Here is something I found super interesting recently. Everyone knows Nash-Moser-De Giorgi is a remarkably deep result, however in 2 dimensions, the proof is a simple consequence of the Cacciopolli inequality. We then discuss briefly how to extend this method to higher dimensions. And yes, WordPress is forever broken.

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I have given up on WordPress, the blog is attached!

]]>The first problem is an application of the Eberlain-Smulyan theorem; let be a countable dense subset of and note that for each there exists a subsequence such that is weakly convergent. We choose the subsequences so that and form a diagonal sequence from these and call it .Let and and let . Choose large enough so that

and large enough so that

Then it follows that

The trick here was to use the definition to write it out rigorously, hence bypassing the need for dealing with limits.

For the second problem:

We first take a subsequence which we do not re-label so that where . We then use Mazur’s lemma to write down a convex combination of the , call them that converge strongly to for every . This is important that we create these sequences truncating off the first elements, since then this will allow us to pass the limit in the inequalities we will later obtain. More precisely, there exists with such that converges strongly to and so for every ,

Then we can first pass the limit and then pass the limit and obtain the result. ]]>

*Claim 1: The operator is a bijection*

*Proof:* To show injectivity we show that this operator has trivial kernel. One inclusion is obvious, we just need to show that . To see this we recall from elliptic regularity theory for the Laplacian that if for some then

Now back to our question which can be written as , and so applying the estimate we can see that and so . Applying the estimate one more time shows that . For surjectivity we begin taking some and let be the unique solution to , and by the regularity theory . Then let be the unique solution . It follows that and . (Super cool right?!)

*Claim 2: If then *

*Proof:* Let , multiplying by and integrating we get

Since we can integrate by parts once with vanishing boundary terms on the left hand side (just replace with a sequence of compactly supported smooth functions, integrate by parts using the definition of weak derivative passing limits by dominated convergence) to get

Now, (since it is the solution to ) and so we can integrate by parts once more, again with vanishing boundary terms and we get

Now if we try to integrate by parts again we will run into a dead end, try it! Instead we play the same game with to get

Equating (1) and (2) gives the claim.

Now we are ready to finish the problem, which is to prove some regularity result on weak solutions to for ; that is . By surjectivity of there exists some such that . We just need to conclude now that . By claim 2 we have that

and since is bijective we have that this is equivalent to

Setting we get that .

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for all . Prove that .

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