If $-A$ is the Laplacian with Dirichlet boundary conditions on a bounded open subset of $\mathbb R^d$, then $\mathcal D(A^{1/2})=H_0^1(\Lambda)$

Question

Let

• $d\in\mathbb N$
• $\Lambda\subseteq\mathbb R^d$ be bounded and open
• $\mathcal D(A):=\left\{u\in H_0^1(\Lambda):u\text{ admits a weak Laplacian }\Delta u\in L^2(\Lambda)\right\}$ and $$A:=-\Delta u\;\;\;\text{for }u\in\mathcal D(A)$$

$A$ is a densely-defined, positive definite and symmetric operator on $L^2(\Lambda)$.There is an ortonormal basis $(e_n)_{n\in\mathbb N}\subseteq\mathcal D(A)$ of $L^2(\Lambda)$ (which is a complete orthogonal system in $H_0^1(\Lambda)$) with $$Ae_n=\lambda_ne_n\tag1$$ for some $(\lambda_n)_{n\in\mathbb N}\subseteq(0,\infty)$ with $$\lambda_n\le\lambda_{n+1}\;\;\;\text{for all }n\in\mathbb N\tag2\;.$$

Let $$\mathcal D(A^\alpha):=\left\{u\in L^2(\Lambda):\sum_{n\in\mathbb N}\lambda_n^{2\alpha}\langle u,e_n\rangle_{L^2(\Lambda)}^2<\infty\right\}$$ for $\alpha\in\mathbb R$.

Question: How can we show that $\mathcal D(A^{1/2})=H_0^1(\Lambda)$?

Note that $$\left|u\right|:=\left\|\nabla u\right\|_{L^2(\Lambda)}\;\;\;\text{for }u\in H_0^1(\Lambda)$$ is a norm equivalent to $\left\|\;\cdot\;\right\|_{H^1(\Lambda)}$ on $H_0^1(\Lambda)$. Moreover, $$\sum_{n\in\mathbb N}\lambda_n\langle u,e_n\rangle_{L^2(\Lambda)}^2=\sum_{n\in\mathbb N}\langle u,Ae_n\rangle_{L^2(\Lambda)}\langle u,e_n\rangle_{L^2(\Lambda)}\stackrel{(\ast)}=\sum_{n\in\mathbb N}\langle Au,e_n\rangle_{L^2(\Lambda)}\langle u,e_n\rangle_{L^2(\Lambda)}=\langle Au,u\rangle_{L^2(\Lambda)}=\left|u\right|^2\tag 3$$ for all $u\in\mathcal D(A)$.

The problematic thing is that $(\ast)$ doesn't hold, unless $u\in\mathcal D(A)\subseteq\mathcal D(A^{1/2})$. Moreover, I don't see why $u\in\mathcal D(A^{1/2})\setminus\mathcal D(A)$ should admit a weak gradient; by definition we only know that $u\in L^2(\Lambda)$ with $\sum_{n\in\mathbb N}\lambda_n\langle u,e_n\rangle_{L^2(\Lambda)}^2<\infty$. So, what am I missing?

Answer 1

Let $\mathcal D(\mathfrak a)$ and $$\mathfrak a(u,v):=\langle\nabla u,\nabla v\rangle_{L^2(\Lambda,\:\mathbb R^d)}\;\;\;\text{for }u,v\in\mathcal D(\mathfrak a)\;.$$

It's easy to verify that $\mathfrak a$ is a densely-defined, accretive, bounded and closed bilinear form on $H:=L^2(\Lambda)$; cf. [Ouhabaz, Definition 1.4].

Now, let $A:\mathcal D(A)\to H$ be the operator associated to $\mathfrak a$, i.e. $$\mathcal D(A)=\left\{u\in\mathcal D(\mathfrak a):\exists x\in H:\langle x,v\rangle_H=\mathfrak a(u,v)\text{ for all }v\in\mathcal D(\mathfrak a)\right\}$$ and $$\langle Au,v\rangle_H=\mathfrak a(u,v)\;\;\;\text{or all }u\in\mathcal D(A)\text{and }v\in\mathcal D(\mathfrak a)\;;\tag9$$ cf. [Ouhabaz, Definition 1.21].

By definition, $$\mathcal D(A)=\left\{u\in H_0^1(\Lambda):u\text{ admits a weak Laplacian in }L^2(\Lambda)\right\}\;.\tag{10}$$

Let $$\langle x,y\rangle_\alpha:=\langle A^\alpha x,A^\alpha y\rangle_H\;\;\;\text{for }x,y\in\mathcal D(A^\alpha)$$ and $$(u,v)_{\alpha}:=\langle x,y\rangle_H+\langle x,y\rangle_\alpha\;\;\;\text{for }u,v\in\mathcal D(A^\alpha)$$ for $\alpha\in\mathbb R$.

We can show that $$V_\alpha:=\overline{\mathcal D(A^\alpha)}^{\left(\;\cdot\;,\;\cdot\;\right)_{\alpha/2}}=\left(\mathcal D(A^{\alpha/2}),\left(\;\cdot\;,\;\cdot\;\right)_{\alpha/2}\right)\tag{11}$$ for all $\alpha\ge0$.

In the question, we're considering the case $\alpha=1$. By definition, $$\left(u,v\right)_{1/2}=\langle u,v\rangle_{H^1(\Lambda)}\;\;\;\text{for all }u,v\in\mathcal D(A)\tag{12}$$ and hence $$V_1\stackrel{\text{def}}=\:\overline{\mathcal D(A)}^{\left\|\;\cdot\;\right\|_{H^1(\Lambda)}}\tag{13}\;.$$

Now, it's easy to see from $(10)$ that $$C_c^\infty(\Lambda)\subseteq\mathcal D(A)\tag{14}\;.$$ Since $$H_0^1(\Lambda)\stackrel{\text{def}}=\:\overline{C_c^\infty(\Lambda)}^{\left\|\;\cdot\;\right\|_{H^1(\Lambda)}}\tag{15}\;,$$ we obtain $$V_1=H_0^1(\Lambda)\tag{16}$$ by $(13)$ and $(14)$ such that we're able to conclude by $(11)$ and $(12)$.