[proofplan]
Write the non-real spectral parameter as $\lambda=a+ib$ with $a,b\in\mathbb R$ and $b\ne0$, and define the shifted self-adjoint operator $B:=A-aI$. The key estimate is the exact identity $\|(B-ibI)x\|_H^2=\|Bx\|_H^2+b^2\|x\|_H^2$, whose mixed terms cancel because $(Bx,x)_H$ is real. This gives a uniform lower bound for $A-\lambda I$, hence injectivity, closed range, and the inverse norm estimate on the range. The same estimate applied to the adjoint $A-\overline{\lambda}I$ shows that the range has zero orthogonal complement, so the closed range is all of $H$.
[/proofplan]
[step:Shift the operator by the real part of $\lambda$]
Fix $\lambda\in\mathbb C\setminus\mathbb R$. Define $a:=\operatorname{Re}\lambda\in\mathbb R$ and $b:=\operatorname{Im}\lambda\in\mathbb R\setminus\{0\}$, so that $\lambda=a+ib$. Define
\begin{align*}
B:=A-aI\in\mathcal L(H).
\end{align*}
Since $A$ is self-adjoint and $a\in\mathbb R$, the operator $B$ is self-adjoint:
\begin{align*}
B^*=(A-aI)^*=A^*-aI=A-aI=B.
\end{align*}
Also,
\begin{align*}
A-\lambda I=B-ibI.
\end{align*}
[/step]
[step:Compute the lower bound from self-adjointness]
For every $x\in H$, expand the norm using the Hilbert-space [inner product](/page/Inner%20Product), which is linear in the first argument:
\begin{align*}
\|(B-ibI)x\|_H^2=(Bx-ibx,Bx-ibx)_H.
\end{align*}
By sesquilinearity,
\begin{align*}
\|(B-ibI)x\|_H^2=\|Bx\|_H^2+ib(Bx,x)_H-ib(x,Bx)_H+b^2\|x\|_H^2.
\end{align*}
Since $B$ is self-adjoint,
\begin{align*}
(Bx,x)_H=(x,Bx)_H.
\end{align*}
Hence the two mixed terms cancel, and therefore
\begin{align*}
\|(A-\lambda I)x\|_H^2=\|(B-ibI)x\|_H^2=\|Bx\|_H^2+b^2\|x\|_H^2.
\end{align*}
It follows that
\begin{align*}
\|(A-\lambda I)x\|_H\geq |b|\|x\|_H=|\operatorname{Im}\lambda|\|x\|_H.
\end{align*}
[guided]
We isolate the real part of $\lambda$ because subtracting a real scalar from a self-adjoint operator preserves self-adjointness. Set $a:=\operatorname{Re}\lambda$, $b:=\operatorname{Im}\lambda$, and $B:=A-aI$. Then $b\ne0$ because $\lambda\notin\mathbb R$, and
\begin{align*}
A-\lambda I=A-(a+ib)I=B-ibI.
\end{align*}
The operator $B$ is self-adjoint because
\begin{align*}
B^*=(A-aI)^*=A^*-aI=A-aI=B,
\end{align*}
where we used $A^*=A$ and $a=\overline a$.
Now fix $x\in H$. The important point is that the imaginary scalar perturbation contributes a strictly positive term $b^2\|x\|_H^2$ and no real mixed term. Expanding with the convention that $(\cdot,\cdot)_H$ is linear in the first argument gives
\begin{align*}
\|(B-ibI)x\|_H^2=(Bx-ibx,Bx-ibx)_H.
\end{align*}
By sesquilinearity,
\begin{align*}
\|(B-ibI)x\|_H^2=\|Bx\|_H^2+ib(Bx,x)_H-ib(x,Bx)_H+b^2\|x\|_H^2.
\end{align*}
Self-adjointness gives $(Bx,x)_H=(x,Bx)_H$, so the two mixed terms are equal with opposite coefficients:
\begin{align*}
ib(Bx,x)_H-ib(x,Bx)_H=0.
\end{align*}
Thus
\begin{align*}
\|(A-\lambda I)x\|_H^2=\|Bx\|_H^2+b^2\|x\|_H^2.
\end{align*}
Since $\|Bx\|_H^2\ge0$, we obtain
\begin{align*}
\|(A-\lambda I)x\|_H^2\ge b^2\|x\|_H^2.
\end{align*}
Taking square roots gives
\begin{align*}
\|(A-\lambda I)x\|_H\ge |b|\|x\|_H=|\operatorname{Im}\lambda|\|x\|_H.
\end{align*}
This is the decisive estimate: it says that $A-\lambda I$ cannot send a nonzero vector close to zero relative to its norm.
[/guided]
[/step]
[step:Use the lower bound to obtain injectivity and closed range]
Let
\begin{align*}
L:=A-\lambda I\in\mathcal L(H).
\end{align*}
By the kernel and range definitions from the theorem statement,
\begin{align*}
\ker(L)=\{x\in H:Lx=0\}
\end{align*}
and
\begin{align*}
\operatorname{Range}(L)=\{Lx:x\in H\}.
\end{align*}
The estimate from the previous step gives
\begin{align*}
\|Lx\|_H\geq |\operatorname{Im}\lambda|\|x\|_H
\end{align*}
for every $x\in H$. If $Lx=0$, then
\begin{align*}
0\geq |\operatorname{Im}\lambda|\|x\|_H.
\end{align*}
Since $|\operatorname{Im}\lambda|>0$, this implies $x=0$. Hence $L$ is injective.
We next show that $\operatorname{Range}(L)$ is closed. Let $(y_n)_{n=1}^{\infty}$ be a sequence in $\operatorname{Range}(L)$ converging in $H$ to some $y\in H$. For each integer $n\geq 1$, choose $x_n\in H$ such that $y_n=Lx_n$. For integers $m,n\geq 1$,
\begin{align*}
\|x_n-x_m\|_H\leq \frac{1}{|\operatorname{Im}\lambda|}\|Lx_n-Lx_m\|_H=\frac{1}{|\operatorname{Im}\lambda|}\|y_n-y_m\|_H.
\end{align*}
Since $(y_n)_{n=1}^{\infty}$ converges, it is Cauchy; hence $(x_n)_{n=1}^{\infty}$ is Cauchy. Because $H$ is complete, there exists $x\in H$ such that $x_n\to x$ in $H$. Since $L$ is bounded, $Lx_n\to Lx$ in $H$. But $Lx_n=y_n\to y$, so [uniqueness of limits](/theorems/625) gives $y=Lx$. Thus $y\in\operatorname{Range}(L)$, and $\operatorname{Range}(L)$ is closed.
[/step]
[step:Show the range is dense by applying the same estimate to the adjoint]
Since $A$ is self-adjoint,
\begin{align*}
L^*=(A-\lambda I)^*=A-\overline{\lambda}I.
\end{align*}
The number $\overline{\lambda}$ is also non-real, and
\begin{align*}
|\operatorname{Im}\overline{\lambda}|=|\operatorname{Im}\lambda|.
\end{align*}
Applying the lower bound already proved with $\overline{\lambda}$ in place of $\lambda$ gives
\begin{align*}
\|L^*y\|_H\geq |\operatorname{Im}\lambda|\|y\|_H
\end{align*}
for every $y\in H$. Therefore $\ker(L^*)=\{0\}$.
Let $y\in\operatorname{Range}(L)^\perp$. Then for every $x\in H$,
\begin{align*}
0=(Lx,y)_H=(x,L^*y)_H.
\end{align*}
Taking $x=L^*y$ gives
\begin{align*}
\|L^*y\|_H^2=0,
\end{align*}
so $L^*y=0$. Since $\ker(L^*)=\{0\}$, we get $y=0$. Hence
\begin{align*}
\operatorname{Range}(L)^\perp=\{0\}.
\end{align*}
We justify density directly. Suppose $z\in H$ satisfies $z\in\overline{\operatorname{Range}(L)}^\perp$. Since $\operatorname{Range}(L)\subset\overline{\operatorname{Range}(L)}$, we have $z\in\operatorname{Range}(L)^\perp$, and hence $z=0$. Thus $\overline{\operatorname{Range}(L)}^\perp=\{0\}$. Because $\overline{\operatorname{Range}(L)}$ is a closed subspace of $H$, the Hilbert-space [orthogonal decomposition](/theorems/436) gives
\begin{align*}
H=\overline{\operatorname{Range}(L)}\oplus \overline{\operatorname{Range}(L)}^\perp.
\end{align*}
Since the orthogonal complement is trivial, this decomposition forces $\overline{\operatorname{Range}(L)}=H$. Since $\operatorname{Range}(L)$ is closed, it follows that
\begin{align*}
\operatorname{Range}(L)=H.
\end{align*}
[/step]
[step:Conclude invertibility and the resolvent estimate]
The operator $L=A-\lambda I$ is injective and has range $H$, so it is bijective. Its inverse
\begin{align*}
L^{-1}:H\to H
\end{align*}
is defined on all of $H$. For any $y\in H$, write $x:=L^{-1}y$, so that $y=Lx$. The lower bound gives
\begin{align*}
\|x\|_H\leq \frac{1}{|\operatorname{Im}\lambda|}\|Lx\|_H=\frac{1}{|\operatorname{Im}\lambda|}\|y\|_H.
\end{align*}
Taking the supremum over all $y\in H$ with $\|y\|_H\leq1$ gives
\begin{align*}
\|(A-\lambda I)^{-1}\|_{\mathcal L(H)}=\|L^{-1}\|_{\mathcal L(H)}\leq \frac{1}{|\operatorname{Im}\lambda|}.
\end{align*}
By the definition of the resolvent set $\rho(A)$ from the theorem statement, this says that every $\lambda\in\mathbb C\setminus\mathbb R$ belongs to $\rho(A)$. Since the spectrum is defined by $\sigma(A):=\mathbb C\setminus\rho(A)$, equivalently,
\begin{align*}
\sigma(A)\subset\mathbb R.
\end{align*}
[/step]
