# Extension of a self adjoint Operator

Suppose we have a open (bounded) domain $\Omega$ in $\mathbb R^d$. And let a plane $\mathcal P$ in $\mathbb R^d$ divides the domain in two (disjoint) open sets. (say $\Omega_1$ and $\Omega_2$)

Hence $\overline{\Omega_1\cup\Omega_2}=\overline{\Omega}$

Consider the Dirichelet Laplacian over $\Omega_1\cup\Omega_2$ (call it $T$) and Dirichelet Laplacian over $\Omega$ call it S.

see that

$D(T)=\{f\in H_0^1\cap H^2(\Omega_1\cup\Omega_2)\}$ and $D(S)=\{f\in H_0^1\cap H^2(\Omega)\}$

$D(T)$ is different from $D(S)$ only for reason $\forall f\in D(T)\;,f(x)=0\;\forall x\in \mathcal P\cap \Omega$ (in Trace Sense)

We can say that $D(T)\subset D(S)$ and $Tu=Su$ wherever both make sense.

I just showed $S$ is extension of $T$.

But I know a result where a self adjoint operator cannot be extended to a self adjoint operator. (Maximality of Self Adjoint Operator).

I am surely wrong in some place, please help me to figure out.

• Who said $T$ is self-adjoint (in $L^2(Omega)$)? – Robert Israel Mar 20 '15 at 20:34
• @RobertIsrael Thanks, Got it. But am I right if I say S is self adjoint extension of T? – Harish Mar 20 '15 at 20:42