Global AnalysisLocal Sobolev spaces

## 2.1 Local Sobolev spaces

To apply functional analytic techniques to PDE problems, we need an appropriate Hilbert space of functions. The right notion is that of a Sobolev space.

Definition

Let $U \subseteq \mathbb {R}^n$ be an open set, $s \in \mathbb {R}$ (possibly negative) and $f, g \in C_c^\infty (U)$. We define

\begin{aligned} (f, g)_s & = \int _{\mathbb {R}^n} (1 + |\xi |^2)^s \overline{\hat{f}(\xi )}\hat{g}(\xi ) \; \mathrm{d}\xi \\ \| f\| _s & = (f, f)_s \end{aligned}

We define $H^s(U)$ to be the Hilbert space completion of $C_c^\infty (U)$ under $\| \cdot \| _s$.

If no confusion arises, we will omit the $(U)$ in $H^s(U)$.

This space is actually usually called $H_0^s(U)$ instead, with $H^s(U)$ reserved for the same definition but without the compactly supported requirement. We will only need the compactly supported version, and will not write the $_0$.

Basic properties of Fourier transforms imply that for $s \in \mathbb {Z}_{>0}$, we have the following more intuitive definition:

Lemma

If $s$ is a non-negative integer, and $f \in C_c^\infty (U)$, then $\| \cdot \| _s$ is equivalent to the norm

$\| f\| _s' = \sum _{|\alpha | \leq s} \| \mathrm{D}^\alpha f\| _{L^2}.$
Lemma

For any $s$, $\mathrm{D}_\alpha$ extends to a continuous map $H^s \to H^{s - |\alpha |}$.

A priori, the above definition does not let us think of elements of $H^s$ as genuine functions. Thankfully, we have the following result:
Lemma
1. The natural map $C_c^\infty (U) \hookrightarrow L^2$ induces an isomorphism $H^0 \cong L^2$.

2. If $s \geq t$ and $f \in C_c^\infty$, then $\| f\| _t \leq \| f\| _s$. Hence there is a continuous map $H^s \to H^t$.

3. The map is injective.

Proof
Only (3) requires proof. We have to show that if $f_n \in C_c^\infty$ is $\| \cdot \| _s$-Cauchy and $\| f_n\| _t \to 0$, then $\| f_n\| _s \to 0$. By assumption, we know $\hat{f}_n^2(1 + |\xi |^2)^t \to 0$ in $L^1$. So it converges to $0$ almost everywhere. So $\hat{f}_n^2 (1 + |\xi |^2)^s \to 0$ almost everywhere. But we know it is Cauchy in $L^1$. So it converges to $0$ in $L^1$.
Proof

This lets us view the $H^s$ as nested subspaces of $L^2$. In fact, we get something better.

Theorem (Sobolev embedding theorem)

Let $k$ an integer such that $s > k + \frac{n}{2}$. Then there exists a constant $A$ such that for any $u \in C^{\infty }(U)$, we have

$\| u\| _{C^k} \leq A \| u\| _s.$

Thus, there is a continuous inclusion $H^s \hookrightarrow C^k$.

Proof
Let $|\alpha | = \ell < k$. For $x \in U$ and $u \in C^\infty _c(U)$, we have

$|\mathrm{D}_\alpha u(x)| \propto \left| \int e^{ix\xi } \xi _\alpha \hat{u}(\xi ) \; \mathrm{d}\xi \right| \leq \int |\xi _\alpha | |\hat{u}(\xi )| \; \mathrm{d}\xi \leq \int |\xi |^\ell |\hat{u}(\xi )|\; \mathrm{d}\xi \leq \left(\int |\xi |^2 (1 + |\xi |^2)^{-s}\; \mathrm{d}\xi \right)^{1/2}\left(\int |\hat{u}(\xi )|^2 (1 + |\xi |^2)^s\; \mathrm{d}\xi \right)^{1/2}.$

The second term is exactly $\| u\| _s$, and the first term is a constant, which by elementary calculus, is finite iff $s > \ell + \frac{n}{2}$.

Proof

Admittedly, the proof above does not make much sense. The following more direct proof of a special case shows why we should expect a result along these lines to be true:

Proof
[Proof of special case] We consider the case $r = 1$, $U = (0, 1)$ and $k = 0$. Then we want to show that

$\| u\| _{C^0} \leq A \| u\| _1$

for some constant $A$. In other words, if $u \in H^1$, we want to be able to make sense of $u$ as a continuous function. The trick is to notice we can make sense of $\frac{\mathrm{d}u}{\mathrm{d}x}$ as an $L^2$ function, and hence we can write

$u(t) = \int _0^t \frac{\mathrm{d}u}{\mathrm{d}x}\; \mathrm{d}x + C$

for some integration constant $C$. This can be determined by

$C = \int _0^1 u(x)\; \mathrm{d}x - \int _0^1 \int _0^t \frac{\mathrm{d}u}{\mathrm{d}x}\; \mathrm{d}x\; \mathrm{d}t.$

This lets us control $\sup u$ in terms of integrals of $u$ and $\frac{\mathrm{d}u}{\mathrm{d}x}$.

Proof
One can in fact come up with a similar proof as long as $s$ is an integer.

Another crucial theorem is the following:

Theorem (Rellich compactness theorem)

If $U \subseteq \mathbb {R}^n$ is precompact and $s > t$, then the natural map $H^s \to H^t$ is compact.

Proof
Let $u_k \in C^\infty _c(U)$ be such that $\| u_k\| _s \leq 1$. We have to produce a subsequence that converges in $H^t$. The key claim is
Claim

$\hat{u}_k$ that equicontinuous and uniformly bounded on compact sets.

Assuming this claim, Arzelá–Ascoli implies there is a subsequence of $\hat{u}_k$ that is uniformly convergent on compact subsets. Thus, we write

$\| u_k - u_\ell \| _t^2 = \left(\int _{|\xi |\leq R} + \int _{|\xi | > R}\right)|\hat{u}_k - \hat{u}_\ell |^2 (1 + |\xi |^2)^t \; \mathrm{d}\xi .$

For any fixed $R$, the first term vanishes in the limit $k, \ell \to \infty$ by uniform convergence. For the second term, we bound

$(1 + |\xi |^2)^t \leq (1 + |\xi |^2)^s (1 + R^2)^{t - s}.$

So we know that

$\int _{|\xi | > R}|\hat{u}_k - \hat{u}_\ell |^2 (1 + |\xi |^2)^t \; \mathrm{d}\xi \leq (1 + R^2)^{t - s}(\| u_k\| _s^2 + \| u_\ell \| _s^2) \leq 2 (1 + R^2)^{t - s}.$

This can be made small with large $R$, and we are done.

To prove the claim, we use the following trick — pick a bump function $a \in C_c^\infty (\mathbb {R}^n)$ such that $a|_U \equiv 1$. Then trivially, $u_k = au_k$, and thus we have

$\hat{u}_k = \hat{a} * \hat{u}_k.$

Then controlling $\hat{u}_k$ would be the same as controlling $\hat{a}$, which is fixed. For example, to show that $\hat{u}_k$ is bounded, we write

$|\hat{u}_k| = |\hat{a} * \hat{u}_k| \leq \int |\hat{a}(\xi - \eta ) \hat{u}_k(\eta )|\; \mathrm{d}\eta \leq \left(\int |\hat{a} (\xi - \eta )|^2 (1 + |\eta |^2)^{-s}\; \mathrm{d}\eta \right)^{1/2} \| u_k\| _s$

by Cauchy–Schwarz. Since $\hat{a}$ is a Schwarz function, the first factor is finite and depends continuously on $\xi$. So $\hat{u}_k$ is uniformly bounded on compact subsets. Equicontinuity follows from similar bounds on $\mathrm{D}_j \hat{u}_k = (\mathrm{D}_j \hat{a}) * \hat{u}_k$.

Proof

If we are lazy, we will often restrict to the case where $s \in \mathbb {Z}$. If further $s \geq 0$, then the definition of the Sobolev norm in terms of derivatives can be very useful. For $s < 0$, we will exploit the following duality:

Lemma

The pairing

\begin{aligned} C_c^\infty \times C_c^\infty & \to \mathbb {C}\\ (f, g) & \mapsto \int \overline{f(x)} g(x) \; \mathrm{d}x. \end{aligned}

satisfies

$|(f, g)| \leq \| f\| _s \| g\| _{-s}.$

Thus, it extends to a map $H^s \times H^{-s} \to \mathbb {C}$. Moreover,

$\| f\| _s = \sup _{g \not= 0} \frac{|(f, g)|}{\| g\| _{-s}}.\tag {\dagger }$

Thus, this pairing exhibits $H^s$ and $H^{-s}$ as duals of each other.

Note that this duality pairing is “canonical”, and doesn't really depend on the Hilbert space structure on $H^s$ (while the canonical isomorphism between $H^s$ and $(H^s)^*$ does). Later, we will see that in the global case, the Hilbert space structure is not quite canonical, but the duality pairing here will still be canonical.

Proof
The first inequality follows from Plancherel and Cauchy–Schwarz

$(f, g) = (\hat{f}, \hat{g}) = \int \overline{\hat{f}(\xi )}(1 + |\xi |^2)^{s/2} \hat{g}(\xi ) (1 + |\xi |^2)^{-s/2}\; \mathrm{d}\xi .$

To show $(\dagger )$, one direction of the inequality is clear. For the other, if $f \in C^\infty$, take $\hat{g} = \hat{f} (1 + |\xi |^2)^s$. For general $f$, approximate.

Proof

A sample application of this is to show that differential operators induce maps between Sobolev spaces. We already said that $\mathrm{D}_\alpha$ tautologically induces a map $H^s \to H^{s - |\alpha |}$. In a general differential operator, we differentiate, then multiply by a smooth function. Thus, we want to show that multiplication by a smooth function is a bounded linear operator. This is easy to show for $s \in \mathbb {Z}_{\geq 0}$, and duality implies the result for negative $s \in \mathbb {Z}$.

Lemma

Let $s \in \mathbb {Z}$, $a \in C_c^\infty (\mathbb {R})$ and $u \in H^s(U)$. Then

$\| au\| _s \leq C\| a\| _{C^{|s|}} \| u\| _s.$

for some constant $C$ independent of $a$. Thus, if $L$ is a differential operator of compact support of order $k$, then it extends to a map $L: H^{s + k} \to H^s$.

Proof
We only have to check this for $u \in C_c^\infty (U)$. For $s \geq 0$, this is straightforward, since $\| au\| _s^2$ is a sum of terms of the form $\| \mathrm{D}_\alpha (au)\| _0^2$ for $|\alpha | \leq k$, and the product rule together with the bound $\| a u\| _0 \leq \| a\| _{C^0} \| u\| _0$ implies the result.

For negative $s$, if $s > 0$, then

$\| au\| _{-s} = \sup _{v \not= 0} \frac{|(au, v)|}{\| v\| _s} = \sup \frac{|(u, \bar{a}v)|}{\| v\| _s} \leq \sup \frac{\| u\| _{-s} \| \bar{a}v\| _s}{\| v\| _s} \leq \| u\| _{-s} \| a\| _{C^s}.$

Proof
When proving elliptic regularity, we will need a slight refinement of the above result.
Lemma

Let $s \in \mathbb {Z}$, $a \in C_c^\infty (\mathbb {R})$ and $u \in H^s(U)$. Then

$\| au\| _s \leq \| a\| _{C^0} \| u\| _s + C \| a\| _{C^{|s| + 1}} \| u\| _{s - 1}$

for some constant $C$ independent of $a$.

Proof
Again the case $s > 0$ is straightforward. If we look at a term $\| \mathrm{D}_\alpha (a u)\| _0$, the terms in the product rule where $|\alpha | = k$ and the derivatives all hit $u$ contribute to the firs term, and the remainder go into the second.

For $s < 0$, we proceed by downward induction on $s$, using the isometry $H^{s + 2} \to H^s$ given by extending $1 + \Delta$. This is an isomorphism, because in the Fourier world, this is just multiplication by $(1 + |\xi |^2)$. Thus, we check the claim for $u$ of the form $Lv$. We have

$\| a Lv\| _s \leq \| [a, L]v\| _s + \| L(av)\| _s \leq \| [a, L]v\| _s + \| av\| _{s + 2}.$

The second term is bounded, by induction, by

$\| a\| _{C^0} \| v\| _{s + 2} + C \| a\| _{C^{|s + 2| + 1}} \| v\| _{s + 1} = \| a\| _{C^0} \| Lv\| _s + C\| a\| _{C^{|s + 2| + 1}} \| Lv\| _{s - 1}$

Observing that $|s + 2| \leq |s|$ since $s < 0$, we are happy with this term. To bound the first term, we have

$[a, L]v = [a, \Delta ]v = -\nabla a \cdot \nabla v.$

So we get a bound (omitting constant multiples)

$\| [a, L]v\| _s = \| \nabla a \cdot \nabla v\| _s \leq \| \nabla a\| _{C^{|s|}} \| \nabla v\| _{s} \leq \| a\| _{C^{|s| + 1}} \| v\| _{s + 1} = \| a\| _{C^{|s| + 1}} \| Lv\| _{s - 1}.$

Proof