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. 
 A: 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).
A: 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.
