Self adjoint extensions 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
 A: 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
