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I do this proof in 3 steps

First I pick a Cauchy sequence of $f_n(x) \in C[a,b]$ and then fix an $\epsilon >0$ and some $x \in [a,b]$ then constitute $f(x)$ s.t $\lim_{n \to \infty} f_n(x)=f(x)$ in absolute value.

then I try to prove that this limit $f(x)$ is continuous and $\lim_{n \to \infty} f_n(x)=f(x)$ in sup-norm.

Which is a very long proof.

The second proof is.

Let $\{f_n\}$ be a Cauchy sequence in $C[a,b]$ then for all $\epsilon >0$ there exists an $N(\epsilon)$ s.t. for all $n,m>N(\epsilon)$ we have $\| f_m -f_n \|_{\infty} < \epsilon$ which implies $| f_m(x) - f_n(x) | <\epsilon$ for all $x \in [a,b]$ then $\{f_n(x)\}$ is uniformly cauchy on $[a,b]$. Then by the following Theorem.

Theorem. A sequence $\{f_n\}$ of functions $f_n : A \to\mathbb R$ converges uniformly on $A$ if and only if it is uniformly Cauchy on $A$.

Then we deduce that the $f_n \to f$ uniformly on $[a,b]$

But given that $f_n$ are continuous functions then by uniform convergence we know that $f$ is continuous on $[a,b]$

then by definition of uniform convergence we have that for all $x \in [a,b]$, for all $\epsilon >0$ $\exists N(\epsilon )$ st $\forall n>N(\epsilon )$ $| f_n(x) - f(x) | < \epsilon$ $\implies$ $\| f_n - f \|_{\infty} < \epsilon$

Does the second proof suffice?

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In your second 'proof' you are assuming what you have to prove. You have to first find a continuous function $f$ and then prove uniform convergence. The statement that $\{f_n\}$ is Cauchy means that given $\epsilon >0$ there exists $k$ (independent of $x$) such that $|f_n(x)-f_m(x)| <\epsilon$ for all $x$ for all $n,m >k$. $\cdots (1)$. In particular, you can fix $x$ an conclude that the sequence $\{f_n(x)\}$ of real numbers is Cauchy, hence convergent. Let $f(x)$ be the limit of this sequence. In (1) if you take limit as $m \to \infty$ you get $|f_n(x)-f(x)| \leq \epsilon$ for all $x$ for all $n >k$. This means $f_n \to f$ uniformly which implies that $f$ is continuous and the convergence takes place in the metric of $C[0,1]$.

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  • $\begingroup$ I used this theorem Theorem. A sequence $\{f_n\}$ of functions $f_n : A \to\mathbb R$ converges uniformly on $A$ if and only if it is uniformly Cauchy on $A$. which as I understood it tells me that the sequence converges uniformly to some limite I called $f$ on $[a,b]$ so the theorem tells that the limit exist but we don't know where it lives then I used another theorem which I edited in the question. Thank you $\endgroup$ – Dreamer123 Jan 8 at 17:16
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    $\begingroup$ @Dreamer123 The question here is what results can be used and what cannot be used. You are using two theorems which make the question trivial, but usually a question like this is supposed to be answered without use of such theorems. I have just followed standard practice in my answer. $\endgroup$ – Kabo Murphy Jan 8 at 23:11

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