Global Analysis — Local 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 URnU \subseteq \mathbb {R}^ n be an open set, sRs \in \mathbb {R} (possibly negative) and f,gCc(U)f, g \in C_ c^\infty (U). We define

(f,g)s=Rn(1+ξ2)sf^(ξ)g^(ξ)  dξfs=(f,f)s\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 Hs(U)H^ s(U) to be the Hilbert space completion of Cc(U)C_ c^\infty (U) under s\| \cdot \| _ s.

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

This space is actually usually called H0s(U)H_0^ s(U) instead, with Hs(U)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_0.

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

Lemma

If ss is a non-negative integer, and fCc(U)f \in C_ c^\infty (U), then s\| \cdot \| _ s is equivalent to the norm

fs=αsDαfL2. \| f\| _ s' = \sum _{|\alpha | \leq s} \| \mathrm{D}^\alpha f\| _{L^2}.\fakeqed

Lemma

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

A priori, the above definition does not let us think of elements of HsH^ s as genuine functions. Thankfully, we have the following result:
Lemma
  1. The natural map Cc(U)L2C_ c^\infty (U) \hookrightarrow L^2 induces an isomorphism H0L2H^0 \cong L^2.

  2. If sts \geq t and fCcf \in C_ c^\infty , then ftfs\| f\| _ t \leq \| f\| _ s. Hence there is a continuous map HsHtH^ s \to H^ t.

  3. The map is injective.

Proof
Only (3) requires proof. We have to show that if fnCcf_ n \in C_ c^\infty is s\| \cdot \| _ s-Cauchy and fnt0\| f_ n\| _ t \to 0, then fns0\| f_ n\| _ s \to 0. By assumption, we know f^n2(1+ξ2)t0\hat{f}_ n^2(1 + |\xi |^2)^ t \to 0 in L1L^1. So it converges to 00 almost everywhere. So f^n2(1+ξ2)s0\hat{f}_ n^2 (1 + |\xi |^2)^ s \to 0 almost everywhere. But we know it is Cauchy in L1L^1. So it converges to 00 in L1L^1.

This lets us view the HsH^ s as nested subspaces of L2L^2. In fact, we get something better.

Theorem (Sobolev embedding theorem)

Let kk an integer such that s>k+n2s > k + \frac{n}{2}. Then there exists a constant AA such that for any uC(U)u \in C^{\infty }(U), we have

uCkAus. \| u\| _{C^ k} \leq A \| u\| _ s.

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

Proof
Let α=<k|\alpha | = \ell < k. For xUx \in U and uCc(U)u \in C^\infty _ c(U), we have
\begin{useimager} 
    \begin{multline*}
      |\D_\alpha u(x)| \propto \left| \int e^{ix\xi} \xi_\alpha \hat{u}(\xi) \;\d \xi\right| \leq \int |\xi_\alpha| |\hat{u}(\xi)| \;\d \xi \leq \int |\xi|^\ell |\hat{u}(\xi)|\;\d \xi \\
      \leq \left(\int |\xi|^2 (1 + |\xi|^2)^{-s}\;\d \xi\right)^{1/2}\left(\int |\hat{u}(\xi)|^2 (1 + |\xi|^2)^s\;\d \xi\right)^{1/2}.
    \end{multline*}
  \end{useimager}
The second term is exactly us\| u\| _ s, and the first term is a constant, which by elementary calculus, is finite iff s>+n2s > \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=1r = 1, U=(0,1)U = (0, 1) and k=0k = 0. Then we want to show that uC0Au1 \| u\| _{C^0} \leq A \| u\| _1 for some constant AA. In other words, if uH1u \in H^1, we want to be able to make sense of uu as a continuous function. The trick is to notice we can make sense of dudx\frac{\mathrm{d}u}{\mathrm{d}x} as an L2L^2 function, and hence we can write u(t)=0tdudx  dx+C u(t) = \int _0^ t \frac{\mathrm{d}u}{\mathrm{d}x}\; \mathrm{d}x + C for some integration constant CC. This can be determined by C=01u(x)  dx010tdudx  dx  dt. 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 supu\sup u in terms of integrals of uu and dudx\frac{\mathrm{d}u}{\mathrm{d}x}.

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

Another crucial theorem is the following:

Theorem (Rellich compactness theorem)

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

Proof

Let ukCc(U)u_ k \in C^\infty _ c(U) be such that uks1\| u_ k\| _ s \leq 1. We have to produce a subsequence that converges in HtH^ t. The key claim is

Claim

u^k\hat{u}_ k that equicontinuous and uniformly bounded on compact sets.

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

ukut2=(ξR+ξ>R)u^ku^2(1+ξ2)t  dξ. \| 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 RR, the first term vanishes in the limit k,k, \ell \to \infty by uniform convergence. For the second term, we bound

(1+ξ2)t(1+ξ2)s(1+R2)ts. (1 + |\xi |^2)^ t \leq (1 + |\xi |^2)^ s (1 + R^2)^{t - s}.

So we know that

ξ>Ru^ku^2(1+ξ2)t  dξ(1+R2)ts(uks2+us2)2(1+R2)ts. \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 RR, and we are done.

To prove the claim, we use the following trick — pick a bump function aCc(Rn)a \in C_ c^\infty (\mathbb {R}^ n) such that aU1a|_ U \equiv 1. Then trivially, uk=auku_ k = au_ k, and thus we have

u^k=a^u^k. \hat{u}_ k = \hat{a} * \hat{u}_ k.

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

u^k=a^u^ka^(ξη)u^k(η)  dη(a^(ξη)2(1+η2)s  dη)1/2uks |\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 a^\hat{a} is a Schwarz function, the first factor is finite and depends continuously on ξ\xi . So u^k\hat{u}_ k is uniformly bounded on compact subsets. Equicontinuity follows from similar bounds on Dju^k=(Dja^)u^k\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 sZs \in \mathbb {Z}. If further s0s \geq 0, then the definition of the Sobolev norm in terms of derivatives can be very useful. For s<0s < 0, we will exploit the following duality:

Lemma

The pairing

Cc×CcC(f,g)f(x)g(x)  dx.\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)fsgs. |(f, g)| \leq \| f\| _ s \| g\| _{-s}.

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

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

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

Note that this duality pairing is “canonical”, and doesn't really depend on the Hilbert space structure on HsH^ s (while the canonical isomorphism between HsH^ s and (Hs)(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)=(f^,g^)=f^(ξ)(1+ξ2)s/2g^(ξ)(1+ξ2)s/2  dξ. (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 fCf \in C^\infty , take g^=f^(1+ξ2)s\hat{g} = \hat{f} (1 + |\xi |^2)^ s. For general ff, approximate.

A sample application of this is to show that differential operators induce maps between Sobolev spaces. We already said that Dα\mathrm{D}_\alpha tautologically induces a map HsHsαH^ 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 sZ0s \in \mathbb {Z}_{\geq 0}, and duality implies the result for negative sZs \in \mathbb {Z}.

Lemma

Let sZs \in \mathbb {Z}, aCc(R)a \in C_ c^\infty (\mathbb {R}) and uHs(U)u \in H^ s(U). Then

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

for some constant CC independent of aa. Thus, if LL is a differential operator of compact support of order kk, then it extends to a map L:Hs+kHsL: H^{s + k} \to H^ s.

Proof

We only have to check this for uCc(U)u \in C_ c^\infty (U). For s0s \geq 0, this is straightforward, since aus2\| au\| _ s^2 is a sum of terms of the form Dα(au)02\| \mathrm{D}_\alpha (au)\| _0^2 for αk|\alpha | \leq k, and the product rule together with the bound au0aC0u0\| a u\| _0 \leq \| a\| _{C^0} \| u\| _0 implies the result.

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

aus=supv̸=0(au,v)vs=sup(u,aˉv)vssupusaˉvsvsusaCs. \| 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}.\fakeqed

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

Lemma

Let sZs \in \mathbb {Z}, aCc(R)a \in C_ c^\infty (\mathbb {R}) and uHs(U)u \in H^ s(U). Then

ausaC0us+CaCs+1us1 \| au\| _ s \leq \| a\| _{C^0} \| u\| _ s + C \| a\| _{C^{|s| + 1}} \| u\| _{s - 1}

for some constant CC independent of aa.

Proof

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

For s<0s < 0, we proceed by downward induction on ss, using the isometry Hs+2HsH^{s + 2} \to H^ s given by extending 1+Δ1 + \Delta . This is an isomorphism, because in the Fourier world, this is just multiplication by (1+ξ2)(1 + |\xi |^2). Thus, we check the claim for uu of the form LvLv. We have

aLvs[a,L]vs+L(av)s[a,L]vs+avs+2. \| 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

aC0vs+2+CaCs+2+1vs+1=aC0Lvs+CaCs+2+1Lvs1 \| 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+2s|s + 2| \leq |s| since s<0s < 0, we are happy with this term. To bound the first term, we have

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

So we get a bound (omitting constant multiples)

[a,L]vs=avsaCsvsaCs+1vs+1=aCs+1Lvs1. \| [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}.\qedhere