If an operator have only Real eigenvalues + symmetric then it's self-adjoint?

Question

I know that if an operator is self-adjoint then has Real eigenvalues but I'm not sure about the converse i.e. if it has only Real eigenvalues and is symmetric then the operator is selfadjoint. Is that true?

----Edit---

The difference between selfadjoint and symmetric being the definition set. Symmetric has an extension which coincide in the original domain, while a Selfadjoint operator has the same domain of definition

I define symmetric as follows. Let be $\mathcal{A}=\left(A,\mathfrak{D}_{A}\right)$ an operator densely defined and $\mathcal{A}^{*}=\left(A^{*},\mathfrak{D}_{A^{*}}\right)$ the adjoint operator, then $\mathcal{A}$ it is called symmetric if \begin{eqnarray} & & \mathfrak{D}_{A^{*}}\supseteq\mathfrak{D}_{A}\\ & & A^{*}\psi=A\psi\qquad\forall\psi\in\mathfrak{D}_{A}. \end{eqnarray}

While I define Self-adjoint like this Let be $\mathcal{A}=\left(A,\mathfrak{D}_{A}\right)$ an operator densely defined and $\mathcal{A}^{*}=\left(A^{*},\mathfrak{D}_{A^{*}}\right)$ the adjoint operator, then $\mathcal{A}$ it is called selfadjoint if \begin{eqnarray} & & \mathfrak{D}_{A^{*}}=\mathfrak{D}_{A}\\ & & A^{*}\psi=A\psi\qquad\forall\psi\in\mathfrak{D}_{A}. \end{eqnarray}

Answer 1

This is far from being true, indeed, every symmetric operator has only real eigenvalues:

If $\psi\in\ker(T-\lambda),\,\psi\neq 0$, then $$ \lambda\|\psi\|^2=\langle T\psi,\psi\rangle=\langle \psi,T\psi\rangle=\bar\lambda\|\psi\|^2, $$ hence $\lambda=\bar\lambda$.

Now every symmetric operator that is not self-adjoint yields a counterexample to your conjecture (if you want to be explicit, take $\Delta$ on $C_c^\infty(\mathbb{R}^n)$ as operator in $L^2(\mathbb{R}^n)$).

To get a criterion for self-adjointness, you have to replace the eigenvalues by the spectrum of the operator. Then the following characterization holds:

A symmetric operator $T$ is self-adjoint if and only if its spectrum is contained in $\mathbb{R}$.

Proof: It suffices to show that $D(T^\ast)\subset D(T)$. Let $z\in\mathbb{C}\setminus\mathbb{R}$. Since $\sigma(T)\subset\mathbb{R}$, the operators $T-z$ and $T-\bar z$ are invertible.

Let $\phi\in D(T^\ast)$ and $\psi:=(T-z)^{-1}(T^\ast -z)\phi\in D(T)$. Then we have $T\psi=T^\ast\psi$ and $(T^\ast-z)\phi=(T-z)\psi$.

It follows that $$ (T^\ast-z)(\phi-\psi)=(T-z)\psi-(T-z)\psi=0, $$ that is, $\phi-\psi\in N(T^\ast-z)=R(T-\bar z)^\perp=\{0\}$. Hence, $\phi=\psi\in D(T)$.

Remark: I used $R(A)$ and $N(A)$ to denote the range and kernel of $A$.