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.
Let be an open set, (possibly negative) and . We define
We define to be the Hilbert space completion of under .
If no confusion arises, we will omit the in .
This space is actually usually called
reserved for the same definition but without the compactly supported requirement. We will only need the compactly supported version, and will not write the
Basic properties of Fourier transforms imply that for , we have the following more intuitive definition:
If is a non-negative integer, and , then is equivalent to the norm
For any , extends to a continuous map .□
A priori, the above definition does not let us think of elements of
as genuine functions. Thankfully, we have the following result:
The natural map induces an isomorphism .
If and , then . Hence there is a continuous map .
The map is injective.
Only (3) requires proof. We have to show that if
. By assumption, we know
. So it converges to
almost everywhere. So
almost everywhere. But we know it is Cauchy in
. So it converges to
This lets us view the as nested subspaces of . In fact, we get something better.
(Sobolev embedding theorem)
Let an integer such that . Then there exists a constant such that for any , we have
Thus, there is a continuous inclusion .
, we have
The second term is exactly
, and the first term is a constant, which by elementary calculus, is finite iff
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
. Then we want to show that
for some constant
. In other words, if
, we want to be able to make sense of
as a continuous function. The trick is to notice we can make sense of
function, and hence we can write
for some integration constant
. This can be determined by
This lets us control
in terms of integrals of
One can in fact come up with a similar proof as long as is an integer.
Another crucial theorem is the following:
(Rellich compactness theorem)
If is precompact and , then the natural map is compact.
Let be such that . We have to produce a subsequence that converges in . The key claim is
that equicontinuous and uniformly bounded on compact sets.
Assuming this claim, Arzelá–Ascoli implies there is a subsequence of
that is uniformly convergent on compact subsets. Thus, we write
For any fixed , the first term vanishes in the limit by uniform convergence. For the second term, we bound
So we know that
This can be made small with large , and we are done.
To prove the claim, we use the following trick — pick a bump function such that . Then trivially, , and thus we have
Then controlling would be the same as controlling , which is fixed. For example, to show that is bounded, we write
by Cauchy–Schwarz. Since is a Schwarz function, the first factor is finite and depends continuously on . So is uniformly bounded on compact subsets. Equicontinuity follows from similar bounds on .
If we are lazy, we will often restrict to the case where . If further , then the definition of the Sobolev norm in terms of derivatives can be very useful. For , we will exploit the following duality:
Thus, it extends to a map . Moreover,
Thus, this pairing exhibits and as duals of each other.
Note that this duality pairing is “canonical”, and doesn't really depend on the Hilbert space structure on
(while the canonical isomorphism between
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
, one direction of the inequality is clear. For the other, if
. For general
A sample application of this is to show that differential operators induce maps between Sobolev spaces. We already said that tautologically induces a map . 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 , and duality implies the result for negative .
Let , and . Then
for some constant independent of . Thus, if is a differential operator of compact support of order , then it extends to a map .
We only have to check this for . For , this is straightforward, since is a sum of terms of the form for , and the product rule together with the bound implies the result.
For negative , if , then
When proving elliptic regularity, we will need a slight refinement of the above result.
Let , and . Then
for some constant independent of .
Again the case is straightforward. If we look at a term , the terms in the product rule where and the derivatives all hit contribute to the firs term, and the remainder go into the second.
For , we proceed by downward induction on , using the isometry given by extending . This is an isomorphism, because in the Fourier world, this is just multiplication by . Thus, we check the claim for of the form . We have
The second term is bounded, by induction, by
Observing that since , we are happy with this term. To bound the first term, we have
So we get a bound (omitting constant multiples)