Let $C([a,b],\mathbb{C})$ be all continuous functions on $[a,b]$ with a complex image. Show that it is a Banach space with respect to $\|f_n\|_\infty=\max_{x\in[a,b]}\{n\in\mathbb{N}\mid|f_n(x)|\}$.
Let $\left(f_n\right)_{n\ge1}$ be a Cauchy sequence in the normed vector space. Then for all $\epsilon>0$ there exists $N\in\mathbb{N}$ such that $|f_n-f_m|<\epsilon$ for all $n,m\ge N$. Next I'll use the lemma
Lemma 1: Let $f_n:A\mapsto\mathbb{C}$ . $\left(f_n\right)_{n\ge1}$ converges uniformly iff $\left(f_n\right)_{n\ge1}$ is a Cauchy sequence.
Since $\left(f_n\right)_{n\ge1}$ is a Cauchy sequence, then $\left(f_n\right)_{n\ge1}$ is also uniformly convergent towards $f$.
Edit: We prove that this $f$ exists. For all $\epsilon>0$ there exists $N$ such that $$|f_m(x)-f_n(x)|\le\|f_n-f_m\|_\infty<\epsilon$$ for all $n,m\ge N$. Since $\left(f_m(x)\right)_{n\ge 1}$ is a complex Cauchy sequence, there exists some convergence point. Let that be $\lim_{m\to\infty}f_m(x)=f(x)$.
Then for all $\epsilon>0$ exists $N\in\mathbb{N}$ such that $\|f-f_n\|_\infty<\epsilon$ for all $n\ge N$. Now that I've shown completeness, I show that $f$ is in the normed vector space.
Lemma 2: Let $\left(f_n\right)_{n\ge1}$ be a sequence of functions defined on $A$ and assume that $\left(f_n\right)_{n\ge1}$ converges uniformly towards $f$ on $A$. Then $f$ is continuous on $A$.
Since $\left(f_n\right)_{n\ge1}$ is defined as continues functions on $[a,b]$ and the convergence towards $f$ is uniformly, we can conclude that $f$ is continuous on $[a,b]$ and thus it is in the normed vector space per definition of the vector space.
Would this suffice as a proof?
Best Answer
No, it doesn't suffice. What is that function $f$ that you mention? You must explain first what $f$ is and then to prove that $(f_n)_{n\in\mathbb N}$ converges uniformly to $f$.