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I am trying to understand the proof that the set of continuous function is complete under uniform/supremum norm.

First, suppose we have a Cauchy sequence of continuous functions ${f_n(t)}$ with respect to the uniform, i.e. $$\forall \epsilon>0, \exists N\text{ such that }||f_n(t)-f_m(t)|| = max_{x\in[a,b]}|f_n(t)-f_m(t)| \leq \epsilon \forall n,m\geq N$$

let $t=t_0$, then $$||f_n(t_0)-f_m(t_0)|| = max_{x\in[a,b]}|f_n(t_0)-f_m(t_0)|=|f_n(t_0)-f_m(t_0)| \leq \epsilon \forall n,m\geq N$$

since $f_n(t_0)$ and $f_m(t_0)$ are in $\mathbb{R}$, then the cauchy sequence is a convergent sequence, i.e. $lim_{n\to\infty}f_n(t_0) = f(t_0)$. do this for each $t_0\in[a,b]$ in order to define $f(t)=lim_{n\to\infty}f_n(t)$. Now we can show that f(t) is continuous by: $$|f(t)-f(t+h)|\leq|f(t)-f_k(t)|+|f_k(t)-f_k(t+h)|+|f_k(t+h)-f(t+h)|<\epsilon/3+\epsilon/3+\epsilon/3=\epsilon$$ I understand that the first $\epsilon/3$ is due to the pointwise limit, and the second is due to the continuity of $f_k(t)$. My problem is in the third term. I think its due to pointwise limit not due to uniform convergence.

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In my opinion, there is one important step missing in the proof you gave: Before you can show that $f$ is continuous, you should first show that $||f - f_k|| \to 0$ (for example by taking $m \to \infty$ in $||f_n - f_m|| < \epsilon$).

Then you know that for each $\varepsilon$ and for large enough $k$, $|f(t) - f_k(t)| < \frac \epsilon 3$ for all $t$. That explains the first and the third term.

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The first and third terms can both be simultaneously controlled independent of $ h $ because of the uniform convergence, provided you have shown that $f_k$ actually does uniformly converge to $f$. Having done that, now that you can freely change $ h $ to control the second term using continuity.

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