Sobolev Space is Banach Space

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Theorem

Let $k \in \Z_{\ge 0}$.

Let $p \in \R_{\ge 1}$ or $p = \infty$.

Let $U \subset \R^n$ be an open set.

Then the Sobolev space $\map {W^{k, p} } U$ equipped with the Sobolev norm is a Banach space.


Proof


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By Sobolev Norm is Norm, it remains to show that $\map {W^{k, p} } U$ is complete.

Assume $\sequence {u_m}$ is a Cauchy sequence in $\map {W^{k, p} } U$.

Then for each multiindex $\size \alpha \le k$, $\sequence {D^\alpha u_m}$ is a Cauchy sequence in $\map {L^p} U$.


From the Riesz-Fischer Theorem, $\map {L^p} U$ is complete.

Hence $\sequence {D^\alpha u_m}$ converges in $L^p$ norm:

$D^\alpha u_m \to u_\alpha\in \map {L^p} U$

for each $\size \alpha \le k$.

In particular, $\sequence{u_m}$ converges in $L^p$ norm:

$u_m \to u_{\tuple {0, \ldots, 0} } =: u$


We now claim:

\(\text {(1)}: \quad\) \(\ds \forall u \in \map {W^{k, p} } U: \forall \size \alpha \le k: \, \) \(\ds D^\alpha u\) \(=\) \(\ds u_\alpha\)


To verify this assertion, fix $\phi \in \map {C_c^\infty} U$.

Then:

\(\ds \int_U u D^\alpha \phi \rd x\) \(=\) \(\ds \lim_{m \mathop \to \infty} \int_U u_m D^\alpha \phi \rd x\) Definition of $u$
\(\ds \) \(=\) \(\ds \lim_{m \mathop \to \infty} \paren {-1}^{\size \alpha} \int_U D^\alpha u_m \phi \rd x\) Integration by Parts
\(\ds \) \(=\) \(\ds \paren {-1}^{\size \alpha} \int_U u_\alpha \phi \rd x\) Definition of $u_\alpha$

By the definition of distributional derivative, $(1)$ holds.

Therefore:

$D^\alpha u_m \to D^\alpha u$ in $\map {L^p} U$

for all $\size \alpha \le k$.

We see that:

$u_m \to u$ in $\map {W^{k, p} } U$

as required.

$\blacksquare$


Sources

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