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I'm taking a course in Functional Analysis using some topics from Kreyszig and Reed & Simon books, I have been asked to solve the following exercise:

Let $A$ be a symmetric operator such that $\rho(A)\neq \emptyset \;$ and $\; \sigma_{res}(A)\neq \emptyset$. Show that if $A \subset T$, i.e. T is an extension of A, then $T\neq T^*$, i.e. T is not a self-adjoint operator.

I have a proof of this excersise but it strongly uses 2 theorems which involves concepts we haven't seen such as connected sets and defect indices, this theorems are from Birdman, Solomjak (theorem 4 pp 83) and Weidmann (theorem 8.6 pp 233-234) books respectively.

Notation:

$\rho (A)$ is the resolvent set of the operator A

$\sigma_{res}(A)$ is the residual spectrum, wich are the $z$ in the spectrum of $A$ such that $z \in \sigma (A)$ and $\overline{Rang(A-zI)} \neq X$, been X the vectorial space.

I was wondering if there is another way to prove this without these tools and definitions, because with the topics I've received I can't do so much.

Any help or reference will be very preciated

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An elementary proof is possible, but it doesn't seem particularly enlightening. First some remarks: If $\lambda$ is in the residual spectrum then part of the definition is that $A-\lambda$ is injective, else you may find counterexamples (for example the $0$ operator). Secondly no self-adjoint operator may have a non-empty residual spectrum, hence any self-adjoint extension is a strict extension.

Finally the last preliminary remark, if $\mu$ is in the resolvent of $A$ and $A$ admits a strict self-adjoint extension $B$, then $\mu$ must be real. For $A-\mu: D(A)\to H$ must be bijective, but if $\mu$ is not real then $B-\mu$ is invertible (since $B$ self-adjoint) and hence $B-\mu : D(B)\to H$ must be bijective, even though it already admits a surjective restriction to a proper subset, contradicting injectivitiy.

Now let $\lambda\in\sigma_{res}(A)$, $\mu\in\rho(A)$ and suppose $B$ is a self-adjoint extension of $A$. Since $\overline{(A-\lambda I)H}\neq H$ is closed, there is a non-zero $z\in H$ so that $z$ is orthogonal to the range of $A-\lambda$, ie $$\langle z , (A-\lambda )y \rangle =0 \quad \text{ for all $y\in D(A)$}.$$ First we remark that $z\notin D(A)$. If $z\in D(A)$ you get that $\langle (A-\overline\lambda )z, y\rangle = \langle z, (A-\lambda)y\rangle= 0$ for all $y\in D(A)$, which is dense in $H$, so $(A-\overline\lambda)z=0$, meaning that $\overline\lambda$ is an eigenvalue of $A$ and $z$ is an eigenvector. Since $A$ is symmetric you immediately find that $\lambda$ must be real. So you get $(A-\lambda)z=0$, contradicting injectivity of $A-\lambda$, finally giving $z\notin D(A)$.

Now we consider two cases: either $z\in D(B)$ or $z\notin D(B)$. Both will yield a contradiction.

If $z\in D(B)$ then by symmetry of $B$ you have $\langle (B-\lambda) z , y\rangle = \langle z , (A-\lambda)y \rangle = 0$ for all $y\in D(A)$. Hence $Bz = \lambda z$. Now let $w\in D(A)$ with $(A-\mu)w= (\lambda-\mu)z$. Then $$\langle z - w , (A-\mu) y \rangle = \langle (\lambda - \mu)z - (\lambda-\mu )z , y\rangle = 0 $$ for all $y\in D(A)$. Since $(A-\mu)D(A)=H$ you then get $z=w$, hence $z\in D(A)$ must already hold, which we have already seen is not allowed.

If $z\notin D(B)$ then by self-adjointness of $B$ you get $z\notin D(B^*)$, hence there must be some sequence of norm one vectors $\xi_n\in D(B)$ with $\langle z, B\xi_n\rangle$ being unbounded. Now let $w_n$ be such that $(A-\mu)w_n = (B-\mu)\xi_n$ and $w$ such that $(A-\mu)w=z$. First note that: $$\langle w, (B-\mu)\xi_n\rangle = \langle z, \xi_n\rangle$$ hence $\langle w, (B-\mu)\xi_n\rangle$ is bounded. On the other hand: $$\langle z, (B-\mu)\xi_n\rangle = \langle z, (A-\lambda)w_n+(\lambda-\mu)w_n\rangle=\langle z, (\lambda-\mu)w_n\rangle =\langle ( A-\mu) w, (\lambda-\mu)w_n\rangle \\ = \langle w,(\lambda -\mu)(A-\mu)w_n\rangle = (\lambda-\mu) \langle w, (B-\mu)\xi_n\rangle$$ whence $\langle w, (B-\mu)\xi_n\rangle$ must be unbounded. Contradiction

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  • $\begingroup$ I have some questions about your proof $\endgroup$
    – PAB
    May 28, 2020 at 20:45
  • $\begingroup$ 1.st: When you take z you say that is orthogonal to the range of $A-\lambda$, I guess you're using the projection theorem, but it should be the closure of the range instead, I'm I right? And this construction is fundamental for the rest of the proof, could you please explain me why what you say would work anyway? \\ 2nd. In the definition we handle of residual spectrum, it is not necesary tha if $\lambda \in \sigma(A)$ then $A-\lambda$ is non-injective, it would only be possible if $\lambda$ is an eigenvalue, which is not the case. $\endgroup$
    – PAB
    May 28, 2020 at 20:52
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    $\begingroup$ Yes, its orthogonal to the closure of the range, in particular also orthogonal to the range. The reason it works is that the closure of the range is a proper closed subspace (by assumption), hence its orthogonal complement is non-zero. As to your second remark, I had a brainfart. What I should have written is that if $\lambda \in \sigma_{res}(A)$ then $A-\lambda$ is injective (not: it is not injective). If $A-\lambda$ is not injective then $\lambda$ is part of the point spectrum, not the residual spectrum. $\endgroup$
    – s.harp
    May 28, 2020 at 20:56
  • $\begingroup$ 3rd. At the end of the second paragraph you say that the fact that $A-\mu$ is a restriction of $B-\mu$ to a proper subset implies contradicts injectiviy, by the way seems ambiguous at the same time because I don't know wich operator's injectivity you are talking about. Could you please clarify this to me? \\ 4th. Why if $\lambda \in \sigma_{res}A$ and $z\in D(A)$ implies $A-\lambda$ is not injective? Any way, I'm very glad that you spend some of your time answering my question. Thanks a lot! $\endgroup$
    – PAB
    May 28, 2020 at 21:00
  • $\begingroup$ $A-\mu$ is the restriction of $B-\mu$ onto $D(A)$ (from the set $D(B)$). But since $A-\mu : D(A)\to H$ must be surjective and there are points in $D(B)\setminus D(B)$ which must be mapped into $H$, the map $B-\mu : D(B)\to H$ cannot be injective as the image of any point in $D(B)\setminus D(A)$ also admits a pre-image in $D(A)$, hence $B-\mu$ cannot be invertible. For the 4th question there is a finer point I overlooked if $\lambda$ is not real, but this can be remedied and I will edit the question. $\endgroup$
    – s.harp
    May 28, 2020 at 21:06

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