Sobolev Spaces are Separable, Uniformly Convex, and Reflexive: A Proof

Theorem¹

When $1\le p <\infty$, the Sobolev space $W^{m, p}$ is separable. Moreover, when $1< p < \infty$, the Sobolev space is reflexive and uniformly convex.

Description

A vector space in which an inner product is defined is called an inner product space, and a complete inner product space is specially called a Hilbert space. Since $W^{m, p}$ is complete, if the inner product is defined as below, $W^{m,\ 2}$ becomes a separable Hilbert space.

$$ \langle u,\ v \rangle_{m} = \sum \limits_{0\le |\alpha | \le m } \left\langle D^\alpha u,\ D^\alpha v \right\rangle $$

Here, $\langle \cdot, \cdot \rangle$ is the inner product in the $L^2$ space.

Proof

Since a norm is defined for $W^{m, p}$, it becomes a metric space. And the $L^{p}$ space is a complete metric space.

Lemma

Let’s call $(X, d)$ a metric space. Let’s call $(Y,d’)$ a complete metric space. Then, there exists an isometric mapping, embedding $f : X \to Y$.

Therefore, according to the theorem, there exists an embedding which is an isometric mapping as follows.

$$ P : W^{m, p} \rightarrow L^{p} $$

Let us define $P(W^{m, p})=W$. Since $P$ is an embedding, $W \subset L^{p}$ holds. Additionally, since $W^{m, p}$ is complete and $P$ is an isometric mapping, $W$ is also complete.

Lemma

Let’s say $M$ is a complete metric space. If $S$ is a subspace of $M$, then the following two propositions hold:

• $S$ is closed in $M$.
• $S$ is complete.

According to the theorem, $W$ is a closed subspace of $L^{p}$.

Lemma

If we call $X$ a Banach space, and $M$ a closed subspace of $X$, then

• $M$ is also a Banach space.
• If $X$ is separable, then $M$ is also separable.
• If $X$ is reflexive, then $M$ is also reflexive.
• If $X$ is uniformly convex, then $M$ is also uniformly convex.

When the $L^{p}$ space is $1 \le p < \infty$, it is separable, and when it is $1 < p < \infty$, it is uniformly convex and reflexive. Therefore, according to the above lemma, $W$ is also $1 \le p < \infty$ when separable, and $1 < p < \infty$ when uniformly convex and reflexive.

Meanwhile, since $P$ is an embedding, $P=(W^{m, p})=W$ and $W^{m, p}$ are topologically isomorphic. Thus, if $W$ is separable, then $W^{m, p}$ is also separable. Furthermore, since $P$ is an isometric mapping, by the definition of uniform convexity, if $W$ is uniformly convex, then $W^{m, p}$ is also uniformly convex.

Lemma

A uniformly convex Banach space is reflexive.

As $W^{m, p}$ is a Banach space, $W^{m, p}$ is reflexive.

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