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If the space $X$ is banach , then I want to show that any linear map $T:X \to X$ is continuous iff the null space is closed. I could show that if $T$ is continuous then the null space is closed. But I am unable to prove the converse. Any hints are appreciated. Thanks

happymath
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  • "I want to show that any continuous map $T:X→X$ is continuous iff the null space is closed". Did you mean "linear map"? – Matias Heikkilä Oct 13 '15 at 08:55
  • @MatiasHeikkilä yes I meant linear map I have made that correction – happymath Oct 13 '15 at 08:55
  • This might be a full solution to your problem? – Matias Heikkilä Oct 13 '15 at 08:57
  • I disagree. You don't even have unbounded operators in the case of a finite-dimensional space so the answer would not make any sense.

    In fact, it also generalizes far beyond Banach spaces (to topological vector spaces).

     – Matias Heikkilä Oct 13 '15 at 09:10
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    @MatiasHeikkilä sorry for my sloppiness but what I actually meant was if $T:X \to Y$ is a linear operator and $Y$ is finite dimensional then I can show that $T$ is continuous iff null space is closed. But notice that for proving this i do not need to assume it is a Banach Space. Here what I am asking is quite more than just for functionals whereas the above link which you provided only works for functionals – happymath Oct 13 '15 at 09:16

2 Answers2

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This is not true. See the $T$ in this answer. Since this $T$ is injective, its kernel is $\{0\}$.

Note that a similar assertion is true for linear functionals $T : X \to \mathbb{R}$.

Community
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gerw
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    @JohnMa I had not realised that mistake initially that is why I accepted your answer. Now I have accepted this answer – happymath Oct 13 '15 at 09:33
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This is not the solution of your problem but its a interesting result in case of $f$ is linear functional.

Here is a sketch of more general result.Try to fill the gaps in the proof.

Lemma: Let $f$ be a linear functional on $X$ .A hyperplane $H= \{ x\in X : f(x) = \alpha\}$ is closed iff $f$ is continuous.

Proof: Clearly if $f$ is continuous then $H$ is closed.Conversely,Assume $H$ is closed therefore $H^c$ is open and since $f$ is nonzero its nonempty.(Why?)

WLOG,assume that $x_0 \in H^c$ is such that $f(x_0) < \alpha$.Since $H^c$ is open there exists $r>0$ such that $B(x_0,2r) \subset H^c$.Now for all $x \in B(x_0,2r)$,we have $f(x) < \alpha$ (Why?)

Thus for any $z \in V$ such that $\|z\| \leq 1$,we get $f(x_0 +rz) \leq \alpha$ or equivalently $$ f(z) \leq \frac { \alpha -f(x_0)}{r}$$.Thus the image of the unit ball is bounded and hence $f$ is bounded (Why?) therefore continuous.

Arpit Kansal
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