I am trying to understand the motivation behind the main theorem of Keraani: See the top of page 356 and which is the motivation behind his Theorem 1.6

We consider $$i\partial_t u + \Delta u =0, u(x,0)=u_0 \in H^{1}(\mathbb R^3)$$ where $u:\mathbb R^{3+1} \to \mathbb C, $ $H^{1}(\mathbb R^3)= {f:\mathbb R^3 \to \mathbb C : \nabla f \in L^{2}(\mathbb R^3)} \}$. Put $\|f\|_{H^1}= \|\nabla f \|_{L^2}.$

By Strichartz estimate and Sobolov's inequality, we have

$$ \|e^{it \Delta}u_0\|_{L^{q}(\mathbb R_t, L^{r}(\mathbb R^3_{x}))} \leq C \|u_0\|_{H^1}$$

where a pair $(q,r)$ is $H^1-$ admissible, that is, $\frac{2}{q}+\frac{3}{r}=\frac{3}{2}$, $r\in [6, \infty).$

My Questions: When can we say the above estimte is not compact, and how to justify this? (What does this mean? How one should interprate, when authors says the estimate is not compact.) More specifically,

Let $\{x_n\}_{n\in \mathbb N}$ be a sequence of $\mathbb R^3$ and $\{t_n\}_{n\in \mathbb N}$ be a sequence of $\mathbb R$ both going to infinity. Also, let $\{h_n\}_{n\in \mathbb N}$ be a sequence of $\mathbb R^{*}_{+}$ going to 0.

Put $\tau_{x_0}\phi (x)=\phi(x-x_0), S_{h}\phi(x)=\frac{1}{\sqrt{h}} \phi (\frac{x}{h}), R_{t_0}\phi (x)= e^{it_0 \Delta} \phi (x), $ where $x, x_0 \in \mathbb R^3, t_0\in \mathbb R, h>0.$

(1) Why for every fixed function $\phi \in H^{1}$, the sequences $\tau_{x_n}\phi (x), S_{h_{n}}\phi(x), R_{t_n}\phi (x) $ converge weakly to 0 ?

(2) Why for every $H^1-$admissible pair $(q, r)$ the $L^{q}L^{r}$ norms of these sequences are equal to $\|e^{it\Delta} \phi \|_{L^{q}(\mathbb R, L^{r}(\mathbb R^3))}$ for every $n$? (These sequnces $\tau_{x_n}\phi (x)$ are functions on $\mathbb R^3,$ how one can take $L^{q}L^{r}$ for these sequnces, as my understnding is that $L^{q}L^{r}$ norm is meant for functions on $\mathbb R \times \mathbb R^3$)

(3) Why one can use (1) and (2) to conlude ``These sequences are not relatively compact in $L^{q}(\mathbb R, L^{r}(\mathbb R^3))$ "? (And what is a meaning by the sequence is not relatively compact in topological space)


1 Answer 1


You have two spaces $X$ and $Y$. You have a linear operator $L: X\to Y$ that is bounded (continuous). (Here $X = H^1(\mathbb{R}^d)$ and $Y$ is the Strichartz norm space on $\mathbb{R}^{d+1}$. The linear operator is $e^{it\Delta}$, the solution operator to the free Schrodinger equation.) So when the author says that the estimate is not compact it is meant that the linear operator $L$ is not a compact operator from $X \to Y$.

A witness to this fact would be a bounded sequence of initial data $u_0^{(k)}$ in $H^1$ such that the sequence of corresponding solutions $u^{(k)}$ has no converging subsequence.

(1) The weak convergence of $\tau_{x_n} \phi$ and $S_{h_n}\phi$ requires zero knowledge of the Schrodinger equation, and you can understand them if you understand what it means for functions to be $L^2$ integrable. If you have trouble figuring out why you need to stop, and re-study your basic real analysis until you understand this, before you try to learn concentration compactness.

The convergence of $R_{t_n} \phi$ is slightly harder, and one way to understand it is through the method of stationary phase by noting formally that the $L^2$ pairing $\langle R_{t_n}\phi, \psi\rangle = \int e^{it|\xi|^2} \hat{\phi}(\xi) \hat{\psi}(\xi) ~\mathrm{d}\xi$.

(2) It is a direct computation: the solutions to the free Schrodinger equations corresponding to $\tau_*\phi$, $S_*\phi$ and $R_*\phi$ (these solutions are the objects whose $L^qL^r$ norm you should compute) can be expressed using changes of variables in terms of the solution to the free Schrodinger equation with initial data $\phi$. Check that these variable changes leave the Strichartz norm invariant.

(3) Relatively compact, as explained by Wikipedia. You should be able to easily show (at least for the case of $R_{t_n}$ and $\tau_{x_n}$) that the corresponding solutions have no converging subsequence in $L^q L^r$. (For sufficiently large $x_n$, for example, the supports of the original solution and the translated solution are mostly disjoint.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.