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Let $\Vert \cdot \Vert$ a norm on $X=C([0,1])$ s.t.

1. $X$ is complete w.r.t. $\Vert \cdot \Vert$;

2. convergence in $\Vert \cdot \Vert$ implies pointwise convegence, i.e. $$ \Vert x_n - x \Vert \to 0 \Rightarrow \forall t \in [0,1], \quad x_n(t) \to x(t). $$

Is $\Vert \cdot \Vert$ equivalent to the usual $\sup$-norm, $\Vert \cdot \Vert_{\infty}$?

By open mapping and 1 (completeness), we just need to prove only one inequality of the two we need to prove equivalence (indeed, the other would follow by open mapping).

Anyway, I think the answer is affirmative; am I right? I do not know how to prove it. Any ideas, please? Thanks.

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Perhaps this is easy from the closed graph theorem... – GEdgar Dec 31 '12 at 18:19
A variation on the same theme. In particular, by perturbing the $\lVert\, \cdot\, \rVert_\infty$-norm with a discontinuous functional it is shown that one can have point 2. for an inequivalent norm. – Martin Dec 31 '12 at 19:41
up vote 5 down vote accepted

The hypothesis implies that for all $t\in I=[0,1]$ the linear functional $\mathrm{ev}_t:X\to\Bbb R, ~f\mapsto f(t)$ is continuous. Also, for any $f\in X$, $$\sup_{t\in I}~|\mathrm{ev}_t(f)|=||f||_{\infty}<+\infty$$ Since $X$ is a Banach space, the uniform boundedness principle applies, and $$ C=\sup_{t\in I}~||\mathrm{ev}_t||_{X'}<+\infty$$ that is, for any continuous function $f$ and any $t\in I$, $|f(t)|\leq C||f||_X$ thus $$||\cdot||_{\infty}\leq C||\cdot||_X$$ As you noted, the open mapping theorem then tells us that both norms are equivalent.

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Thanks, nice application of uniform boundedness. Very useful, thanks again for your help. – Romeo Dec 31 '12 at 19:43
My pleasure Romeo : ) happy new year btw – Olivier Bégassat Dec 31 '12 at 19:53
Happy new year to you, too :-) – Romeo Dec 31 '12 at 19:54

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