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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 $||(x,y)||=||x||+||y||$.

user90189
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1 Answers1

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This is not true even for bounded operators. Let $H=l^2(\mathbb N),\ \{e_k,\ k\in\mathbb N\}$ be the standard base. Define $$A((x_1,x_2,x_3,\dots))=(0,x_2,x_3,\dots),$$ and $$D(A)=lin\{f_1,e_2,e_3,\dots\},$$ where $f_1=(1,\frac 1 2,\frac 1 3,\dots).$ Then $A$ is densely defined, $e_1\notin D(A),$ hence $\overline{\ker A}=\ker A=\{0\}$ but $\ker\overline A=\mathbb R e_1.$

Yurii Savchuk
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