Kernel of the closure of an unbounded operator

Kernel of the closure of an unbounded operator

Let $T:\mathcal{D}(T)\subset X\to Y$ be a densely defined closable
operator. Define $\text{ker}\,T=\{(x,0)\in \text{graph}\,T\}$ where
$\text{graph}\,T\subset X\times Y$. My question is
$\overline{\text{ker}\,T}=\text{ker}\,\overline{T}$?
I recall that $\overline{T}$ is the closure of $T$.
I know that $\overline{\text{ker}\,T}\subset\text{ker}\,\overline{T}$,
what about the reverse inclusion? So far I have proved that
$\text{ker}\,T^{\perp}\subset R(T^*)$ implies the inclusion
$\overline{\text{ker}\,T}\supset\text{ker}\,\overline{T}$, but this
inclusion seems to be false in general.
Someone has any clue about the kernel of the closure of an operator?
Thanks in advance.
Remarks: $M^{\perp}=\{x'\in X'\mid x'(x)=0 \text{ for every } x\in M\}$
where $X'$ is the dual space and $T^*:\mathcal{D}(T^*)\subset Y'\to X'$ is
the adjoint. The norm in $X\times Y$ is $||(x,y)||=||x||+||y||$.