Second derivative operator closed on $AC^2[0,1]$

Question

Consider the Hilbert space $\mathcal H=L^2(0,1)$ and let $D:=AC^2[0,1]$ be the space of functions $f\in C^1[0,1]$ such that $f'$ is absolutely continuous and the weak derivative $f''$ is in $ L^{2}(0,1)$. Define the operator $$T:D\to \mathcal H; \, T= -\frac{d^2}{dx^2}.$$
I want to prove that $T$ is closed, i.e. the graph of $T, \Gamma(T)$ is closed. To that end, let $(\varphi, \psi) \in \overline{\Gamma(T)}$ and $(\varphi_n)_{n\in \mathbb N}$ such that $\varphi_n \stackrel{n\to \infty}{\longrightarrow} \varphi$ and $T\varphi_n \stackrel{n\to \infty}{\longrightarrow} \psi$ in $L^2$. I need to prove $\varphi \in D$ and $T\varphi = \psi$. Here I am stuck. I tried a lot playing with integration by parts and other stuff. I am also not sure what I can say about the first derivative of $\varphi$. Any help appreciated!

Answer 1

The operator $Tf = -f''$ subject to $f(0)=f(1)=0$ has an inverse given by $$ Sf=(1-x)\int_{0}^{x}f(t)tdt+x\int_{x}^{1}f(t)(1-t)dt. $$ To see this, note that, for all $f \in L^2$, the Lebesgue differentiation theorem shows that $Sf$ is absolutely continuous on $[0,1]$, vanishes at $x=0,1$, and satisfies the following a.e.: \begin{align} (Sf)' &= (1-x)f(x)x-\int_0^xf(t)tdt-xf(x)(1-x)+\int_x^1f(t)(1-t)dt \\ &= -\int_0^x f(t)tdt+\int_x^1f(t)(1-t)dt. \end{align} Therefore, $(Sf)'$ is absolutely continuous with $(Sf)'\in L^2$ and $$ (Sf)'' = -f(x)x-f(x)(1-x)=-f(x) \;\;\; a.e.. $$ Therefore $Sf\in\mathcal{D}(T)$ for all $f\in L^2$, and $TSf=f$ a.e., which means that $T$ is a left inverse of $S$. And it can be directly shown that $STf=f$ for all $f\in \mathcal{D}(T)$. So $S=T^{-1}$. And $S : L^2[0,1]\rightarrow AC^2[0,1]$ is bounded on $L^2[0,1]$, which implies that $S$ is closed. So $T=S^{-1}$ is also closed because of how the graphs of $S$ and $T$ are related.

Consider $T_ef=-f''$ on the domain of $T$ except without the endpoint condition requirements. The point evaluations $E_0 f=f(0)$ and $E_1 f=f(1)$ are continuous linear functionals on the graph of $T_e$. For example, consider \begin{align} \langle T_e f,g\rangle -\langle f,T_e g\rangle &= -\int_0^1 (f''g-fg'')dx \\ & = -(f'g-fg')|_{0}^{1}. \end{align} By choosing $g$ carefully, you can represent point evaluations of $f,f'$ at $0$ or $1$ as continuous linear functionals on the graph of $T_e$. So $E_0,E_1$ are continuous on the graph of $T_e$. Then, for $f\in\mathcal{D}(T_e)$, one has $$ f-E_0(f)x-E_1(f)(1-x) \in\mathcal{D}(T) $$ and \begin{align} T_e f = T(f- &E_0(f)x-E_1(f)(1-x))\\ +&E_0(f)T_e x+E_1(f)T_e(1-x) \\ = S^{-1}(f- &E_0(f)x-E_1(f)(1-x)) \\ +&E_0(f)T_e x+E_1(f)T_e(1-x) \end{align} Hence $T_e$ is closed because $S^{-1}$ is closed and $E_0,E_1$ are bounded on the graph of $T_e$.