# Compactness modulo symmetries of critical NLS solution

I was referred to the paper C. Kenig and F. Merle, "Global well-posedness, scattering, and blow-up for the energy critical, focusing non-linear Schrodinger equation in the radial case", in which one of the propositions (Proposition 4.2) concerns compactness modulo symmetries of the flow of a certain critical element to the energy-critical NLS

$$\begin{cases}iu_{t}+\Delta u \pm |u|^{\frac{4}{N-2}}u=0 & {(x,t)\in\mathbb{R}^{N}\times\mathbb{R}}\\ u|_{t=0}=u_{0}\in\dot{H}^{1}(\mathbb{R}^{N}) & {}\end{cases} \tag{CP}$$

Proposition 4.1, which I have omitted because I do not think it essential for my question, gives the existence of a critical initial datum $u_{0,C}\in\dot{H}^{1}$ with corresponding solution $u_{C}$ to the Cauchy problem, such that $u_{C}$ has maximal time interval of existence $I$ containing the origin, and $\|u_{C}\|_{S(I)}=+\infty$, where $S(I)$ is a certain space-time norm.

Proposition 4.2 Assume $u_{C}$ is as in Proposition 4.1 and that $\|u_{C}\|_{S(I_{+})}=+\infty$, where $I_{+}=(0,+\infty)\cap I$. Then there exists $x(t)\in \mathbb{R}^{N}$ and $\lambda(t)\in\mathbb{R}^{+}$, for $t\in I_{+}$, such that $$K = \{v(x,t) : v(x,t) = \frac{1}{\lambda(t)^{(N-2)/2}}u_{C}(\frac{x-x(t)}{\lambda(t)},t)\}$$ has the property that $\overline{K}$ is compact in $\dot{H}^{1}$.

I took a look at the proof on pg. 13, and I am having trouble arriving at the first claim the authors present, assuming that Proposition 4.2 is false. The aforementioned claim is reproduced below.

Using the notation $u(x,t)=u_{C}(x,t)$, the authors claim that if Proposition 4.2 is false, then there exists an $\eta_{0}>0$ and a sequence $\{t_{n}\}_{n=1}^{\infty}$ of times $t_{n}\geq 0$, such that, for all $\lambda_{0}\in\mathbb{R}^{+}$, $x_{0}\in\mathbb{R}^{N}$, we have $$\|\frac{1}{\lambda_{0}^{(N-2)/2}}u(\frac{x-x_{0}}{\lambda_{0}}, t_{n})-u(x,t_{n}')\|_{\dot{H}_{x}^{1}}\geq \eta_{0}, \quad n\neq n' \tag{*}$$

If Proposition 4.2 is false, then we know that for all functions $x:I_{+}\rightarrow\mathbb{R}^{N}$ and $\lambda:I_{+}\rightarrow\mathbb{R}^{+}$, there exist a sequence of times $\{t_{n}\}$ such that $$\{\frac{1}{\lambda(t_{n})^{(N-2)/2}}u(\frac{x-x(t_{n})}{\lambda(t_{n})},t_{n})\}_{n=1}^{\infty}$$ has no convergent subsequence. However, it's mot clear to me how to obtain what Kenig-Merle write. Rather, it seems to me that (*) follows from assuming that that flow $\{u(t) : t\in I\}$ is not relatively compact in $G \backslash\dot{H}^{1}$, where $G$ is the group of translations and dilations associated to the equation and we equip this space with the quotient metric.

Perhaps something related to what I am asking is the following from pg. 6 of the paper T. Tao, M. Visan, and X. Zhang, "Minimal-mass blowup solutions of the mass-critical NLS".

A function $u\in C_{t,loc}^{0}L_{x}^{2}(I\times\mathbb{R}^{N})$ is almost periodic modulo $G$ if the quotiented orbit $\{Gu(t) : t\in I\}$ is a precompact subset of $G\backslash L_{x}^{2}(\mathbb{R}^{N})$. Equivalently, $u$ is almost periodic modulo $G$ if there exists a compact subset $K\subset L_{x}^{2}(\mathbb{R}^{N})$ such that $u(t)\in GK$, $t\in I$.

Above, $G$ is the group of symmetries now associated to the mass-critical NLS. Proving the equivalence assertion seems relevant to what I am asking in regards to the Kenig-Merle paper. I feel like I'm missing some standard functional analysis/topology argument implicit in the cited works, but it's not clear to me what such an argument is. Any help would be greatly appreciated.

• If $\{u(t): t \in I\}$ were totally bounded in $G \backslash \dot H^1$, then there would exist $g(t) \in G$ such that $\{ g(t) u(t): t \in I \}$ was totally bounded in $\dot H^1$. Now take contrapositives (and use Heine-Borel). – Terry Tao Apr 28 '17 at 5:59
• @TerryTao Dear Professor Tao, thanks for your comment. My difficulty with your suggestion, which I think is just a reformulation of my difficulty as stated in my original question, is that I do not know how to show the proposition contained in your first sentence. It's clear to me that for each, say $2^{-n}$, I can find a function $g_{n}(t)\in G$ such that $\{g_{n}(t)u(t):t\in I\}$ is contained in finitely many balls of radius $2^{-n}$ in $\dot{H}^{1}$. But I do not see why I can extract a limiting $g$ such that that $\{g(t)u(t) :t \in I\}$ is totally bounded in $\dot{H}^{1}$. – Matt Rosenzweig Apr 28 '17 at 6:43
• Given any two vectors $f_1, f_2$ in $\dot H^1$, $\|g(t) f_1 - f_2\|_{\dot H^1}$ will converge to $(\|f_1\|_{\dot H^1}^2 + \|f_2\|_{\dot H^1}^2)^{1/2}$ as $t \to \infty$. As a consequence, if $g_n(t) u(t)$ is within $2^{-n}$ of some fixed $f_n$, and $g_{n+1}(t) u(t)$ is within $2^{-n-1}$ of some fixed $f_{n+1}$, and $u(t)$ has norm larger than, say, $10 \times 2^{-n}$, then $g_{n+1}(t)^{-1} g_n(t)$ is constrained to a compact set that depends only on $f_n$ and $f_{n+1}$. One can use this to adjust the $g_{n+1}$ and $f_{n+1}$ so that $g_{n+1}(t)$ is within, say, $2^{-n}$ of $g_n(t)$ ... – Terry Tao Apr 28 '17 at 16:06
• (increasing the number of balls needed to cover at the $2^{-n-1}$ stage if necessary). Iterating this, the $g_n(t)$ now converge to a usable limit $g(t)$. – Terry Tao Apr 28 '17 at 16:07
• [Correction: it is $g_{n+1}(t) g_n(t)^{-1}$ that is constrained, not $g_{n+1}(t)^{-1} g_n(t)$.] – Terry Tao Apr 28 '17 at 16:12

The following is based on Professor Tao's very helpful comments. I have included them in a separate answer, rather than editing the original question, due to length, not for claim of originality.

Abbreviate the critical element $u_{C}(x,t)=u(x,t)$. By the small data theory and the property $\|u\|_{S([0,T_{+}(u_{0})))}=+\infty$, we know that there exists an $A_{0}>0$ such that $\inf_{t\in[0,T_{+}(u_{0}))}\|u(t)\|_{\dot{H}^{1}}\geq A_{0}$. Set $I:=[0,T_{+}(u_{0}))$.

Suppose that we have constructed the functions $g_{1},\ldots,g_{n}: I \rightarrow G$ such that for $1\leq j\leq n$, there exists $f_{j,1},\ldots,f_{j,N_{j}}\in\dot{H}^{1}$ such that for every $t\in I:=[0,T_{+}(u_{0}))$, there exists $k$ such that $$|g_{j}(t)u(t)-f_{j,k}\|_{\dot{H}^{1}}\leq \epsilon(A_{0})2^{-j};$$ $$\sup_{t\in I}\|g_{j+1}u(t)-g_{j}(t)u(t)\|_{\dot{H}^{1}} \leq C\epsilon(A_{0})2^{-j}, \quad \forall j=1,\ldots,n-1;$$ and $$|g_{j+1}(t)-g_{j}(t)|\leq \epsilon(A_{0})2^{-j}, \quad \forall t\in I, \enspace \forall j=1,\ldots,n-1$$ where $\epsilon(A_{0})$ is some small constant depending only on the lower bound $A_{0}$.

We proceed to construct $g_{n+1}: I\rightarrow G$. Let $f_{n,1},\ldots,f_{n,N_{n}}$ be the elements of $\dot{H}^{1}$ such that for all $t\in I$, there exists $j\in\{1,\ldots,N_{n}\}$ such that $$\|g_{n}(t)u(t)-f_{n,j}\|_{\dot{H}^{1}}\leq 2^{-n}$$ By compactness of the orbits, we know that there exists a function $\tilde{g}_{n+1}:I\rightarrow G$ and elements $\tilde{f}_{n+1,1},\ldots,\tilde{f}_{n+1,\tilde{N}_{n+1}}$ such that for all $t\in I$, there exists $j\{1,\ldots,\tilde{N}_{n+1}\}$ such that $$\|\tilde{g}_{n+1}(t)u(t)-\tilde{f}_{n+1,j}\|_{\dot{H}^{1}}\leq 2^{-n-3}$$ Divide $I$ into finitely many intervals $I(f_{n,j},\tilde{f}_{n+1,k})$ such that $$\forall t\in I(f_{n,j},\tilde{f}_{n+1,k}), \quad \|g_{n}(t)u(t)-f_{n,j}\|_{\dot{H}^{1}}\leq 2^{-n}, \enspace \|\tilde{g}_{n+1}(t)u(t)-\tilde{f}_{n+1,k}\|_{\dot{H}^{1}}\leq 2^{-n-3}$$

We claim that for all $t\in I(f_{n,j},\tilde{f}_{n+1,k})$, $\tilde{g}_{n+1}(t)g_{n}(t)^{-1}$ lies in a compact subset depending only on $f_{n,j}$ and $\tilde{f}_{n+1,k}$. Indeed, otherwise there is a subsequence of times $t_{m}$ such that $\tilde{g}_{n+1}(t_{m})g_{n}(t_{m})\rightharpoonup 0$ and therefore \begin{align*} 2^{-n+1}&\geq \|\tilde{g}_{n+1}(t_{m})u(t_{m})-\tilde{f}_{n+1,k}\|_{\dot{H}^{1}}+\|g_{n}(t_{m})u(t_{m})-f_{n,j}\|_{\dot{H}^{1}}\\ &\geq \|g_{n}(t_{m})^{-1}f_{n,j}-\tilde{g}_{n+1}(t_{m})^{-1}\tilde{f}_{n+1,k}\|_{\dot{H}^{1}}\\ &=\|\tilde{g}_{n+1}(t_{m})g_{n}(t_{m})^{-1}f_{n,j}-\tilde{f}_{n+1,k}\|_{\dot{H}^{1}}\\ &\stackrel{m\rightarrow\infty}\longrightarrow (\|f_{n,j}\|_{\dot{H}^{1}}^{2}+\|\tilde{f}_{n+1,k}\|_{\dot{H}^{1}}^{2})^{1/2}\\ &\geq 2^{-n+2}, \end{align*} provided $n$ is sufficiently big depending on $A_{0}$ (this isn't an issue provided we choose $\epsilon(A_{0})$). Since the set $\{\tilde{g}_{n+1}(t)g_{n}(t)^{-1} : t\in I(f_{n,j},\tilde{f}_{n+1,k})\}$ lies in a compact set. We can cover it by finitely many balls $B_{1},\ldots,B_{M}$ of diameter at most $\min_{j,k}\delta(f_{n,j},\tilde{f}_{n+1},k)>0$, where $\delta(f_{n}-f_{n+1})$ is chosen so that for any $g\in G$ with $|g-id|\leq \delta(f_{n},f_{n+1})$, $$\|gf_{n,j}-f_{n,j}\|_{\dot{H}^{1}}\leq \epsilon(A_{0})2^{-n}, \quad \delta(f_{n,j},\tilde{f}_{n+1,k})\leq \epsilon(A_{0})2^{-n}$$ For each $l=1,\ldots,M$, let $g_{l}^{*}$ denote the center of $B_{l}$. Define the intervals $$I(f_{n,j},\tilde{f}_{n+1,k},l) := \{t\in I(f_{n,j},\tilde{f}_{n+1,k}) : \tilde{g}_{n+1}(t)g_{n}(t) \in B_{l}\},$$ where we make a choice in the event of overlap so that the $I(f_{n,j},\tilde{f}_{n+1,k},l)$ form a partition of $I(f_{n,j},\tilde{f}_{n+1,k})$. Now define the new function $g_{n+1}$ on $I(f_{n,j}, \tilde{f}_{n+1,k},l)$ by $$g_{n+1}(t) := ({g}_{l}^{*})^{-1}\tilde{g}_{n+1}(t), \quad t\in I(f_{n,j}, \tilde{f}_{n+1,k},l)$$ Next, define new elements $f_{n+1,k,l}$ of $\dot{H}^{1}$ by $$f_{n+1,k,l} := (g_{l}^{*})^{-1}f_{n+1}$$ so that for every $t\in I(f_{n,j},\tilde{f}_{n+1,k},l)$, $$\|g_{n+1}(t)u(t)-{f}_{n+1,k,l}\|_{\dot{H}^{1}}\leq \epsilon(A_{0})2^{-n-1}$$ Additionally, we have that for all $t\in I(f_{n,j},\tilde{f}_{n+1,k},l)$, \begin{align*} \|g_{n+1}(t)u(t)-g_{n}(t)u(t)\|_{\dot{H}^{1}} &= \|(g_{n+1}(t)g_{n}(t)^{-1})g_{n}(t)u(t)-g_{n}(t)u(t)\|_{\dot{H}^{1}}\\ &\leq \|(g_{n+1}(t)g_{n}(t)^{-1})g_{n}(t)u(t)-f_{n,j}\|_{\dot{H}^{1}} + \|f_{n,j}-g_{n}(t)u(t)\|_{\dot{H}^{1}}\\ &=\|g_{n}(t)u(t)-g_{n}(t)g_{n+1}(t)^{-1}f_{n,j}\|_{\dot{H}^{1}} + \epsilon(A_{0})2^{-n}\\ &\leq \epsilon(A_{0})2^{-n+1}+\|((g_{l}^{*})^{-1}-g_{n}(t)g_{n+1}(t)^{-1})f_{n,j}\|_{\dot{H}^{1}}\\ &\leq 3\epsilon(A_{0})2^{-n} \end{align*}

We now repeat this procedure on each of the finitely remaining intervals $I(f_{n,j'},\tilde{f}_{n+1,k'})$, each time adjoining finitely many new elements $f_{n+1,k',1},\ldots,f_{n+1,k',M_{j',k'}}$ to the net. In the end, we have constructed the function $g_{n+1}$ on the entire interval $I$ and adjoined finitely many elements to the $(n+1)$-th stage net.

Now define $$g(t) := \lim_{n\rightarrow\infty}g_{n}(t), \quad t\in I$$ To see that the set $\{g(t)u(t) : t\in I\}$ is totally bounded, we note that the sequence $g_{n}(t)u(t)\rightarrow g(t)u(t)$ in $\dot{H}^{1}$, uniformly in $t\in I$.