Sobolev Spaces are Banach Spaces: A Proof

Theorem¹

The Sobolev space $W^{m, p}$ is a Banach space.

Description

A Banach space is defined as a space where a norm is defined and is complete. As the norm is also defined when defining the Sobolev space, we only need to verify that it is complete. Therefore, it suffices to show that the Cauchy sequence within $W^{m, p}$ converges within $W^{m, p}$. The proof is relatively straightforward.

Proof

Let $\Omega \subset \mathbb{R}^{n}$ be an open set. Let $\left\{ u_{n} \right\}$ be a Cauchy sequence within $W^{m, p}$.

Definition of Sobolev Space

$$W^{m, p}(\Omega):=\left\{ u \in L^{p}(\Omega) : D^\alpha u \in L^{p}(\Omega),\ 0\le |\alpha | \le m \right\}$$

Here, $D^\alpha u$ is the weak derivative of $u$.

Then, by the definition of $W^{m, p}$, $\left\{ D^\alpha u_{n} \right\}$ is a Cauchy sequence in the $L^{p}$ space with respect to $0\le |\alpha| \le m$. Since $L^{p}$ is a complete space, both Cauchy sequences converge. Let these limits be $u$ and $u_\alpha$, respectively.

$$u_{n} \rightarrow u \quad \mathrm{in}\ L^{p}$$

$$D^\alpha u_{n} \rightarrow u_\alpha\quad \mathrm{for}\ 0\le |\alpha | \le m \quad \mathrm{in} \ L^{p}$$

Moreover, $u_{n} \in L^{p}(\Omega) \subset L_{\text{loc}}^{1}(\Omega)$ is locally integrable, so there exists the corresponding distributional $T_{u_{n}} \in D^{\prime}(\Omega)$.

$$T_{u_{n}}(\phi)=\int_{\Omega} u_{n}(x)\phi (x)dx,\quad \phi \in D(\Omega)$$

Then,

$$\left| T_{u_{n}}(\phi)-T_{u}(\phi) \right| \le \int |u_{n}(x)-u(x)||\phi (x)|dx \le \|u_{n}-u\|_{p}\ \|\phi\|_{p^{\prime}}$$

The first inequality is due to the properties of absolute values, and the second inequality holds by the Hölder’s inequality. $p^{\prime}$ is the conjugate exponent of $p$. And since $u_{n} \rightarrow u$, the above expression converges to $0$.

$$\begin{equation} T_{u_{n}}(\phi) \rightarrow T_{u}(\phi), \quad \forall \phi\in D(\Omega)\ \mathrm{as}\ n\rightarrow \infty \end{equation}$$

In the same manner, it can be verified that the equation below holds as well.

$$\begin{equation} T_{D^\alpha u_{n}}(\phi) \rightarrow T_{u_\alpha}(\phi) \end{equation}$$

Now, let’s compute $T_{u_\alpha}$.

$$\begin{align*} T_{u_\alpha}(\phi) =&\ \lim \limits_{n\rightarrow \infty}T_{D^\alpha u_{n}}(\phi) \\ =&\ \lim \limits_{n\rightarrow \infty} (-1)^{|\alpha|}T_{u_{n}}(D^\alpha \phi) \\ =&\ (-1)^{|\alpha|}T_{u}(D^\alpha \phi) \end{align*}$$

The first equation holds by $(2)$. As when defining the differentiation of distributions, using integration by parts shows that it amounts to $T_{D^\alpha u_{n}}(\phi)=T_{u_{n}}(D^\alpha \phi)$, thus validating the second equation. The third equation is true due to $(1)$. By the definition of weak derivatives, $u_\alpha$ and $D^\alpha u$ are the same with respect to $0 \le |\alpha | \le m$ in a distributional sense. Therefore, it is $D^\alpha u_{n} \rightarrow u_\alpha=D^\alpha u$. Now, we check if $\| u_{n} -u\|_{m, p}$ converges to $0$. When $1 \le p < \infty$,

$$\begin{align*} \lim \limits_{n \rightarrow \infty} \|u_{n}-u\|_{m, p}^p =&\ \lim \limits_{n \rightarrow \infty} \sum\limits_{0\le |\alpha| \le m } \|D^\alpha u_{n}-D^\alpha u \|^p_{p} \\ =&\ \sum\limits_{0\le |\alpha| \le m } \|u_\alpha-D^\alpha u \|^p_{p} \\ =&\ \sum\limits_{0\le |\alpha| \le m } \|D^\alpha u-D^\alpha u \|^p_{p} \\ =&\ 0 \end{align*}$$

Similarly, when $p=\infty$,

$$\begin{align*} \lim \limits_{n \rightarrow \infty} \|u_{n}-u\|_{m, p} =&\ \lim \limits_{n \rightarrow \infty}\max\limits_{0\le |\alpha| \le m} \|D^\alpha u_{n} - D^\alpha u\|_\infty \\ =&\ \lim \limits_{n \rightarrow \infty}\max\limits_{0\le |\alpha| \le m} \|u_\alpha - D^\alpha u\|_\infty \\ =&\ \lim \limits_{n \rightarrow \infty}\max\limits_{0\le |\alpha| \le m} \|D^\alpha u - D^\alpha u\|_\infty \\ =&\ 0 \end{align*}$$

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