Exercise 3 of Lecture notes for 18.155 On concentration compactness and soliton solutions for the NLS equation

Question

As picture below, I want to prove the Proposition 2.1. I want to use Poincare inequality, but fail. How should I do it ? Thanks for any answer or hint.

Answer 1

Poincare's inequality only holds in bounded domains. You have to use Gagliardo-Nirenberg inequality, which says that$$ \Vert u\Vert_{L^{q}(\mathbb{R}^{d})}\leq c_{1}\Vert u\Vert_{L^{2}% (\mathbb{R}^{d})}^{1-\theta}\Vert\nabla u\Vert_{L^{2}(\mathbb{R}^{d})} ^{\theta}, $$ where $\theta\in\lbrack0,1)$ and $\frac{1}{q}=\frac{1}{2}-\frac{\theta}{d}$. This gives $$ \frac{d(q-2)}{2q}=\theta\in\lbrack0,1) $$ and so you need $2\leq q$ and $\frac{d(q-2)}{2q}<1$ that is, $q(d-2)<2d$. Hence, if $d=2$ any $q$ will do, while if $d>2$, then you need $q<\frac {2d}{d-2}=2^{\ast}$ the critical exponent. In your case you have $q=p+1$ so $2<p+1<2+\frac{4}{d}=\frac{2(d+2)}{d}$ and so for $d>2$, $$ q=p+1<\frac{2(d+2)}{d}\leq\frac{2d}{d-2} $$ which holds because $(d+2)(d-2)\leq d^{2}$.

Update If you take any function $v\in H^{1}(\mathbb{R}^{d})$ with $\Vert v\Vert_{L^{2}}=\lambda$ and for $t>0$ define $u_{t}(x)=v(tx)t^{d/2}$, if you make the change of variables $y=tx$ you see that \begin{align*} \int_{\mathbb{R}^{d}}|u(x)|^{2}dx & =\int_{\mathbb{R}^{d}}|v(tx)|^{2}% t^{d}dx=\int_{\mathbb{R}^{d}}|v(y)|^{2}dy,\\ \int_{\mathbb{R}^{d}}|u(x)|^{p+1}dx & =\int_{\mathbb{R}^{d}}|v(tx)|^{p+1}% t^{d(p+1)/2}dx=t^{d(p+1)/2-d}\int_{\mathbb{R}^{d}}|v(y)|^{p+1}dy\\ \int_{\mathbb{R}^{d}}|\nabla u(x)|^{2}dx & =\int_{\mathbb{R}^{d}}|\nabla v(tx)|^{2}t^{2+d}dx=t^{2}\int_{\mathbb{R}^{d}}|\nabla v(y)|^{2}dy, \end{align*} and so \begin{align*} E(u_{t}) & =t^{2}\int_{\mathbb{R}^{d}}|\nabla v(y)|^{2}dy-\frac{1}% {p+1}t^{d(p+1)/2-d}\int_{\mathbb{R}^{d}}|v(y)|^{p+1}dy\\ & =t^{2}A-Bt^{d(p+1)/2-d}=g(t). \end{align*} Since $d(p+1)/2-d<2$ exactly for $p<1+\frac{4}{d}$ you have that $\inf g<0$ since for $t$ small $t^{d(p+1)/2-d}$ dominates.