# Kernel of the Extension of a Bounded Linear Operator

Suppose $T\colon E\to F$ is a bounded linear operator between Banach spaces. Moreover let $i\colon E\to E’$ be a dense, compact inclusion of $E$ into some other Banach space $E’$. Finally assume that there exists a positive constant $c\in\mathbb{R}$ such that $\|T(x)\|_F\leq c\|x\|_{E’}$ for all $x\in E'$.

By uniform continuity there exists a unique bounded linear operator $T’\colon E’\to F$ such that $T=T’\circ i$. How is $\operatorname{ker}T$ related to $\operatorname{ker}T’$? My guess is that $i(\operatorname{ker}T)$ is in general not dense in $\operatorname{ker}T’$. Can someone give an example where it is not dense? Or a proof that it is dense? Does the situation change if the spaces are assumed to be Hilbert spaces?

As you suspected $i($ker $T)$ needn't be dense in ker $T'$. Take $E=\ell^2$ and $E'=\lbrace (x_n)_{n\in\mathbb N}: (x_n/n)\in \ell^2\rbrace$ endowed with the obvious Hilbert norm. The inclusion $E\hookrightarrow E'$ is dense and compact. Define $T':E'\to E'$, $(x_n)_{n\in\mathbb N} \mapsto (x_n-x_{n+1})_{n\in\mathbb N}$. This is a continuous linear operator with $(1,1,1,\ldots)\in$ ker $T'$ and the restriction $T=T'|_{\ell^2}$ is injective.