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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?

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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}

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  • $\begingroup$ sorry I forgot to add symmetric, now I edited the question $\endgroup$
    – Picard
    Jan 19 '16 at 8:36
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    $\begingroup$ How do you define symmetric for an arbitrary operator? $\endgroup$
    – gerw
    Jan 19 '16 at 8:37
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    $\begingroup$ ...what's is the difference between symmetric and self-adjoint? $\endgroup$ Jan 19 '16 at 8:38
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    $\begingroup$ Sorry I added the definitions I used... $\endgroup$
    – Picard
    Jan 19 '16 at 8:45
  • $\begingroup$ Could be indeed interesting! But right now I'm interested in this problem how is it related? $\endgroup$
    – Picard
    Jan 19 '16 at 8:49
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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 $(T^\ast-z)\psi=(T-z)\psi$.

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 $A$.

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  • $\begingroup$ Thank You Very Much!!! Could you also give me a sketch of the proof of if the spectrum is contained in R then it's self-adjoint? $\endgroup$
    – Picard
    Jan 19 '16 at 10:46
  • $\begingroup$ Thank you again!!! Why does $\psi$ belong to D(T)? $\endgroup$
    – Picard
    Jan 19 '16 at 11:38
  • $\begingroup$ Because $(T-z)^{-1}$ maps to $D(T)$ by definition. $\endgroup$
    – MaoWao
    Jan 19 '16 at 11:38
  • $\begingroup$ True :D :D :D Perfect! Thank You $\endgroup$
    – Picard
    Jan 19 '16 at 11:47
  • $\begingroup$ @MaoWao Could you please check whether the last edit to your post is correct? If yes: Could you please explain from where you get $T \psi = T^\ast \psi$, $(T^\ast - z) \psi = (T - z) \psi$ and $(T^\ast - z) (-\psi) = (T - z) \psi$? I can't see how you arrive there. $\endgroup$
    – Jan
    Jun 11 '20 at 11:49

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