**Motivation**: I'm trying to understand the proof of Theorem 3.1 in [Antonelli, Saut, and Sparber - Well-Posedness and averaging of NLS with time-periodic dispersion management](https://arxiv.org/abs/1204.4468). Though in the following I'm raising kind of abstract question.


Consider non-linear Schrödinger equation (NLS):
$$i\partial\_tu +\Delta u = |u|^{2}u, \quad u(t\_0, x)= \varphi(x).$$


Suppose $X$ (space of functions on $\mathbb R^d$) is Banach space and assume that
$$\|u\|\_{L\_I^{\infty}X} \lesssim \|\varphi\|\_{X} + |I|\|u\|^3\_{L\_{I}^{\infty}X}$$
here $I$ is small time interval.


**Questions**: (1) How to use the standard continuity argument to say that: there exists a solution $u$ to NLS in $I \times \mathbb R^d$ such that
$$\|u\|\_{L^{\infty}\_{I} X} \leq C \|\varphi \|\_{X}$$
for sufficiently small $I$? (2) What is standard continuity argument?



What is standard continuity argument for well-posedness?
I understand continuity argument a bit differently than the other post. Suppose you've shown the inequality
$$\|u\|\_{L^\infty(I,X)} \leq C(\|\varphi\|\_{X} + |I|\|u\|\_{L^\infty(I,X)}^3) \tag{1},$$
for some absolute constant $C>0$ and all intervals $I$ with $|I|\leq\delta$.


Since $u(t\_0)=\varphi$ and your solution $t\mapsto u(t)$ is continuous with values in $X$, there exists a $t\_\*>t\_0$ such that for $t\in [t\_0,t\_\*]$, $\|u(t)\|\_{X}\leq \frac{3C}{2}\|\varphi\|\_{X}$. Let $T\_\*>t\_0$ be the maximal such $t\_\*$, that is
$$\forall t\_0\leq t\leq T\_\*, \enspace \|u(t)\|\_{X}\leq \frac{3C}{2}\|\varphi\|\_{X}$$
and there exists a sequence $t\_n\rightarrow T\_\*^{+}$ such that $\|u(t\_n)\|\_{X}>\frac{3C}{2}\|\varphi\|\_{X}$. Note this implies $\|u(T\_\*)\|\_{X} = \frac{3C}{2}\|\varphi\|\_{X}$. If no such $T\_\*$ exists (i.e., $\|u(t)\|\_{L^\infty([t\_0,\infty),X)}\leq \frac{3C}{2}\|\varphi\|\_{X}$), then there is nothing to prove, so we may assume otherwise.


We can use inequality (1) to get a lower bound for $T\_\*-t\_0$. Indeed, setting $I=[t\_0,T\_\*]$, if $|I|>\delta$, then there is nothing to prove. If $|I|\leq \delta$, then
$$\|u\|\_{L^\infty(I,X)} \leq C\|\varphi\|\_{X} + \frac{27C^4(T\_\*-t\_0)}{8}\|\varphi\|\_{X}^3.$$
This implies that $\frac{27C^3(T\_\*-t\_0)}{8}\|\varphi\|\_{X}^2 \geq \frac{1}{2}$, otherwise we would have
$$\|u\|\_{L^\infty(I,X)} < \frac{3C}{2}\|\varphi\|\_{X} \Longrightarrow \|u(T\_\*)\|\_{X} < \frac{3C}{2}\|\varphi\|\_{X},$$
which contradicts our choice of $T\_\*$.


