Has the distributional Laplacian $\Delta f:C_c^\infty(\Omega)'\to C_c^\infty(\Omega)'$ a unique extension in $H_0^1(\Omega)'$?

Question

Let

• $d\in\mathbb N$
• $\Omega\subseteq\mathbb R^d$ be open
• $\mathcal D:=C_c^\infty(\Omega)$ and $$H=\overline{\mathcal D}^{\langle\;\cdot\;,\;\cdot\;\rangle_H}\tag 1$$ with $$\langle\phi,\psi\rangle_H:=\langle\phi,\psi\rangle_{L^2(\Omega)}+\langle\nabla\phi,\nabla\psi\rangle_{L^2(\Omega,\:\mathbb R^d)}\;\;\;\text{for }\phi,\psi\in\mathcal D$$

Let $u\in L^1_{\text{loc}}(\Omega)$ be weakly differentiable (with $\nabla u\in L^1_{\text{loc}}(\Omega,\mathbb R^d)$). Then, $v\in L^1_{\text{loc}}(\Omega)$ is called weak Laplacian of $u$ $:\Leftrightarrow$ $$\langle\phi,v\rangle_{L^2(\Omega)}=-\langle\nabla u,\nabla\phi\rangle_{L^2(\Omega,\:\mathbb R^d)}\;\;\;\text{for all }\phi\in\mathcal D\tag 2\;.$$ In that case, we write $\Delta u:=v$.

Let $f\in L^2(\Omega)$. Can we show that $\Delta f$ with $$(\Delta f)(u):=\langle\Delta u,f\rangle_{L^2(\Omega)}\;\;\;\text{for }u\in H_0^1(\Omega)$$ is an element of $H_0^1(\Omega)'$?

Answer 1

let's take $\Omega = [0,1]$, $\varphi,f \in C^\infty_c((0,1))$.

Then $$\langle f'',\varphi \rangle = \int_0^1 f''(x) \varphi (x) dx = f'(1)\varphi (1)-f'(0)\varphi (0) - \int_0^{1} f'(x) \varphi '(x) dx$$

$\varphi,f$ have their support strictly inside $(0,1)$ so $f'(0)\varphi (0) = f'(1)\varphi (1)= 0$ and

$$\langle f'',\varphi \rangle =-\langle f',\varphi' \rangle$$

Then with $\|u\|_{H^1_0} = \|u\|_{L^2}+\|u'\|_{L^2} \ge \|u'\|_{L^2}$ and the Cauchy inequality you have $$|\langle f',\varphi' \rangle| \le \|f'\|_{L^2}\|\varphi'\|_{L^2}\le \|f'\|_{L^2}\|\varphi\|_{H^1_0} \le \|f\|_{H^1_0}\|\varphi\|_{H^1_0}$$

The conclusion is that since $C^\infty_c(\Omega)$ is dense in $H^1_0(\Omega)$, all this stays true for any $\varphi,f \in H^1_0([0,1])$ (this is where $H^1_0$ makes a huge difference to $H^1$)

Replacing $'$ by $\nabla$ it extends to any domain $\Omega \subset \mathbb{R}^d$ and

• $u \mapsto \langle \Delta f,u \rangle$ is a bounded operator $H^1_0(\Omega) \to \mathbb{C}$ whenever $f \in H^1_0(\Omega)$.

• $f \mapsto \langle \Delta f,. \rangle$ is bounded when seen as an operator $H^1_0(\Omega) \to H^1_0(\Omega)'$