# Definition of the Laplacian as an operator from $H_0^1(\Omega)$ to $H_0^1(\Omega)'$

Let

• $d\in\mathbb N$
• $\Omega\subseteq\mathbb R^d$ be open
• $\mathcal D(\Omega):=C_c^\infty(\Omega)$
• $f\in L^2(\Omega)$ and $$\langle f\rangle:=\left.\langle\;\cdot\;,f\rangle_{L^2(\Omega)}\right|_{\mathcal D(\Omega)}\in\mathcal D'(\Omega)$$
• $i\in\left\{1,\ldots,d\right\}$

Since $\mathcal D(\Omega)$ is dense in the Sobolev space $H_0^1(\Omega)$, it's easy to see that the distributional derivative $$\frac{\partial\langle f\rangle}{\partial x_i}(\phi)=-\langle\frac{\partial\phi}{\partial x_i},f\rangle_{L^2(\Omega)}\;\;\;\text{for all }\phi\in\mathcal D(\Omega)\tag 1$$ has a unique extension $\overline{\frac{\partial\langle f\rangle}{\partial x_i}}\in H_0^1(\Omega)'$ to $H$ with $$\overline{\frac{\partial\langle f\rangle}{\partial x_i}}(u)=-\langle\frac{\partial u}{\partial x_i},f\rangle_{L^2(\Omega)}\;\;\;\text{for all }u\in H_0^1(\Omega)\;,\tag 2$$ where $\frac{\partial u}{\partial x_i}$ denotes the weak derivative. The mapping $$\overline{\frac\partial{\partial x_i}}:L^2(\Omega)\to H_0^1(\Omega)'\tag 3$$ is linear and continuous.

Now, I've read that we can define $$\Delta:=\sum_{j=1}^d\overline{\frac\partial{\partial x_i}}\:\overline{\frac\partial{\partial x_i}}\tag 4$$ as a continuous operator $H_0^1(\Omega)\to H_0^1(\Omega)'$. Why is the composition of the operators in $(4)$ well-defined? That's not clear to me, cause each operator is $H_0^1(\Omega)'$-valued and takes values from $L^2(\Omega)$.

I've noticed that $$\overline{\frac{\partial\langle f\rangle}{\partial x_i}}=\left.\langle\;\cdot\;,\frac{\partial f}{\partial x_i}\rangle_{L^2(\Omega)}\right|_{L^2(\Omega)}\;,$$ if $f\in H_0^1(\Omega)$. So, maybe $\overline{\frac{\partial\langle f\rangle}{\partial x_i}}$ is identified with the weak derivative $\frac{\partial f}{\partial x_i}$ (which is an element of $L^2(\Omega)$ in that case.

• you are over-complicating things. Start with $\Omega = [0,1]$ and integrate by parts $\int_0^1 f'(x) g'(x) dx$, see the 4 lines answer below. – reuns Jul 10 '16 at 19:06

## 1 Answer

One can define $\Delta$ in $H^1$ because $\Delta$ is in divergence form

$$\Delta u = \text{div}\nabla u$$

Thus we define $\Delta : H^1_0(\Omega) \to H^1_0(\Omega)^*$ by

$$\Delta f (\phi) := -\int_\Omega \nabla f \cdot \nabla \phi.$$

The right hand side is well defined for $f , \phi \in H^1_0(\Omega)$. If $f$ is in $H^2_0$, then integration by part gives

$$\Delta f(\phi) = \int_\Omega (\Delta f) \phi$$

• why $H^2_0$ ? $\int_\Omega \nabla f . \nabla \phi$ is well-defined whenever $f \in H^1, g \in H^1$, and if $f,g\in H^1_0$ it is equal to $\int_\Omega \phi \Delta f$, no ? – reuns Jul 10 '16 at 19:08
• To make sense of $\Delta f$ in $\int_\Omega \phi \Delta f$, , you need some sort of twice differentiability @user1952009 – user99914 Jul 10 '16 at 19:10
• not if by definition $\int_\Omega \phi \Delta f = -\int_\Omega \nabla \phi . \nabla f$ ? (the distributional derivative), but for showing it makes sense we need to show it is valid on a dense subset of $H^1_0$ (for example $C^\infty_c$) or something like that – reuns Jul 10 '16 at 19:11
• As distribution for all $f\in L^2$ you can define $\Delta f$ as a distribution $\Delta f(\phi) = \int f \Delta \phi$. But this $\Delta f$ is not in $H^1_0(\Omega)^*$ @user1952009 – user99914 Jul 10 '16 at 19:14
• that's my question, that it is in $(H^1_0(\Omega))^*$ depends on if $\int_\Omega \phi \Delta f = -\int_\Omega \nabla \phi . \nabla f$ – reuns Jul 10 '16 at 19:15