Invertibility of second derivative operator 
Let $H$ be the Sobolev space of twice $L^2$-differentiable functions, and define the operator $T:H\to L^2(\mathbb R/2\pi\mathbb Z)$ by
$$
 T=-\frac{d^2}{dx^2}.
 $$

*

*Prove that if $\lambda\in\mathbb C$ is not the square of any integer, then the map $\lambda I-T$ is invertible.

*In particular, prove that if also $\lambda<0$ then $\|(\lambda I-T)^{-1}\|\leq1/|\lambda|$.


I'm not sure how to attempt this problem. I began by writing the definitions I know:

*

*$L^2(\mathbb R/2\pi\mathbb Z)$ is the space of square integrable, $2\pi$-periodic functions

*$h_2\subset\ell^2(\mathbb Z)$ is the space of sequences $(x_n)_{n\in\mathbb Z}$ such that $\sum_{n\in\mathbb Z}(1+n^2)^2|x_n|^2<\infty$.

*$F:L^2(\mathbb R/2\pi\mathbb Z)\to\ell^2(\mathbb Z)$ is the Fourier map defined by

$$
Ff=\left(\langle f(x),(2\pi)^{-1/2}e^{inx}\rangle_{L^2}\right)_{n\in\mathbb Z}=\left(\frac1{\sqrt{2\pi}}\int f(x)e^{-inx}dx\right)_{n\in\mathbb Z}
$$

*

*$H\subset L^2(\mathbb R/2\pi\mathbb Z)$ is the function space defined by $H=\{f\in L^2(\mathbb R/2\pi\mathbb Z):Ff\in h_2\}$.

Help or hints on this question? Thanks!
 A: Suppose that $g=(\lambda I-T)f$. This means that
$$g=\lambda f+f'',\ f(0)=f(2\pi),\ f'(0)=f'(2\pi).$$
Then
$$
f(x)=c_1\,e^{i\,\sqrt{\lambda}\,x}+c_2\,e^{-i\,\sqrt{\lambda}\,x}-\frac1{2\lambda}\,\int_0^{x}g(t)\,e^{i\,\sqrt{\lambda}\,(t-x)}\,dt-\frac1{2\lambda}\,\int_x^{2\pi}g(t)\,e^{-i\,\sqrt{\lambda}\,(t-x)}\,dt.
$$
The boundary conditions give two linear equations in $c_1$ and $c_2$, of the form
\begin{align}
(1-e^{2\pi i\sqrt\lambda})\,c_1+(1-e^{-2\pi i \sqrt\lambda})\,c_2&=\text{stuff}\\
i\sqrt\lambda\,(1-e^{2\pi i\sqrt\lambda})\,c_1-i\sqrt\lambda\,(1-e^{-2\pi i \sqrt\lambda})\,c_2&=\text{other stuff}\\
\end{align}
For this to have unique solution, we need the determinant to be nonzero, that is
$$
-2i\sqrt\lambda\,(1-e^{2\pi i\sqrt\lambda})(1-e^{-2\pi i \sqrt\lambda})\ne0.
$$
This expression is $0$ precisely when $\sqrt\lambda$ is an integer, that is to say when $\lambda$ is the square of an integer. When $\lambda$ is not the square of an integer, we have found that $\lambda I -t$ is invertible and that
\begin{align}
(\lambda I-T)^{-1}g&=\frac{e^{i\sqrt\lambda x}}{4\lambda(1-e^{2\pi i \sqrt\lambda})}\biggl(\int_0^{2\pi}g(t)\,(\cos\sqrt\lambda(t-2\pi)-i\sin\sqrt\lambda t)\,dt\biggr)\\[0.3cm]
&+\frac{e^{-i\sqrt\lambda x}}{4\lambda(1-e^{-2\pi i \sqrt\lambda})}\biggl(\int_0^{2\pi}g(t)\,(\cos\sqrt\lambda(t-2\pi)-i\sin\sqrt\lambda t)\,dt\biggr)\\[0.3cm]
&-\frac1{2\lambda}\,\int_0^{x}g(t)\,e^{i\,\sqrt{\lambda}\,(t-x)}\,dt-\frac1{2\lambda}\,\int_x^{2\pi}g(t)\,e^{-i\,\sqrt{\lambda}\,(t-x)}\,dt.
\end{align}
Unless I've made a mistake (as often happens), an estimate like the one you want seems obvious when $\lambda>0$. In that case,
$$
\|(\lambda I-T)^{-1}g\|^2\leq\frac1{\lambda^2}\,\int_0^{2\pi}\left|\int_0^{2\pi}|g(t)|\,dt\right|^2\,dx\leq\frac{4\pi}\lambda^2\,\|g\|^2,
$$
so
$$
\|(\lambda I-T)^{-1}\|\leq\frac{2\pi}\lambda. 
$$
I used fairly crude estimates, so probably something better can be achieved.
A: Let $Tf=if'$ on $\mathcal{D}(T)$ consisting of all absolutely continuous functions $f : [0,2\pi]\rightarrow\mathbb{C}$ with $f'\in L^2[0,2\pi]$, and suppose $g\in L^2[0,2\pi]$ is in the domain of $\mathcal{D}(T^*)$. Then $h=T^*g$ is in $L^2[0,2\pi]$ and satisfies
$$
            \int_0^{2\pi}if'(\theta)\overline{g(\theta)}d\theta = \int_0^{2\pi}f(\theta)\overline{h(\theta)}d\theta,\;\;\; h\in\mathcal{D}(T).
$$
For example, let $f$ be $0$ on $[0,x-h]$, be $1$ on $[x,y]$, and be $0$ on $[y+k,2\pi]$, and be linear on $[x-h,x]$ and $[y,y+k]$ so that $f$ is continuous on $[0,2\pi]$. Then
$$
                 f'=\frac{1}{h}\chi_{[x-h,x]}-\frac{1}{k}\chi_{[y,y+k]}
$$
and
$$
         \langle if',g\rangle=\langle f,ig'\rangle
$$
Letting $h,k\downarrow 0$ gives
$$
    \lim_{h,k\downarrow 0} \left(  \frac{1}{h}\int_{x-h}^{x}g(t)-\frac{1}{k}\int_{y}^{y+k}g(t)dt \right)=-i\int_x^y f(t)dt, \\
            g(x)-g(y)=-i\int_x^y g(t)dt
$$
Therefore, $g$ is absolutely continuous with $g'=if$ a.e. on $[0,2\pi]$. So $T$ is self-adjoint on $\mathcal{D}(T)$.
The estimate that you want is a general consequence of having a positive self-adjoint operator $T$, meaning that $T^*=T$ and $\langle Tx,x\rangle \ge 0$ for all $x\in\mathcal{D}(T)$. For $\lambda < 0$, and $x\in\mathcal{D}(T)$,
$$
             \langle (T-\lambda I)x,x\rangle \ge -\lambda\langle x,x\rangle \\
          (-\lambda)\|x\|^2 \le \|(T-\lambda I)x\|\|x\| \\
             (-\lambda)\|x\| \le \|(T-\lambda I)x\|.
$$
Setting $x=(T-\lambda I)^{-1}y$ gives
$$
             \|(T-\lambda I)^{-1}y\| \le \frac{1}{(-\lambda)}\|y\|.
$$
Therefore, $\|(T-\lambda I)^{-1}\| \le 1/(-\lambda)$ for $\lambda\in\mathbb{R}$, $\lambda < 0$.
