Notes on Lewin's QM book - Appendix

Appendix A: Sobolev Spaces

Definition

Definition of is given, and the Meyer-Serrin theorem is stated without proof.

Sobolev spaces on the interval

Considering with distributional derivative . Then is continuous¹ on and in , so is a.e. equal to a continuous function on , and we identify with this continuous representative. This lets us write . This implies that , which after integrating, implies the estimate

If , we can use this estimate on with the identity

to get the variant inequality or after the subadditivity of the square root,

Exercise A.3 - Density of smooth functions

Let . (1) Construct in with in . (2) If we further assume that , then do the same with . (3) Finally, extend this result to . Solutions

1. Extend by zero outside to get a function in . Then convolve with a mollifier for with to get the desired sequence of smooth functions - one gets

   and similarly with the derivative, whence the result.

2. We use a cutoff with to define and . Then . We focus on ; how to treat will be clear. We extend to a function on that is zero for or . By the given boundary conditions, . Indeed, the derivative of the extension by zero will have the term unless . We next need that converge in to . To do this we modify the convolution formula by setting

   Then and is zero if or , and for the convergence we have

   The first term converges by Young’s inequality for convolutions and the second by good kernel properties. One can then repeat this for and then symmetrically for to get the result.

3. Exactly the same method works for .

Elliptic Regularity on

This is the statement that if then with the estimate . We recap the proof given on . Put . First, we know that implies that . Then we write , and integrate it against :

Integrating by parts we get . The boundary term vanishes since . For the second term on the right, we write

This gives the identity

This leads to a bound of using only and . We see that was introduced so that we could use (and ).

Exercise A.4 - Half line

1. Show that
2. Show that is dense in , and that is the space of functions in that vanish at zero.
3. Prove the elliptic estimate (bound by and .)

Solutions:

1. All the estimates trivially carry over as we can just restrict to an interval and use the previous estimates. The only new thing we need to show is that as . But this follows from the fact that and the estimate

2. Same proof works.
3. The result follows immediately by restricting to intervals. But we could also redo the proof with some weight such that and .

The Sobolev Space

The usual norm for is . Note that we have the trivial inequality , and also by multinomial theorem that . This implies by Fourier transform the equivalence of norms. This is also equivalent to the ‘elliptic estimate’ .

Trace, Lifting and Extension on a Bounded Open Set

First the half-space is treated. For general spaces the and spaces are defined by local charts and partitions of unity, so the half-space case is the main one to understand.

Recall that if then a.e. So if , then for a.e. , the function belongs to by Fubini’s theorem -

So for a.e. , coincides a.e. in with a continuous function on that vanishes at infinity by earlier exercise. We can then define the trace of at to be this continuous representative, which is well-defined for a.e. . This gives us a well-defined and bounded (again, by earlier exercise) trace operator .

For , , the trace map is , mapping

with the bound

Stated without proof:

• iff and .
• Theorem A.10 (“smooth lifting”): Given continuously differentiable boundary values we can find such that equals the given boundary values.
• Fractional Sobolev space defined by weighted integral of difference quotient to state that this space is the image of under the restriction map, and that the trace map is (or rather can be) bounded from to .

Sobolev Embedding and Rellich-Kondrachov Compactness

Interesting version of Sobolev embedding is proven- let and . For any such that the superlevel sets have finite measure for all , then

where is the number such that . There is a version for Bessel potential spaces with , but the proof given only extends to with the Hausdorff-Young inequality.

Then a version o the RK Compactness theorem is given - if is a bounded open set, then the embedding is compact for any . (Here is the space of restrictions of functions in .) The proof uses the Fourier transform and the fact that the Fourier transform of a compactly supported function is smooth and decays at infinity. In particular this lemma is used- Let decay at infinity, and let . Then is compact on .

Here it is used that compactness means upgrading from from weak convergence to strong convergence. This is equivalent to the usual extraction of convergent subsequences on reflexive spaces (so not ). In one direction it is used that every subsequence has a further subsequence that converges, and in the other direction Banach-Alaoglu is used to get a weakly convergent subsequence, and then the compactness is used to upgrade to strong convergence.

Evans has a slightly sharper version of the theorem, namely that is compactly embedded in for .

Elliptic Regularity on bounded domains

Without conditions on at the boundary, the result is false. The example given is the Fundamental solution of the Laplacian translated to the boundary, which is in and has , but is not in if the boundary supports a cone. The elliptic regularity result holds under Robin boundary conditions.

1. ¹In fact, . the space of absolutely continuous functions is given by the following modification of uniform continuity – instead of saying that we have control on any interval, we require this control on arbitrary finite collections of disjoint intervals, whose total length is at most . This helps rule out counterexamples like the Cantor function, where the control is maintained on each small interval, but not on a null set (which can be approximated by a collection of intervals of small total length).

   One direction is trivial-Given , as in this note we have . Thus given and any sequence of disjoint intervals with , we have

   The converse is harder. Sketch - AC implies obviously. It also implies BV and hence the distributional derivative is a finite signed Radon measure . We can write where and is singular, supported on some Lebesgue null set . By the regularity of Radon measures, we can find a cover of by disjoint open intervals with arbitrarily small total length . Then , showing . Then one can check that , proving the result.