Does a symmetric operator on a Hilbert space have a symmetric adjoint? Suppose we have a linear operator $T$, densely-defined on some Hilbert space. If $T$ is symmetric (i.e., $T^*$ extends $T$: notationally, $T\subseteq T^*$) does it follow that $T^*$ is also symmetric (and therefore, in fact, self-adjoint)?
If so, where could I find a proof of this? If not, what is a counter-example showing this?
 A: Consider the differentiation operator, which is a classical example of a closed, densely-defined, symmetric, linear operator on the Hilbert Space which is not selfadjoint.
Specifically, let $X=L^{2}[0,1]$, and let $T=\frac{1}{i}\frac{d}{dx}$ on the domain $\mathcal{D}(T)$ consisting of all $f \in X$ such that $f$ is equal a.e. to an absolutely continuous $\tilde{f}$ on $[0,1]$ such that $\tilde{f}(0)=\tilde{f}(1)=0$. Then $T$ is symmetric with adjoint $T^{\star}$ equal to $\frac{1}{i}\frac{d}{dx}$ on the domain $\mathcal{D}(T^{\star})$ consisting of all $f \in X$ such that $f$ is equal a.e. to an absolutely continuous function $\tilde{f}$ on $(0,1)$ for which $\int_{0}^{1}|\tilde{f}'|^{2}dx < \infty$.
You know that $T^{\star}$ is not symmetric because $e^{-x}\in\mathcal{D}(T^{\star})$ satisfies $T^{\star}e^{-x}=ie^{-x}$, and no symmetric operator can have an imaginary eigenvalue. Indeed, if $A$ is symmetric with $Af=\lambda f$, then
$(\lambda-\overline{\lambda})\|f\|^{2}=(Af,f)-(f,Af)=0$ implies $f=0$ or $\lambda$ is real.
One person asked about showing that $T^{\star}$ is as stated. Actually, it's very instructive to see how this works. So, let $T$ be as stated above, and let $V$ be $\frac{1}{i}\frac{d}{dx}$ on the domain $\mathcal{D}(V)$ consisting of all absolutely continuous functions $f$ on $[0,1]$ with $f'\in L^{2}$. It's easy to verify that
$$    (Tf,g) = (f,Vg),\;\;\; f \in \mathcal{D}(T), g\in\mathcal{D}(V). $$
So, $T^{\star}$ extends or equals $V$, written as $V\preceq T^{\star}$. To show that $g \in \mathcal{D}(T^{\star})$ implies $g \in \mathcal{D}(V)$, assume that $g\in\mathcal{D}(T^{\star})$. Equivalently, there exists $m \in L^{2}$ such that
$$      (Tf,g) = (f,m),\;\;\; f \in \mathcal{D}(T). $$
Let $\chi_{[a,b]}$ the characteristic function of the interval $[a,b]$. We construct functions $f$ to use in the above:
$$       f_{a,h,b,k}(x) = \int_{0}^{x}\left(\frac{1}{h}\chi_{[a-h,a]}(t)-\frac{1}{k}\chi_{[b,b+k]}(t)\right)dt $$
For $0 \le a-h < a \le b < b+h \le 1$ this function is in $\mathcal{D}(T)$ with derivative equal a.e. to the integrand of the above integral. Plugging into the adjoint equation gives
$$     (Tf_{a,h,b,k},g)=-\frac{i}{h}\int_{a-h}^{a}\overline{g(t)}\,dt+\frac{i}{k}\int_{b}^{b+k}\overline{g(t)}\,dt = (f_{a,h,b,k},m)=\int_{0}^{1}f_{a,h,b,k}(t)\overline{m(t)}\,dt. $$
As $h \downarrow 0$  $f_{a,h,b,k}$ converges pointwise a.e. and remains bounded in doing so. The same thing is true as $k\downarrow 0$. So the left- and right- derivatives of $G(x)=\int_{0}^{x}g(t)dt$ exists at all $x \in (0,1)$, and the two are equal as can be seen by the above. Furthermore, the derivative is continuous, with
$$     \left.\left[ \frac{d}{dx}i\int_{a}^{x}\overline{g(t)}dt\right]\right|_{a}^{b}
   = \int_{a}^{b}\overline{m(t)}\,dt,\;\;\; 0 < a \le b < 1. $$
However, the integral of $g$ is absolutely continuous. Therefore, the above derivative equals $\overline{g}$ a.e., which means that $g$ is equal a.e. to a continuous function $\tilde{g}$ and
$$
              -i\{ \tilde{g}(b)-\tilde{g}(a)\} = \int_{a}^{b}m(t)dt,\;\;\; 0 < a < b < 1.
$$
It follows that $\tilde{g}$ is absolutely continuous and $-i\tilde{g}'=m$ a.e.. By assumption $m\in L^{2}[0,1]$, which proves that $g \in \mathcal{D}(V)$ and $T^{\star}g=h=Vg$. Therefore $T^{\star}=V$, by domain and by action.
A: If $T^*$ is symmetric then we would have that a symmetric linear operator $T$ is ad-joint,because $T^{**}=T$.
But this is false in general.
Check here when this can be true here
