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.

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:

A priori, the above definition does not let us think of elements of $H^s$ as genuine functions. Thankfully, we have the following result:

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 $L^1$.

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

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 $s > \ell + \frac{n}{2}$.

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 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 $\frac{\mathrm{d}u}{\mathrm{d}x}$.

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

Another crucial theorem is the following:

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 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 $\mathrm{D}_j \hat{u}_k = (\mathrm{D}_j \hat{a}) * \hat{u}_k$.

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:

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.

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.

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 $s \in \mathbb {Z}$.

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

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

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

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)

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