** Concentration compactness approach to Gagliardo–Nirenberg in 1-dimensions **

We have with an optimizing sequence; in particular, . Choose (Lebesgue points of ) so that

Note that ; we proved . These functions are in fact continuous; indeed, they are Hölder continuous uniformly in :

Observe that cannot tend to as since otherwise, by interpolation with , we would have that , contradicting the optimizing hypothesis. Henceforth we restrict to a subsequence so that . By Alaoglu and Rellich–Kondrachov, we can pass to a further subsequence so that

weakly in ( weakly in and in ) and in the norm on any compact set.

We observe that . As and is Hölder continuous (uniformly in ), there is a small number so that

Thus, .

For the sake of discussion, suppose (i.e., is capturing all the norm). Then, by Radon–Riesz, in and we can finish the proof just as above. The remaining alternative is that . In that case write

Note that and that as because . This shows that and are “asymptotically orthogonal in the Pythagorean sense

Similarly,

Moreover,

where the first term is due to in con compact sets and the second term is due to going to as . That is,

The remaining goal is to show that , because then , which then implies .

The key to showing this is sub-additivity:

and so

Because we also had

Therefore, because , the equalities

must imply or (this can be seen by drawing the unit balls). Namely, and implies or . As and , we must have and hence in , and thus also in .

In the original vocabulary of concentration compactness, there are three scenarios for sequences ,

- vanish, .
- compactness (co-compactness) [modulo translations in .
- dichotomy strongly on compact sets bu .

at least after passing to a subsequence.

In the proof, we had a sequence and either or it contained (after passing to a subsequence and translating in space), a “bubble of concentration” , leaving behind a remainder . We can proceed inductively and look inside for a further bubble of concentration and so

TheoremFix . Let be a bounded sequence in . Then, after passing to a subsequence, we can decompose asfor each where is the “total number of bubbles” such that

- for all and , weakly in and strongly in on compact sets.
- If , then .
- .
- For all , as . Furthermore and .

RemarkFor assertion 2, if was the minimal pair such that does not diverge, then, along a subsequence, it converges, say to and so

Thus, , which is to say we are removing the same “bubble” twice, a troubling fact. In the proof, we find by applying our “bubble-finding lemma” to , which converges to zero when translated by . Any limit along that sequence was removed by and hence would be identically .

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