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Does ($C([0,1]), \left\|\cdot\right\|_2)$) has a subspace that is isomorphic to ($L^2([0,1]), \left\|\cdot\right\|_2)$)?

I think the answer is no but I have no idea to prove it.

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  • $\begingroup$ Isomorphic as Banach spaces? Does that mean we would need an isometry between them? $\endgroup$ – Omnomnomnom May 6 '17 at 5:54
  • $\begingroup$ @Omnomnomnom Yes. $\endgroup$ – Pluviophile May 6 '17 at 5:56
  • $\begingroup$ What do you mean by isomorphic? Preserve norms? $\endgroup$ – copper.hat May 6 '17 at 6:02
  • $\begingroup$ @copper.hat yeah that's it $\endgroup$ – Pluviophile May 6 '17 at 6:10
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    $\begingroup$ @copper.hat: I do mean complete, or closed in $L^2$, which is the same thing. $\endgroup$ – user138530 May 6 '17 at 22:32
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As you suspected, this is not possible. Suppose we had a subspace $H\subseteq C[0,1]$ that is closed in $L^2$. Then the identity map $(H,\|\cdot\|_{\infty})\to (H,\|\cdot\|_2)$ is a continuous bijection. Moreover, $(H,\|\cdot\|_{\infty})$ is a closed subspace of $C[0,1]$ with the same norm because if $f_n\in H$, $f\in C$, $\|f_n-f\|_{\infty}\to 0$, then we also have convergence in $L^2$ norm, so $f\in H$ since $H$ with this norm was closed by assumption. Thus the open mapping theorem shows that the two norms are equivalent on $H$.

Now take an ONB $f_n$ of $H$. Since, as we just saw, $\sup_n\max |f_n|<\infty$, we can find a $\delta>0$ so that the sets $$ A_n =\{ x: |f_n(x)|\ge 1/2 \} . $$ have measure $|A_n|\ge\delta$. This means that given $N/\delta$ such sets, there will be a point $x$ that is contained in at least $N$ of them.

This means that we can always make $$ \left\| \sum_{n=1}^{N/\delta} e^{i\alpha_n} f_n \right\|_{\infty} \ge N/2 $$ by choosing suitable phases $\alpha_n$. However, the $L^2$ norm of this function is only $\sqrt{N/\delta}$. It follows that $H$ can not be infinite-dimensional.

Remark: The fact that $[0,1]$ is compact is crucial here. On the real line (or $(0,1)$), we can easily find such subspaces: the Paley-Wiener space works.

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    $\begingroup$ In case anyone is slow like me, ONB = orthonormal basis :-) $\endgroup$ – copper.hat May 6 '17 at 22:37

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