Let $H$ be a Hilbert space, and $T:D(T)\subset H\rightarrow H$ and $S:D(S)\subset H\rightarrow H$ be unbounded self-adjoint operators.


Is $T+S:D(T)\cap D(S)\rightarrow H$ self-adjoint?



Sum of two self-adjoint unbounded operators
The answer is no. $\newcommand{\bR}{\mathbb{R}}$ Consider the Hilbert space of $L^2$-functions


$$ u:[0,1]\to \bR^2,\;\;u(t)=(x(t), y(t)) . $$


Define 


$$ D(T)=\Bigl\lbrace u\in H;\;\int\_0^1 \bigl\vert u'(t)\bigr\vert^2 dt <\infty,\;;\;x(0)=x(1)=0\;\Bigr\rbrace, $$


$$ D(S)=\Bigl\lbrace u\in H;\;\int\_0^1 \bigl|u'(t)\bigr|^2 dt <\infty,\;;\;y(0)=y(1)=0\;\Bigr\rbrace. $$


Denote by $J:\bR^2\to\bR^2$ the linear operator $(x,y)\mapsto (-y,x)$. For $u\in D(T)$ defines


$$ Tu=J\frac{du}{dt}, $$


while for $ u\in D(S)$ define


$$ Su=J\frac{du}{dt}. $$


Both operators $S$ and $T$ are selfadjoint. Note that


$$ D(T)\cap D(S)= \Bigl\lbrace u\in H;\;\int\_0^1 \bigl|u'(t)\bigr|^2 dt <\infty,\;\;u(0)=u(1)=0\,\Bigr\rbrace. $$


The operator $A:=T+S:D(A)= D(T)\cap D(S)\to H$ is symmetric, but not selfadjoint. Indeed, $ v\in D(A^\*)$ if and only if, there exists $C>0$ such that


$$ \bigl\vert\;(Au, v)\_{H} \;\bigr\vert \leq C\Vert u\Vert\_H,\;\;\forall u\in D(A). $$


The constant function $t\mapsto v(t)= (1,1)$ satisfies this condition because for any $u\in D(A)$ we have


$$ (Au, v)\_H=\int\_0^1 \bigl(\; 2Ju'(t), v(t)\;\bigr) dt = 2\int\_0^1\frac{d}{dt} \bigl(\; Ju(t), v(t)\;\bigr) =0. $$


We have produced a function $v\in D(A^\*)\setminus D(A)$.


