# If $A$ is the Laplacian on $H^2(0,1)∩H_0^1(D)$, then the fractional power space $\mathfrak D(A^{r/2})=H_0^r(D)$ for all $r\in\mathbb R$

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 $$A\phi_n=\underbrace{\pi^2n^2}_{=:\lambda_n}\phi_n\;\;\;\text{for all }n\in\mathbb N\tag 1\;.$$

Let $r\in\mathbb R$, $$\mathfrak D(A^r):=\left\{u\in U:A^ru:=\sum_{n\in\mathbb N}\lambda_n^\alpha\langle u,\phi_n\rangle\phi_n\in U\right\}$$ and $$\left\|u\right\|_r:=\left\|A^ru\right\|\;\;\;\text{for }u\in\mathfrak D(A^r)\;.$$ I want to show that $$\left\|u\right\|_r=\sum_{n\in\mathbb N}(n\pi)^{4r}\langle u,\phi_n\rangle^2\tag 2\;.$$ We need somehow to use that $$u=\sum_{n\in\mathbb N}\langle u,\phi_n\rangle\phi_n\;\;\;\text{for all }u\in U\;,$$ but I don't know how I need to proceed in $$\left\|u\right\|_r=\int_D\left|\sum_{n\in\mathbb N}(n\pi)^{2r}\langle u,\phi_n\rangle\phi_n\right|^2{\rm d}\lambda\;\;\;\text{for all }u\in U\;.\tag 3$$

How can we prove $(2)$ and why (and that is the main part of the question) can we conclude that $\mathfrak D(A^{r/2})=H_0^r(D)$?

I've found the last statement in An Introduction to Computational Stochastic PDEs, Example 10.9, and absolutely don't understand why it holds.

• @MoritzDoll You're right! I've confused the symbols used in the mentioned book with mine. Do you know why $\mathfrak D(A^{r/2})=H_0^r(D)$? – 0xbadf00d Feb 24 '16 at 16:26

$\|u\|_r = \|A^r u\| = \|\sum \langle u,ϕ_j\rangle λ_j^r ϕ_j\|$. Now using Parseval (and that the $ϕ_j$ are an orthonormal basis) we get $\|u\|_r = \sum λ_j^{2r}|\langle u, ϕ_j\rangle|^2$.
• That was easy! Do you know why $\mathfrak D(A^{r/2})=H_0^r(D)$? – 0xbadf00d Feb 24 '16 at 16:18
• Apply $A^{r/2}$ again and then use elliptic regularity, ie $∂^2$ is an isomorphisms between $H^s$ and $H^{s-2}$. – mcd Feb 24 '16 at 16:29
• Apply $A^{r/2}$ to what? Unfortunately, I'm not aware of the fact that $\partial^2/\partial x^2$ is an isomorphism between $H^s$ and $H^{s-2}$. Is there an easier proof for $\mathfrak D(A^{r/2})=H_0^r(D)$ if $r\in\mathbb N$? – 0xbadf00d Feb 24 '16 at 16:35
• For even $r$ its clear (because H^{2r} is defined such that $∂^{2r}u ∈ L^2$). For not-even $r$ it is not obvious what $A^{r/2}$ (really) is, so you need some abstract argument. – mcd Feb 24 '16 at 16:42
• If $W^k(D)$ is the space of $k$-times weakly differentiable functions $D\to\mathbb R$, then $$H^{2r}(D)\stackrel{\text{def}}=\left\{u\in W^{2r}(D):\frac{{\rm d}^k}{{\rm d}x^k}u\in U\text{ for all }k\in\left\{0,\ldots,2r\right\}\right\}\;.$$ Could you please explain in more detail why this set is equal to $\mathfrak D(A^r)=H_0^r(D)$, for all $r\in\mathbb N$? – 0xbadf00d Feb 24 '16 at 17:32