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Given $f\in C_b(\mathbb{R})$, let $Tf(x) = e^{-|x|}f(x)$. Show that $T$ defines a bounded linear map into itself such that $\ker T' \neq {0}$.

My try: look at the space $A = \{ f \in C_b : \exists \lim_{x\rightarrow \infty} f(x)\}$ define $\ell_0 \in A' : \ell_0 (f) = \lim_{x\rightarrow \infty} f(x)$. Then $|\ell_0| (f) \leq \|f\|$ so we can expand $\tilde{\ell_0}$ to $C_b$. I have some trouble with the finish, how do I get $\tilde{\ell_0} (Tf) = 0 \;\; \forall f \in C_b$?

Edit: It would work if $\lim_{x\rightarrow \infty} e^{-|x|} f(x) = 0$ I guess.

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1 Answer 1

up vote 2 down vote accepted

The idea in the OP is good. Let $E:=C_b(\Bbb R)$ and $F:=\mathcal C_0(\Bbb R)$. Then $T(E)\subset F$. Let $\widetilde \ell_0$ an extension of $\ell_0$ to $E$. Then for all $f\in E$, $\langle \widetilde\ell_0,Tf\rangle_{E',E}=0,$ as $Tf\in F$, so $T'(\widetilde\ell_0)=0$. As $\ell_0$ is not identically $0$, so is $\widetilde \ell_0$ so $T'$ is not injective.

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