# Welcome.

## 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

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

Similarly,

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

Moreover,

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

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

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

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

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 ${0}$.