# 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?

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If $f_n$ is Cauchy in sobolev space $W^{m,p}$, then for each $|\alpha|\le m$, $D^{\alpha}f_n$ is Cauchy in $L^p$ and converges to some $f_{\alpha}\in L^p$. The Crucial thing is to show $f_{\alpha}=D^{\alpha}f$. – Hui Yu Nov 8 '12 at 16:27

Here I present a standard argument.

Assume we know the following three things:

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

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

3. 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.

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@Shushao the conclude steps are: Than $$\int_{\Omega} f D^{\alpha}\phi=(-1)^{|\alpha|}\int_{\Omega}g\phi$$ I will be able to say that $g=D^{\alpha}f$? Why? – user46060 Nov 8 '12 at 19:25
@user46060 This is exactly the definition of weak derivative of $f$. – Shuhao Cao Nov 8 '12 at 19:30
understand, and the previous steps $$\int_{\Omega} (f_n-f)\,D^{\alpha}\phi \leq \|f_n -f\|_{L^p} \,\|D^{\alpha}\phi\|_{L^q}\to 0$$ $$\int_{\Omega} (g_n-g)\phi \leq \|g_n -g\|_{L^p} \,\|\phi\|_{L^q}\to 0$$ Why wrote? – user46060 Nov 8 '12 at 19:33
@user46060 Please see my edit. – Shuhao Cao Nov 8 '12 at 19:49
@user46060 $\displaystyle\int_{\Omega} (f_n -f)D^{\alpha} \phi \to 0$ is equivalent to say $\displaystyle\int_{\Omega} f D^{\alpha}\phi= \lim\limits_{n\to\infty} \displaystyle\int_{\Omega} f_n D^{\alpha}\phi$, otherwise if we don't have this, the last step would be invalid. – Shuhao Cao Nov 8 '12 at 22:57