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Let $\phi \in C[0,1]$ and $T_{\phi}: C[0,1] \to C[0,1]$ to be the multiplication operator such that $T_{\phi}(f) = \phi f$, then either the range of it is the whole space or it is of the first Baire catergory.

Well, by Baire category theorem, if $T_{\phi}$ is surjective, then its image cannot be of first Baire Category. Now for the other direction, suppose $T$ is not surjective, and its image is of the first Baire category, how to get a contradiction?

Bernard
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z.z
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  • This dichotomy holds for the range of any continuous linear operator between Banach (or Frechet) spaces. This is a version of the open mapping theorem. – Jochen Sep 20 '18 at 06:57

2 Answers2

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This is pretty easy to do explicitly, without much machinery.

Hint: there are two cases. Either there exists $x_0 \in [0,1]$ such that $\phi(x_0) = 0$, or there does not. In the first case you should be able to show that the range of $T_\phi$ is nowhere dense, by showing it is contained in a closed set with empty interior. In the second case, you should be able to show $T_\phi$ is surjective: for any given $f\in C[0,1]$, explicitly construct $g \in C[0,1]$ such that $T_\phi g = f$.

Nate Eldredge
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  • Indeed simple, I was thinking of countable union of nowhere dense set at the first sight which stumbled me. – z.z Sep 19 '18 at 20:24
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It is know that the spectrum of the multiplication operator is $\sigma(T_\phi) = \phi([0,1])$.

If $T_\phi$ is not surjective then $0 \in \sigma(T_\phi)$ so $\exists x_0 \in [0,1]$ such that $\phi(x_0) = 0$.

Therefore $$\operatorname{Im} T_\phi \subseteq \{f \in C[0,1] : f(x_0) = 0\}$$

The latter is a proper closed subspace of $C[0,1]$ so in particular it is nowhere dense.

mechanodroid
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