Proving that a linear operator $T$ is continuous. 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
 A: 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 $T : X \to \mathbb{R}$.
A: 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.
