Let
- $D:=(0,1)$
- $U:=L^2(D)$ and $$\phi_n(x):=\sqrt 2\sin(n\pi x)\;\;\;\text{for }n\in\mathbb N\text{ and }x\in D$$
- $H:=H^2(D)\cap H_0^1(D)$ and $$A:=-\frac{\partial^2}{\partial x^2}u\;\;\;\text{for }u\in H$$
Since $(\phi_n)_{n\in\mathbb N}\subseteq C^\infty(D)$ is an orthonormal basis of $U$ with $$\left.\phi_n\right|_{\partial D}=0\;\;\;\text{for all }n\in\mathbb N\;,$$ $(\phi_n)_{n\in\mathbb N}\subseteq H$ with $$A\phi_n=\underbrace{\pi^2n^2}_{=:\lambda_n}\phi_n\;\;\;\text{for all }n\in\mathbb N\tag 1$$ and $$\left\|A\phi_n\right\|_U\stackrel{(1)}=\lambda_n\stackrel{n\to\infty}\infty\;.\tag 2$$ Thus, $A$ is an unbounded operator.
Now, let $y\in D$ and $G:D\times D\to\mathbb R$ be the solution of $$\left\{\begin{matrix}AG(\;\cdot\;,y)&=&\delta(\;\cdot\;-y)&&\text{in}&D\\ G(\;\cdot\;,y)&=&0&&\text{on}&\partial D\end{matrix}\right.\;.\tag 3$$ We can show that $$$$ $$G(x,y)=\begin{cases}x(1-y)&\text{, if }x<y\\y(1-x)&\text{, if }x\ge y\end{cases}\;.\tag 4$$ Moreover, let $L$ be the integral operator on $U$ with kernel $G$, i.e. $$Lu:=\int_DG(\;\cdot\;,y)u(y)\;{\rm d}\lambda(y)\;\;\;\text{for }u\in U\;.$$ We can show that $L$ is Hilbert-Schmidt.
I want to show that $$ALu=\int_DAG(\;\cdot\;,y)u(y)\;{\rm d}\lambda(y)\;\;\;\text{for all }u\in U\;.\tag 5$$ I'm aware of the rule of differentiation under the integral sign. However, that rule cannot be applied here since it is stated for continuously partially differentiable integrands. So, how can we show $(5)$?