# Closed graph theorem and equivalence of norms in $C[0,1]$

Consider $$id:(C[0,1],||.||_\sup) \mapsto (C[0,1], ||.||_{*}$$ Show that up to an equivalence of norms, the supremum norm is the only norm on $$C[0,1]$$ which makes C[0,1] complete and which also implies the point-wise convergence.

Sketch:

I consider $$id:(C[0,1],||.||_\sup) \mapsto (C[0,1], ||.||_{*}$$ and his graph $$G(T)$$, I need show that graph is closed in $$X \times Y$$. So show that for every sequence on $$G(T)$$ have limit in $$G(T)$$, but I am not sure how to prove this. One of my intuition suggestion that $$f_{n}$$ is uniformly convergence to $$f \in (C[0,1],||*||_{sup})$$. Id is continuous.

Next is simple, because from Closed Graph Theorem we have implication that id are bounded so it $$||id(f)||_{\sup} < C||f||_{*}$$ and $$||id^{-1}(f)||_{*} < C_{1}||f||_{supp}$$.

Let $$(f_n)_{n\in \mathbb{N}}$$ be a sequence in $$C([0,1])$$ which converges in both $$||\cdot||_{\textrm{sup}}$$ and $$||\cdot||_{*}$$. So let $$(f,g)$$ be a pair of functions such that $$f_n\xrightarrow{||\cdot||_{\textrm{sup}}} f$$ and $$f_n\xrightarrow{||\cdot||_{*}} g$$. By assumption, this implies that both $$f$$ and $$g$$ are the pointwise limits of $$f_n$$, and hence $$f=g$$. This implies that the graph of $$id$$ (and hence, the graph of $$id^{-1}$$ as well) is closed. By the closed graph theorem, we conclude that the identity is a bounded isomorphism of $$C([0,1])$$ with these two norms and hence, the norms are equivalent.
• $f_n\xrightarrow{||\cdot||_{\textrm{sup}}} f$ I am a bit confusing, because we need to show that implies the pointwise convergence. It's not my assumption? – Blabla Nov 8 '19 at 14:57
• Sory I have mistake $f_n\xrightarrow{||\cdot||_{*}} g$, for assumptions we have that $||.||_{*}$ are convergence uniformly ? – Blabla Nov 8 '19 at 15:07
• Our assumption is that $||\cdot||_{*}$ is a complete norm on $C([0,1])$ such that $f_n\xrightarrow{||\cdot||_*} g$ implies $f_n(x)\to g(x)$ pointwise. – WoolierThanThou Nov 8 '19 at 15:13