Given $\Omega$ an open in $\mathbb{R}^n$ and $m\in \mathbb{N}$, the sobolev space of order $m$ is defined as $$H^m(\Omega)=\{u\in D'(\Omega): \partial^{\alpha}u\in L^2(\Omega) \forall \alpha, |\alpha|\leq m \}$$ which is provided with the norm $||u||_{H^m(\Omega)}=\{\sum_{|\alpha|\leq m}||\partial^{\alpha}u||^2_{L^2(\Omega)}\}^{1/2}$. Assume that $L^2(\Omega)$ with the usual scalar product $\left \langle u,v \right \rangle=\int_{\Omega}uv dx$ is a Hilbert space, and show that $H^m(\Omega)$ also it is.
I have tried to do the following:
Let's take $\{u_n\}_n\subseteq H^m(\Omega)$ a Cauchy sequence, so we can define $\{\partial^\alpha u_n\}_n$, which is a Cauchy sequence in $L^2(\Omega)$ (This follows from the fact that $\{u_n\}_n\subseteq H^m(\Omega)$ is a Cauchy sequence), so, as $L^2(\Omega)$ is a Banach space, then there is $u\in L^2(\Omega)$ such that $\lim_{n\to \infty} \partial^\alpha u_n=u $. I suspect $\lim_{n\to \infty} u_n=\partial^{\alpha}u $ but I don't know how to prove this, I also don't know how to prove $\partial^{\alpha}u\in H^m(\Omega)$, any ideas? Thank you!
$u_{n}\rightarrow u$ in $L^{2}$. Now we claim that $\partial^{\alpha}u$ exists for $|\alpha|\leq m$ and $\partial^{\alpha}u_{n}\rightarrow\partial^{\alpha}u$ in $L^{2}$.
For $\varphi\in\mathcal{D}$, we have \begin{align*} \left<\partial^{\alpha}u_{n},\varphi\right>&=(-1)^{|\alpha|}\left<u_{n},\partial^{\alpha}\varphi\right>\\ &\rightarrow(-1)^{|\alpha|}\left<u,\partial^{\alpha}\varphi\right> \end{align*} since $u_{n}\rightarrow u$ in $L^{2}$. On the other hand, we know that $\partial^{\alpha}u_{n}\rightarrow v_{\alpha}$ in $L^{2}$, this gives \begin{align*} \left<\partial^{\alpha}u_{n},\varphi\right>\rightarrow\left<v_{\alpha},\varphi\right>, \end{align*} so $\left<v_{\alpha},\varphi\right>=(-1)^{|\alpha|}\left<u,\partial^{\alpha}\varphi\right>$, this shows that $v_{\alpha}=\partial^{\alpha}u$. Note that this also means that $\partial^{\alpha}u_{n}\rightarrow\partial^{\alpha}u$ in $L^{2}$.