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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 $(5)$?

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  • $\begingroup$ Throw a test function on it, $$\begin{aligned}\int Lu(x) A\psi(x)\,dx &= \iint G(x,y)u(y)A\psi(x)\,dy\,dx = \int u(y) \int G(x,y)A\psi(x)\,dx\,dy\\ &= \int u(y) \int AG(x,y)\psi(x)\,dx\,dy = \int u(y)\psi(y)\,dy.\end{aligned}$$ $\endgroup$
    – Daniel Fischer
    Feb 24, 2016 at 12:00
  • $\begingroup$ @DanielFischer You've used that $A$ is symmetric in order to obtain $$\langle ALu,\psi\rangle=\langle Lu,A\psi\rangle\;,$$ right? Then you've applied Fubini's theorem in order to obtain $$\langle Lu,A\psi\rangle=\int_Du(y)\int_DG(x,y)A\psi(x)\;{\rm d}\lambda(x){\rm d}\lambda(y)\;,$$ right? But what have you done to obtain the third equality? Moreover, at the end you've shown that $$\langle ALu,\psi\rangle=\langle u,\psi\rangle$$ for all $\psi\in C_c^\infty(D)$, but why should this imply $(5)$? $\endgroup$
    – 0xbadf00d
    Feb 24, 2016 at 12:39
  • $\begingroup$ I used the definition of distributional derivatives rather than the symmetry. At the end, we see that $u$ and $ALu$ define the same distribution, so $u = ALu$ in that sense, and if you know that $Lu$ is regular enough to apply a semiclassical Laplacian to it, you're done. The third equality is sloppy notation, it's again the definition of distributional derivatives, for the distribution $\psi \mapsto \int G(x,y)\psi(x)\,dx$. $\endgroup$
    – Daniel Fischer
    Feb 24, 2016 at 12:49
  • $\begingroup$ Ah I see, but the symmetry should follow from the definition of the weak derivative. What do you mean by "if you know that Lu is regular enough to apply a semiclassical Laplacian to it" Simply $Lu\in H$ for all $u\in U$? I suppose that you say that this is enough, cause the $u\in U$ are approximated by functions $\psi\in C_c^\infty(D)$ and the identity follows for $u$ by taking the limit. Right? I don't get what you say about the third equality. It seems like that you're using $$G(x,y)A\psi(x)=AG(x,y)\psi(x)\;.$$ $\endgroup$
    – 0xbadf00d
    Feb 24, 2016 at 13:29
  • $\begingroup$ $Lu\in H$ would be sufficient to apply $A$ (which is a semiclassical Laplacian, as opposed to the distributional Laplacian, which I used, despite also denoting it by $A$). Point is, my calculation gives $-\Delta (Lu) = u$ in the sense of distributions. But distributional derivatives and more classical derivatives (e.g. weak derivatives) are compatible in the sense that if the more classical derivative exists (and gives a locally integrable function), then that is also the distributional derivative. So "classical exists and distributional gives foo implies classical gives foo". $\endgroup$
    – Daniel Fischer
    Feb 24, 2016 at 13:40

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