\delta}(x_s)$ for some $s\in \{1,2,\dots,p\}$ $\Rightarrow$ $d(x,x_s)<\delta$ for some $s\in \{1,2,\dots,p\}$ and for any $n,m\in \mathbb{N}$ we have $|f_n(x)-f_n(x_s)|<\epsilon/3$ and $|f_m(x)-f_m(x_s)|<\epsilon/3$.

Finally we get: Let $\epsilon>0$ be given then $\exists N_0=\max\{N_1, N_2, \dots, N_p\}$ such that $\forall n,m\ge N_0$ and $\forall x\in K$ we have $$|f_n(x)-f_m(x)|\leqslant |f_n(x)-f_n(x_s)|+|f_n(x_s)-f_m(x_s)|+|f_m(x_s)-f_m(x)|<\epsilon.$$ Thus $\{f_n\}$ converges uniformly on a compact set $K$.

Dear colleagues! Can anyone chech my proof plese? I would be grateful for your help!

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    $\begingroup$ I would move the phrase "Let $\epsilon>0$ be given" (but not the part with the word "then") to the beginning of the proof. Other than that your text makes good sense. $\endgroup$ – Justpassingby Dec 11 '15 at 7:21
  • $\begingroup$ @Justpassingby, Can you explain your remark more extended? I don't understand you :( $\endgroup$ – ZFR Dec 11 '15 at 7:25
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This proof has all the correct elements in it. Just one remark about the order in which you place them.

The definition of uniform convergence requires the existence of a certain $\delta$ (with certain properties) for any given $\epsilon.$

You are starting your proof by constructing such a $\delta,$ and then halfway down the proof you say "let $\epsilon>0$ be given". It would make more sense to assume the given $\epsilon$ before you start the construction.

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    $\begingroup$ So first of all i must fix any $\epsilon>0$. Right? $\endgroup$ – ZFR Dec 11 '15 at 7:35
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    $\begingroup$ That's what I meant. It's just cosmetics, though: if I were grading this as an exam you would have had full credit first time around. $\endgroup$ – Justpassingby Dec 11 '15 at 7:39
  • $\begingroup$ Thanks a lot for your valuable remark! $\endgroup$ – ZFR Dec 11 '15 at 7:44

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