Space Sobolev $W^{m,p}$ complete Show that Sobolev space is complete. I am trying
Than $L^p(\Omega)$ is complete then If $f_n \in L^p(\Omega)$ then $f_n \to f \in L^p(\Omega)$. But rest show that $D^{\alpha}f \in L^p(\Omega)$. How I will be able to show this?
 A: Here I present a standard argument.
Assume we know the following three things:


*

*$L^p(\Omega)$ is complete.

*$\{f_n\}$ is Cauchy in $\|\cdot\|_{W^{m,p}}$, this implies $\{f_n\}$ is Cauchy in $\|\cdot\|_{L^p}$, hence $f_n\to f$ for some $f\in L^p$. 

*For any $|\alpha|\leq m$, $\{D^{\alpha}f_n\}$ is Cauchy in $\|\cdot\|_{L^p}$, hence $D^{\alpha}f_n=g_n\to g$ for some $g\in L^p$.
We want to prove the following claim: 

If $\{f_n\}$ is Cauchy in $\|\cdot\|_{W^{m,p}}$, then $ f_n\to f$ for some $f\in W^{m,p}(\Omega).$ 

Basically what is needed to show is that the limit of any weak derivative of the sequence coincides with the weak derivative of the limit in $L^p$, i.e., for any $\alpha$:
  $$g=D^{\alpha}f,$$
so that $D^{\alpha}f \in L^p$ as well.
Proof: First use the definition of weak derivative:
$$
\int_{\Omega} f_n\,D^{\alpha}\phi = (-1)^{|\alpha|}\int_{\Omega} g_n\phi,
$$
for $\phi \in C^{\infty}_c$. Applying the Hölder's inequality gives:
$$
\int_{\Omega} (f_n-f)\,D^{\alpha}\phi \leq \|f_n -f\|_{L^p} \,\|D^{\alpha}\phi\|_{L^q}\to 0,
\\
\text{and }\int_{\Omega} (g_n-g)\phi \leq \|g_n -g\|_{L^p} \,\|\phi\|_{L^q}\to 0.
$$
Above are true because $\|\phi\|_{L^q}$ and $\|D^{\alpha}\phi\|_{L^q}$ are bounded for any smooth test function, and $g_n\to g$ in $L^p$ and $f_n\to f$ in $L^p$. These two limits give us the ability to interchange of the limit and the integral:
$$
\lim_{n\to \infty}\int_{\Omega} (f_n-f)\,D^{\alpha}\phi =0 \implies \lim_{n\to \infty}\int_{\Omega} f_n\,D^{\alpha}\phi =\int_{\Omega} f\,D^{\alpha}\phi.
\\
\lim_{n\to \infty}\int_{\Omega}  (g_n-g)\phi = 0\implies 
\lim_{n\to \infty}\int_{\Omega} g_n\phi =  \int_{\Omega} g\phi .
$$
Therefore: 
$$
\int_{\Omega} f\,D^{\alpha}\phi = \lim_{n\to \infty}\int_{\Omega} f_n\,D^{\alpha}\phi =\lim_{n\to \infty}(-1)^{|\alpha|}\int_{\Omega} g_n\phi = (-1)^{|\alpha|}\int_{\Omega} g\phi 
$$
This is the result we want.
