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## May 4, 2011

### Harmonic Analysis Lecture Notes 21

Filed under: 2011 Harmonic Analysis Course — xuhmath @ 5:56 pm

In this discussion, we will prove the following theorem.

Theorem (Sharp Sobolev Embedding, Aubin & Talenti) Fix ${d \geq 3}$. The inequality

$\displaystyle \|f\|_{L^\frac{2d}{d-2}} \leq C_d \| \nabla f\|_{L^2}$

where the constant ${c_d}$ is determined by inserting ${W(x) = (1 + \frac{|x|^2}{d(d-2)})^{-\frac{d-1}{2}}}$. Moreover, equality occurs in the above if and only if ${f(x) = \alpha W(\frac{x-x_0}{\lambda})}$ for some ${\alpha \in {\mathbb C}}$, ${x_0 \in {\mathbb R}^d}$, and ${\lambda > 0}$.

The most difficult part of the proof is to show that there is indeed an optimizer (cf previous discussion with the sharp Gagliardo–Nirenberg inequality). Moreover, once we have the correct “bubble decomposition’, the existence of an optimizer is then a simple sub-additivity argument.

Theorem Let ${f_n}$ be a bounded sequence in ${\dot H^1({\mathbb R}^d)}$ with ${d \geq 3}$. Then there exist ${J_{\max} \in {\mathbb N} \cup \{ \infty\}}$, ${\{ \phi^j\}_{j=1}^{J_{\max}} \subseteq \dot H^1({\mathbb R}^d)}$, ${\{x_n^j\}_{j=1}^{J_{\max}}}$, ${\{\lambda_n^j\}_{j=1}^{J_{\max}} \in (0,\infty)}$ so that along a subsequence in ${n}$ we can write

$\displaystyle f_n(x) = \sum_{j=1}^J (\lambda_n^j)^\frac{2-d}{2}\phi\left(\frac{x-x_n^j}{\lambda_n^j}\right) + r_n^J$

for each finite ${J \geq J_{\max}}$, with the following properties

1. ${\phi^j = \text{w-lim}_{n\rightarrow \infty} (\lambda_n^j)^\frac{d-2}{2}\phi\left(\lambda_n^j[x-x_n^j]\right)}$ in ${\dot H^1}$.
2. ${\lim_{j \rightarrow J_{\max}} \lim_{n\rightarrow\infty} \| r_n^J\|_{L^\frac{2d}{d-2}} = 0}$.
3. For all ${J}$, ${\lim_{n\rightarrow\infty} \Big| \|f_n\|_{\dot H^1}^2 - \sum_{j=1}^J \|\phi^j\|_{\dot H^1}^2 - \|r_n^J\|_{\dot H^1}^2 \Big| = 0}$.
4. ${\lim_{J \rightarrow J_{\max}} \lim_{n\rightarrow\infty} \Big| \|f_n\|_{L^\frac{2d}{d-2}}^\frac{2d}{d-2} - \sum_{j=1}^J \|\phi_j\|_{L^\frac{2d}{d-2}}^\frac{2d}{d-2} \Big| = 0}$.
5. ${\frac{|x_n^j x_n^{j'}|}{\sqrt{\lambda_n^j\lambda_n^{j'}}} + \big| \log \left(\frac{\lambda_n^j}{\lambda_n^{j'}}\right) \big| \rightarrow \infty}$.