# A question on the completeness of $(C[a,b], \sup)$ where $[a,b] \subset \mathbb{R}$

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?

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 $$C[0,1]$$.
• 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 – Dreamer123 Jan 8 at 17:16