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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 $||id^{-1}(f)||_{*} < C_{1}||f||_{supp}$.

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Well, you use your assumption.

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.

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  • $\begingroup$ $f_n\xrightarrow{||\cdot||_{\textrm{sup}}} f$ I am a bit confusing, because we need to show that implies the pointwise convergence. It's not my assumption? $\endgroup$ – Blabla Nov 8 '19 at 14:57
  • $\begingroup$ @Blabla: This is just the trivial fact that a sequence of functions that converges uniformly must, in particular, converge pointwise to the same function. $\endgroup$ – Nate Eldredge Nov 8 '19 at 15:06
  • $\begingroup$ Sory I have mistake $f_n\xrightarrow{||\cdot||_{*}} g$, for assumptions we have that $||.||_{*}$ are convergence uniformly ? $\endgroup$ – Blabla Nov 8 '19 at 15:07
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    $\begingroup$ 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. $\endgroup$ – WoolierThanThou Nov 8 '19 at 15:13

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