April 27, 2011

Harmonic Analysis Lecture Notes 20

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 5:53 pm

Concentration compactness approach to Gagliardo–Nirenberg in 1-dimensions

We have {\|f_n\|_{L^2} = \| \nabla f \|_{L^2} = 1} with {f_n} an optimizing sequence; in particular, {J(f_n) = \|f_n\|_{L^p} \rightarrow J_{\max} > 0}. Choose {x_n} (Lebesgue points of {f_n}) so that

\displaystyle  	|f(x_n)| \geq \frac{1}{2} \| f_n\|_{L^\infty}.

Note that {f_n \in L^\infty({\mathbb R})}; we proved {H^1 \hookrightarrow L^\infty}. These functions are in fact continuous; indeed, they are Hölder {\tfrac{1}{2}} continuous uniformly in {n}:

\displaystyle  	|f_n(x) - f_n(y)| \leq \int_x^y |f_n'(t)| \; dt \leq \sqrt{\int_x^y 1 \; dt} \sqrt{\int_x^y |f_n'(t)|^2 \; dt} \leq |y-x|^\frac{1}{2}.

Observe that {\|f_n\|_{L^\infty}} cannot tend to {0} as {n \rightarrow \infty} since otherwise, by interpolation with {L^2}, we would have that {\|f_n\|_{L^p} \rightarrow 0}, contradicting the optimizing hypothesis. Henceforth we restrict {n} to a subsequence so that {\|f_n\|_{L^\infty} \rightarrow B > 0}. By Alaoglu and Rellich–Kondrachov, we can pass to a further subsequence so that

\displaystyle  	f(x + x_n) \rightarrow \phi

weakly in {H^1} ({\equiv} weakly in {L^2} and in {\dot H^1}) and in the {L^p} norm on any compact set.

We observe that {\phi \not \equiv 0}. As {|f_n(x_n)| \gtrsim 1} and {f_n} is Hölder {\frac{1}{2}} continuous (uniformly in {n}), there is a small number {\delta} so that

\displaystyle  	|f_n(x + x_n)| \geq \delta \hbox{\hskip 18pt} \forall |x| < \delta.

Thus, {\|\phi\|_{L^p} \geq \left(\int_{-\delta}^\delta \delta^p \; dx \right)^\frac{1}{p} \gtrsim \delta^{1+\frac{1}{p}} > 0}.

For the sake of discussion, suppose {\|f_n\|_{L^p} \rightarrow \|\phi\|_{L^p}} (i.e., {\phi} is capturing all the {L^p} norm). Then, by Radon–Riesz, {f_n \rightarrow \phi} in {L^p} and we can finish the proof just as above. The remaining alternative is that {0 < \|\phi\|_{L^p} < \lim_{n \rightarrow \infty} \|f_n\|_{L^p} = J_{\max}}. In that case write

\displaystyle  	f_n(x) = \phi(x - x_n) + \underbrace{r_n(x)}_{\text{remainder}}.

Note that {1 = \|f_n\|_{L^2} = \langle \phi, \phi \rangle + \langle r_n, r_n \rangle + 2 \text{ Re }\langle \phi(x) , f_n(x-x_n) - \phi(x) \rangle} and that {\langle \phi(x) , f_n(x-x_n) - \phi(x) \rangle \rightarrow 0} as {n \rightarrow \infty} because {f_x(x-x_n) \rightharpoonup \phi}. This shows that {\phi(x-x_n)} and {r_n} are “asymptotically orthogonal in the Pythagorean sense

\displaystyle  	\lim \|r_n\|_{L^2}^2 + \|\phi\|_{L^2}^2 = 1 \hbox{\hskip 18pt but \hskip 18pt } 1= \|\phi(x-x_n) + r_n(x) \|_{L^2}.


\displaystyle  	J_{\max} = \lim \int |f_n|^p \; dx.


\displaystyle  	J_{\max} = \lim_{n \rightarrow \infty} \int|f_n|^p \; dx = \lim_{n \rightarrow \infty} |\phi(x) + r_n(x+x_n)|^p \; dx = \lim_{R \rightarrow \infty} \lim_{n \rightarrow \infty} \int_{|x| \leq R} + \int_{|x| \geq R} = \lim_{n \rightarrow \infty} \int |\phi|^p + \int |r_n(x+x_n)|^p \; dx,

where the first term is due to {r_n(x+x_n) \rightarrow 0} in {L^p} con compact sets and the second term is due to {\int_{|x| \geq R} |\phi|^2 \; dx} going to {0} as {R \rightarrow \infty}. That is,

\displaystyle  	\lim_{n \rightarrow \infty} \|\phi\|_{L^p}^p + \|r_n\|_{L^p}^p = \lim_n \|f_n\|_{L^p}^p = J_{\max}

The remaining goal is to show that {\|r_n\|_{L^p}}, because then {\|f_n\|_{L^p} \rightarrow \| \phi \|_{L^p}}, which then implies {\int |f_n(x+x_n) - \phi(x)|^p \rightarrow 0}.

The key to showing this is sub-additivity:

\displaystyle  \begin{array}{rcl}  	J_{\max} &\longleftarrow& \displaystyle \Big(\|\phi\|_{L^p}^p + \|r_n\|_{L^p}^p\Big)^{1/p} \leq J_{\max} \Big( \|\phi\|_{L^2}^{p(1-\theta)} \|\nabla \phi\|_{L^2}^{p\theta} + \|r_n\|_{L^2}^{p(1-\theta)} \|\nabla r_n\|_{L^2}^{p\theta}\Big)^{1/p} \\ \\ 		&\qquad& \displaystyle \leq J_{\max} \Big( (1-\theta) \|\phi\|_{L^2}^p + \theta \|\nabla \phi\|_{L^2}^p + (1-\theta) \|r_n\|_{L^2}^p + \theta \|\nabla r_n\|_{L^2}^p\Big)^{1/p} 	\end{array}

and so

\displaystyle  	(1-\theta)\Big(\|\phi\|_{L^2}^p + \|r_n\|_{L^2}^p\Big) + \theta \Big( \|\nabla \phi\|_{L^2}^p + \|\nabla r_n\|_{L^2}^p \Big) \rightarrow 0.

Because we also had

\displaystyle  \begin{array}{rcl}  	\|\phi\|_{L^2}^2 + \|r_n\|_{L^2}^2 &\rightarrow& 1 = \|f_n\|_{L^2} \\ \\ 	\|\nabla \phi\|_{L^2}^2 + \|\nabla r_n\|_{L^2}^2 &\rightarrow& = \|\nabla f_n\|_{L^2}. 	\end{array}

Therefore, because {p >2}, the equalities

\displaystyle  	a^2 + b^2 = 1, \hbox{\hskip 18pt and \hskip 18pt} a^p + b^p = 1

must imply {a = 0} or {b = 0} (this can be seen by drawing the unit balls). Namely, {a_n^2 + b_n^2 \rightarrow 1} and {a_n^p + b_n^p \rightarrow 1} implies {a_n \rightarrow 0} or {b_n \rightarrow 0}. As {\|\phi\|_{L^2} \neq 0 } and {\|\nabla \phi\|_{L^2} \neq 0}, we must have {\|r_n\|_{L^2} + \|\nabla r_n\|_{L^2} \rightarrow 0} and hence {f_n(x+x_n) \rightarrow \phi} in {H^1}, and thus also in {L^p}. {\square}

In the original vocabulary of concentration compactness, there are three scenarios for sequences {\|f_n\|_{L^2} = \|\nabla f_n\|_{L^2} = 1},

  1. vanish, {\|f_n\|_{L^p} \rightarrow 0}.
  2. compactness (co-compactness) [modulo translations {f_n(x+x_n) \rightarrow \phi} in {L^p}.
  3. dichotomy {f_n \rightarrow \phi} strongly on compact sets bu {0 < \|\phi\|_{L^p} < \limsup \|f_n\|_{L^p}}.

at least after passing to a subsequence.

In the proof, we had a sequence {f_n} and either {\|f_n\|_{L^p} \rightarrow 0} or it contained (after passing to a subsequence and translating in space), a “bubble of concentration” {\phi}, leaving behind a remainder {r_n = f_n(x) - \phi(x - x_n)}. We can proceed inductively and look inside {r_n} for a further bubble of concentration and so

Theorem Fix {2 < p < \infty}. Let {f_n} be a bounded sequence in {H^1({\mathbb R}^)}. Then, after passing to a subsequence, we can decompose {f_n} as

\displaystyle  	f_n(x) = \sum_{j=1}^J \phi^j (x - x_n^j) + r_n^J(x)

for each {0 \leq J \leq M_\phi \in {\mathbb N} \cup \{\infty\}} where {M_\phi} is the “total number of bubbles” such that

  1. for all {j} and {\phi^j \not \equiv 0}, {f_n(x+x_n^j) \rightarrow \phi^j(x)} weakly in {H^1} and strongly in {L^p} on compact sets.
  2. If {j < j'}, then {|x_n - x_n^{j'}| \rightarrow \infty}.
  3. {\lim_{J \rightarrow M_\phi} \lim_{n \rightarrow \infty} \| r_n^J\|_{L^p} = 0}.
  4. For all {J}, {\|f_n\|_{L^2}^2 - \left( \sum_{j=1}^J \|\phi^j\|_{L^2}^2 + \|r_n^J\|_{L^2}^2\right) \rightarrow 0} as {n \rightarrow \infty}. Furthermore {\|\nabla f_n\|_{L^2}^2 - \left( \sum_{j=1}^J \|\nabla \phi^j\|_{L^2}^2 + \|\nabla r_n^J\|_{L^2}^2\right) \rightarrow 0} and {\lim_{J \rightarrow M_\phi} \limsup_{n \rightarrow \infty} \Big| \|f_n\|_{L^2}^p - \sum_{j=1}^J \|\phi^j\|_{L^2}^p\Big| = 0}.

Remark For assertion 2, if {(j,j')} was the minimal pair such that {x_n^j - x_n^{j'}} does not diverge, then, along a subsequence, it converges, say to {y} and so

\displaystyle  	f_n(x + x_n^j) \rightarrow \phi^j \hbox{\hskip 18pt and \hskip 18pt} \phi^{j'} \longleftarrow f(x+x_n^{j'}) = f(x + x_n^{j} -(x_n^{j} - x_n^{j'})).

Thus, {\phi^{j'}(x) = \phi^j(x-y)}, which is to say we are removing the same “bubble” twice, a troubling fact. In the proof, we find {\phi^{j'}} by applying our “bubble-finding lemma” to {r_n^{j'-1}}, which converges to zero when translated by {x_n^j}. Any limit along that sequence was removed by {\phi^j} and hence {\phi^{j'}} would be identically {0}.


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