Equivalence of two norms on $C[0,1]$

Question

Assume that $C[0,1]$ is complete under the norm ||.||. If the norm convergence implies pointwise convergence, show that ||.|| is equivalent to the $||.||_{\infty}$.

Attempt:

Two norms are equivalent if convergence of sequence in one implies convergence of sequence in other.

Let $f_n$ be a sequence in $C[0,1]$ which converges to a function $f\in C[0,1]$. i.e. $\lim_{n \rightarrow \infty} ||f_n-f||=0 $. Since convergence in norm implies that pointwise convergence i.e. $\lim_{n \rightarrow \infty} |f_n(x)-f(x)|=0 $. Now how to show that it converges uniformly? Any help is appreciated. Thank you.

Answer 1

Let $E:=C([0,1])$. The linear functionals $\phi_t:(E,\|\cdot\|) \to \mathbb{R}$, $\phi_t(f)=f(t)$ are continuous (since convergence in $\|\cdot\|$ implies pointwise convergence) and the family $\{\phi_t:t \in [0,1]\}$ is pointwise bounded: For each $f \in E$ we have $$ \forall t \in [0,1]:~ |\phi_t(f)|=|f(t)| \le\|f\|_\infty. $$ By the theorem of uniform boundedness $\|\phi_t\|_\ast \le c$ $(t \in [0,1])$ for some $c\ge 0$ (where the functional norm is meant with respect to $\|\cdot\|$). Thus $$ |\phi_t(f)|=|f(t)| \le c \|f\| \quad (t \in [0,1], f \in E), $$ hence $\|f\|_\infty \le c \|f\|$ $(f\in E)$.

Answer 2

Consider $\textrm{id}: (C[0,1], ||\cdot||) \to (C[0,1], ||\cdot||_{\infty})$. We have to show that it is continuous because then by open mapping theorem it is a linear isomorphism and thus the norms are equivalent. But by closed graph theorem it is enough to show that it is a closed map. This is easy: if $f_n \to f$ in $(C[0,1], ||\cdot||)$ and $f_n \to g$ in $(C[0,1], ||\cdot||_{\infty})$ then $f_n(x) \to f(x)$ by assumption and $f_n(x) \to g(x)$ for all $x \in [0,1]$, so $f(x) = g(x)$ for all $x \in [0,1]$ and therefore $\textrm{id}$ is closed, hence continuous.