I have been reading the paper on nonlinear Klein-Gordon equation(NLKG) for initial data in modulation space: For detail please see the paper "Klein-Gordon Equations on Modulation Spaces (2014)" (side remark: modulation spaces are closely related to Besov spaces and very useful for PDE view point.)
And I have been trying to figure out the proof of the following theorem.
Theorem 6.(persistence of regularity). Let $1\leq p_{1}<p_{2}\leq \infty$ and $k\in \mathbb N$, $0\leq s_{1} \leq s_{2} < \infty $ and let $u$ be a strong $M^{s_{2}}_{p_{2},1}$ solution to NLKG with its maximal existence interval $I.$ If $u(t_{0})\in M^{s_{1}}_{p_{1},1}$, and $u_{t}(t_{0})\in M^{s_{1}-1}_{p_{1},1}$ for some $t_{0}\in I$ then $u$ is also a strong $M^{s_{1}}_{p_{1},1}$ solution to NLKG with the same maximal interval.
My Question: How to write the rigorous proof or the above Theorem 6?
[The authors says it follows by the following two Lemmas 31 and 32: but I am unable to see how does it follows? My little effort, $p_{1}<p_{2}\implies M_{p_{1}, 1} \subset M_{p_{2},1}$; and so $\|\cdot\|_{M_{p_{2},1}} \lesssim \|\cdot\|_{M_{p_{1,1}}}$; so my confusion: if we have strong $M_{p_{2},1}$ solution how does it gives strong-$M_{p_{1},1}$ solution. Why the interval is SAME? How to use the following lemmas effectively? See also Remark 7 in the above paper(for getting the flavour how it is useful for global well posedness)]
Lemma 31. Let $I$ be a bounded time interval containing $t_{0}$, let $p\in [1, \infty]$, and let $u\in C(I, M^{s}_{p,1})$ be a strong $M^{s}_{p,1}$ solution to NLKG. If the quantity $\|u\|_{L_{t}^{k} (I, M^{s}_{\infty,1})}$ is finite, then one has\ $\|u(t)\|_{L_{t}^{\infty}(I, M^{s}_{p,1})} \lesssim C(|I|) \left( \|u(t_{0}\|_{M^{s}_{p,1}} +\|u_{t}(t_{0}\|_{M^{s-1}_{p,1}} \right) \exp \left( \int_{I} C(|I| \|u(\tau)\|^{k}_{M^{s}_{\infty,1}} d\tau \right)$\ where $C(|I|)$ is some positive constant associated with the length of $I.$ [For the proof see the above paper, page no. 11; it is simple and follows as an application of Gronwall inequality]
Lemma 32. Let $I$ be a bounded time interval containing $t_{0}$, let $p\in [1, \infty]$, and let $u\in C(I, M^{s}_{\infty,1})$ be a strong $M^{s}_{\infty,1}$ solution to NLKG. If the quantity $\|u\|_{L_{t}^{k} (I, M^{0}_{\infty,1})}$ is finite, then one has\ $\|u(t)\|_{L_{t}^{\infty}(I, M^{s}_{p,1})} \lesssim C(|I|) \left( \|u(t_{0}\|_{M^{s}_{\infty,1}} +\|u_{t}(t_{0}\|_{M^{s-1}_{\infty,1}} \right) \exp \left( \int_{I} C(|I| \|u(\tau)\|^{k}_{M^{0}_{\infty,1}} d\tau \right)$\ where $C(|I|)$ is some positive constant associated with the length of $I.$ [For the proof see the above paper see, page no. 11]
EDIT: [In what follows I am trying to figure out the answer in line of comments given below by Willie Wong.]
I used the Theorem 1 (local well posedness) from the same paper). Let $I_{1}$ be compact interval containing $t_{0}$ and $u(t_{0})\in M^{s_{1}}_{p_{1},1}$, and $u_{t}(t_{0})\in M^{s_{1}-1}_{p_{1},1}$; and therefore by Theorem 1, it follows that, $u$ is a strong $M^{s_{1}}_{p_{1}, 1}$ solution to NLKG, in the sense that, $u\in C(I_{1}, M^{s_{1}}_{p_{1},1}) \cap C^{1}(I_{1}, M^{s_{1}-1}_{p_{1},1}).$ And now as an application of Lemma 31; we get, $\|u(t)\|_{L_{t}^{\infty}(I_{1}, M^{s}_{p_{1},1})} \lesssim C(|I_{1}|) \left( \|u(t_{0}\|_{M^{s}_{p_{1},1}} +\|u_{t}(t_{0}\|_{M^{s-1}_{p_{1},1}} \right) \exp \left( \int_{I} C(|I_{1}| \|u(\tau)\|^{k}_{M^{s}_{\infty,1}} d\tau \right);$\ as you have pointed out: since $M^{p_{2},1}\subset M^{\infty, 1};$ it follows that, $\|u(t)\|_{M_{\infty,1}} \leq \|u(t)\|_{M_{p_{2,1}}};$ and so $\|u\|_{L^{\infty}(I_{1}, M_{\infty, 1})} \leq \|u\|_{L^{\infty}(I_{1}, M_{p_{2},1})}$ (is finite by hypothesis of Theorem 6, since $I_{1}$ is compact). And therefore $\|u(t)\|_{L_{t}^{\infty}(I_{1}, M^{s}_{p_{1},1})} < \infty.$
But why the $I_{1}$ and $I$ are same? For instance, what will happen if I start $I=\mathbb R$ in Theorem 6? So in that case how should I show: $u\in C(\mathbb R, M^{s_{1}}_{p_{1},1}) \cap C^{1}(\mathbb R, M^{s_{1}-1}_{p_{1},1})$?($I_{1}$ could be arbitrary large compact interval but from this how to extend solution continuously to open $\mathbb R$(full time)) Don't we need to use Lemma 32 somewhere? Where does it comes into the picture? Please tell me if I have been doing some mistakes; thanks a lot.
