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Let $T$ be a linear operator between two infinite dimensional normed spaces $X$ and $Y$ whose kernel is a closed subset of its domain.

Does it imply that $T$ is bounded or not necessary ?

If yes I would be very grateful if one could mention the proof.

If no it would be better if a counter-example were provided.

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    $\begingroup$ Possible duplicate of $T$ is continuous if and only if $\ker T$ is closed $\endgroup$ – Hirak Sep 29 '18 at 15:41
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    $\begingroup$ @Hirak the thread you linked only considers the case where either $X$ or $Y$ is finite dimensional. In particular, the answer is different. $\endgroup$ – Lorenzo Quarisa Sep 29 '18 at 15:48
  • $\begingroup$ Infinite dimensional normed spaces $\endgroup$ – user344374 Sep 29 '18 at 15:49
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That would imply that all injective linear operators are continuous, which is not true. Namely, extend the canonical Hilbert basis $\{e_i\}_{i\in\omega}$ of $\ell^2$ to a basis $\{e_i\}_{i\in\beth_1}$ and consider the one and only linear map $T:\ell^2\to\ell^2$ such that $T(e_i)=e_i$ for all $i\in\omega$ and $T(e_i)=2e_i$ for all $i\in\beth_1\setminus\omega$. Since it sendsa basis to another basis, it's bijective. However, it agree with $id$ on a dense subset while not being $id$ itself.

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  • $\begingroup$ So it not necessary that $T$ is bounded $\endgroup$ – user344374 Sep 29 '18 at 15:47
  • $\begingroup$ $T$ isn't necessarily continuous. $\endgroup$ – Saucy O'Path Sep 29 '18 at 15:52
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Saucy O'Path's answer looks OK, but maybe derivatives give a more familiar counterexample. Let $$ Af(x)=f'(x), \qquad f\in C^1([0, 1]) $$ denote a linear operator $A\colon C^1\subset L^2\to L^2$. This operator is not continuous; for example, $$ \lVert \sin(2\pi n\cdot)\rVert_2=\sqrt{\frac{\pi}{2}},\quad \forall n\in\mathbb N,\ \text{ but }\lim_{n\to \infty}\lVert A\sin(2\pi n\cdot) \rVert_2=\infty.$$ However, the kernel of $A$ consists of the subspace of constant functions, which is closed in $L^2$. (Proof: if a sequence of constants is Cauchy with respect to the $L^2(0,1)$ norm, then it is Cauchy as a sequence of constants, hence converges to a constant limit).

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