The Sobolev Space $W^{1,p}(I)$ is complete

Question

I am trying to verify that Sobolev Space $W^{1,p}(I)$ is complete. This is the definition of $W^{1,p}(I)$:

$W^{1,p}(I)= \{ u\in L^{p}(I) | \exists g \in L^p(I) :\int_{I}^{}u\varphi'= - \int_{I}^{}g \varphi \} \subset L^p(I)$

For every $\varphi \in C^{1}_c(I)$.

With the norm $||u||=||u||_{L^p}+||u'||_{L^p}$.

We call $g$=$u'$, the weak derivative of $u$.

For a Cauchy sequence $(u_n)$ in $W^{1,p}(I)$ we need to show that $(u_n) \rightarrow u \in W^{1,p}(I)$. If $(u_n)$ is a Cauchy sequence in $W^{1,p}(I)$ then it is obvious both $(u_n)$ and $(u'_n)$ are Cauchy sequences, from the definition of the norm $||·||$, so both the limits $u$ of $(u_n)$ and $\hat{u}$ of $(u'_n)$ are in $L^p(I)$.

We need to show that $ \hat{u}=u'$. I don't know how to proceed now. It is clear that:

$$\int_{I}^{}u_n\varphi'= - \int_{I}^{}u'_n \varphi $$

And that what I need is:

$$\int_{I}^{}u\varphi'= - \int_{I}^{} \hat{u} \varphi$$

So I need to take the limit, but How can I put the limit inside the integral so that the last equation holds?

Answer 1

Use Hölder's inequality to get $$\left|\int_{I}^{}u_n\varphi'-\int_{I}^{}u\varphi'\right|\leqslant \int_{I}^{}\left|(u_n-u)\varphi'\right| \leqslant \lVert u_n-u\rVert_p\rVert \varphi'\rVert_{p'},$$ where $p'$ is the conjugate exponent of $p$. Since $\varphi'$ is continuous and compactly supported, its norm in $\mathbb L^{p'}$ is finite. From the convergence of $u_n\to u$ in $\mathbb L^p$, we derive that $\lim_{n\to \infty}\int_{I}^{}u_n\varphi'=\int_{I}^{}u\varphi'$. Use a similar argument to prove the convergence of $\int_{I}^{}u'_n \varphi$ to $\int_{I}^{}\widehat u \varphi$.

Answer 2

Since the $L^{p}(I)$ space is complete, there exist functions $u$ and $\hat{u}$ s.t. $u_{n}\rightarrow u$ and $u_{n}'\rightarrow \hat{u}$ in $L^{p}(I)$. We have (for $\varphi\in C_{0}^{\infty}(I)$): $$ \int_{I}u'\varphi=-\int_{I}u\varphi'=-\lim_{n\rightarrow\infty}\int_{I}u_{n}\varphi'\\ =\lim_{n\rightarrow\infty}\int_{I}u_{n}'\varphi=\int_{I}\hat{u}\varphi. $$ Therefore, $u'=\hat{u}$.