Theorem (Gagliado–Nirenberg Inequality)We haveprovided

*Proof:* From Bernstein’s inequality (because )

When

Moreover, by using the “wrong ” above (e.g. ) and then interpolate with (actually, genuine Hölder by letting ) we see that for ,

with some determined by .

We treat the remainder of the proof in the case—the other region is similar. Now,

To finish, we figure out when the minimum switches and then sum the geometric series. The details are left as an exercise.

Now we ask

**Question:** If , is

positive, or even bounded from below? Think, stars.

Well, if , then we do have positivity, by Gagliardo–Nirenberg

(note that with , ). Thus if is small

Conversely, if we take with , then

which will be negative once is large enough, though this makes large.

Thus there is a transition size at which we switch from always to sometimes. What happens if we rescale via to keep constant? We obtain

Therefore

and so as soon as has negative energy, a re-scaled version has arbitrarily negative energy.

Theorem (Rellich–Kondrachov)LetThen for all ,

is compact in , where .

RemarkNote that, without the , this result would be false. For example, consider with fixed .

*Proof:* From the previous proof, we know that

and thus

Therefore, the question reduces to showing that

is compact for all . There are many approaches to prove this.

The first approach is to recall that a is pre-compact if and only if is

- Bounded: such that for all , we have .
- Tight: such that for all , we have .
- Equicontinuous: such that for all and for all , we have .

In our case, property 1 follows by the Gagliardo–Nirenberg inequality, property 2 follows by the analysis of , and property 3 follows by the analysis of .

The second approach is to note that, because we are on a compact/bounded region (namely, ), compactness of as a set of continuous functions in the supremum norm implies compactness in . By Arzelà–Ascoli, we need only to check boundedness and equicontinuity:

which follow by Bernstein’s inequalities.

Theorem (Radial Gagliardo–Nirenberg)If and is spherically symmetric, then

*Proof:* Let us write as a function of radius . By the fundamental theorem of calculus,

Therefore,

namely,

CorollaryThe set

is compact in when , where and are as above.

RemarkThis result is false when (c.f. two bump functions who are symmetric with respect to each other about 0 and letting them move to infinity).

*Proof:* We just need to check tightness, so that we can apply Rellich–Kondrachov, which we can do via interpolation:

Finally, this leads us to

Theorem (Sharp Gagliardo–Nirenberg)Let and . The functionachieves its maximum on . Moreover, any such maximizer can be transformed via to a solution of

In particular, this partial differential equation has a solution.

RemarkOne can show (we will not do so here) that there is a unique radial non-negative solution to this PDE in . Moreover, one can deduce that every optimizer has the form and even that the Hessian at these points is positive definite perpendicular to the minimizing manifold.

*Proof:* By the Gagliardo–Nirenberg inequality, is bounded. Let be a sequence in so that converges to the extremal value. Replacing by their radial symmetric decreasing rearrangements will only increase (or retain) the value of , hence is also an optimizing sequence.

Next we consider replacing by . It is easy to check that this does not affect the value of . We choose so that . Henceforth, we call .

By Alaoglu, we can pass to a subsequence so that in and , also weak- in . Note that since is defined via testing against , we are guaranteed that . Lower semi-continuity of norms under weak- limits shows that and, similarly, that . [Note that Alaoglu in leads nowhere.]

By Rellich–Kondrachov and tightness for radial functions, we know that a further subsequence converges strongly in . [Actually, we do not need to pass to a further subsequence because testing with identifies all subsequential limits as being —the weak- limit.] Note that as , we thus have , which is not zero as . Noting that

therefore and ( cannot beat the optimal value). By the Radon–Riesz theorem, we see that in .

Now, let be an optimizer such that . Consider for some . This must have . Furthermore

and

Therefore, using the fact that ,

valid for every . Choosing and and canceling , we obtain

for all . Thus is a distributional solution to the PDE

which is called the Euler–Lagrange equation for our extremal problem. By rescaling , with determined by alone, we get

To recap, every optimizer will, after suitable rescaling, solve the above PDE. Moreover, there is at least one optimizer that has .

As a sketch of “uniqueness” of optimizers: given an optimizer , note that is also an optimizer. Moreover, we must have . In general, this is not enough to guarantee that (see submarine remark above); however this conclusion does follow if we can show that the set has zero measure for . Switching to polar coordinates, we see that radial optimizers obey an ODE (in a distributional sense)

after rescaling. As such, we can see that is actually real analytic (at least away from ) and so has no positive measure level set — we can also realize this from the fact that radially decreasing means that a plateau would force our ODE to have a locally constant solution, but uniqueness for ODES implies this would be a globally constant solution , which fails to be in . Lastly, we need the uniqueness of non-negative solutions to our ODE.

The proof failed to cover the case and because is not tight in — for example, consider . Nevertheless, is indeed tight in , which allows us to adapt the preceding proof.

*Proof of tightness: * We compute

because is decreasing in . Thus

and so by Hölder’s inequality

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