Counter example to show that if $\ker(T)$ is closed and $W$ is not finite dimensional $T$ is not bounded.

Question

Let operator $T:(V,\|-\|_v)\rightarrow(W,\|-\|_w)$.

I know that if $\ker(T)$ is closed and $W$ is finite dimensional, then $T$ is bounded. But is there an easy counter example to show that the finite-dimensional criterion is critical?

I know in the proof of the continuity of $T$ the finite dimensional property of $W$ is used to show that the quotient space of $V$ is finite-dim as well and the proof proceeds. But I can't derive a counter example from there.

Link to the proof of continuity :$T$ is continuous if and only if $\ker T$ is closed

Tsemo Aristide · Accepted Answer · 2019-03-01T10:10:45.153

5

Let $V$ be the space of sequences with finitely many non zero terms with $Sup$ norm and the endomorphism of $V$ defined by $T(e_n)=ne_n$.

edited Mar 01 '19 at 10:10

answered Mar 01 '19 at 09:55

Tsemo Aristide

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By $l^2$ do you mean $\ell _2$ – Jhon Doe Mar 01 '19 at 10:01
What is the codomain of your map? How do you extend your map to all $l^2$ – Federico Fallucca Mar 01 '19 at 10:05

Counter example to show that if $\ker(T)$ is closed and $W$ is not finite dimensional $T$ is not bounded.

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