# Is a linear operator continuous if its kernel is closed?

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.

• Possible duplicate of $T$ is continuous if and only if $\ker T$ is closed – Hirak Sep 29 '18 at 15:41
• @Hirak the thread you linked only considers the case where either $X$ or $Y$ is finite dimensional. In particular, the answer is different. – Lorenzo Quarisa Sep 29 '18 at 15:48
• Infinite dimensional normed spaces – user344374 Sep 29 '18 at 15:49

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.
• So it not necessary that $T$ is bounded – user344374 Sep 29 '18 at 15:47
• $T$ isn't necessarily continuous. – Saucy O'Path Sep 29 '18 at 15:52
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).