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(May be this is very basic question for MO)

(For details or this question you may see the paper page no. 9, MR2506839, Local well-posedness of nonlinear dispersive equations on modulation spaces; Bull. Lond. Math. Soc. 41 (2009), no. 3, 549-558)

Let $u_{0}$ be in the modulation space $M^{p,1}(\mathbb R^{d})= M^{p,1}, \ (1\leq p < \infty).$

Theorem 1. If $u_{0}\in M^{p,1}$ then $||e^{it\bigtriangleup}u_{0}||_{M^{p,1}}\leq C (t^{2}+4\pi^{2})^{\frac{d}{4}} ||u_{0}||_{M^{p,1}},$ where $C$ is universal constant depending only on $d$. Therefore, $||e^{it\bigtriangleup}||_{M^{p,1}(\mathbb R^{d})}\leq C_{T} ||u_{0}||_{M^{p,1}(\mathbb R^{d})},$ where $C_{T}= \sup_{t\in [0, T)} C(t^{2}+ 4\pi^{2})^{d/4}.$ Consider the mapping, $$\mathcal{T}(u)=e^{it\bigtriangleup}(u_{0})-i \int_{0}^{t}e^{i(t-s)\bigtriangleup}( |u|^{2}u) (s) ds.$$ Let $B_{M}$ denote the closed ball of radius $M=2C_{T}\left\|u_{0}\right\|_{M^{p,1}}$ , centred at the origin in the space $C([0, T]; M^{p,1}).$

Theorem 2. If $u\in M^{p,1}$, then $\left\||u|^{2k}u \right\|_{M^{p,1}}\lesssim\left\|u\right \|_{M^{p,1}}^{2k+1}.$

By using, theorem 1 and theorem 2, we notice, \begin{eqnarray*} \left \|\int_{0}^{t}e^{i(t-s)\bigtriangleup}(\lambda |u|^{2k}u)(s)ds \right \|_{M^{p,1}} &\leq & \int_{0}^{t}||e^{i(t-s)\bigtriangleup}(\lambda |u|^{2k}u)(s) ||_{M^{p,1}} ds\\ &\leq & T C_{T} \sup_{t\in [0, T]} \left \|\lambda |u|^{2k}u \right\|_{M^{p,1} (\mathbb R^{d})}\\ & \lesssim & C_{T} T \left \| u(t)\right \|_{M^{p,1}}^{2k+1}. \end{eqnarray*} So,$\left\|\mathcal{T}(u) \right \|_{C([0, T]; M^{p,1})}\leq C_{T}(||u_{0}||_{M^{p,1}}+cT ||u||^{2k+1}_{M^{p,1}}),$ for some universal positive constant $c$. Let $u\in B_{M}$, we obtain,$\left\|T(u)\right \|_{C([0, T]; M^{p,1})} \leq \frac{M}{2}+cC_{T}T M^{2k+1}.$ Now, let $T$ be such that $cC_{T}M^{2k} \leq \frac{1}{2},$ that is, $T\leq \tilde{T}(||u_{0}||_{M^{p,1}}).$ We obtain, $\mathcal{T}(u) \in B_{M},$ and so $\mathcal{T}:B_{M}\to B_{M}.$

My Question: How to prove $$\left\|\mathcal {T}(u)-\mathcal {T} (v) \right \|_{C([0, T]; M^{p,1})}\leq \frac{1}{2}\left\|u-v\right \|_{C([0, T]; M^{p,1})} ? $$

The paper (p.9) I have been reading gives hint that one can use this identity, $(|u|^{2}u - |v|^{2}v)(s)= (u-v)|u|^{2}(s)+ v(|u|^{2}-|v|^{2}) (s)$; to prove the contraction, but I am unable to see, and I am wondering how such identity will be useful ? or may be some different method is available ?)

Any suggestions or comments or any reference will be helpful to me; thanks a lot;

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You also need a version of Theorem 2.1 for the product of two functions. This kind of extension is usually obtained through minor modifications in the proof of the case of a single power – Piero D'Ancona Mar 5 '14 at 12:23
    
@PD; what is known so far is; $||uv||_{M^{p,1}}\leq ||u||_{M^{p,1}} ||v||_{M^{p,1}}$ for $u,v \in M^{p,1}$; that is $M^{p,1}$ is an Banach algebra.; thanks for the attention; but I don't see how this is useful? – Inquisitive Mar 5 '14 at 12:38
    
Look at Formula (12) page 8 of the first paper you mention. – username Mar 5 '14 at 12:43
    
@AW; For $\frac{1}{p_{1}}+...+\frac{1}{p_{m}}=\frac{1}{p_{0}}$ and $\frac{1}{q_{1}}+...+\frac{1}{q_{m}} =m-1+\frac{1}{q_{0}},$; then $||u_{1}\cdot\cdot\cdot u_{m}||_{M^{p_{0},q_{0}}}\leq ||u_{1}||_{M^{p_{1},q_{1}}}\cdot \cdot \cdot ||u_{m}||_{M^{p_{m}, q_{m}}}$; but how this will be useful here ? – Inquisitive Mar 5 '14 at 13:08
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So \begin{eqnarray*} &&\left \|\int_{0}^{t}e^{i(t-s)\bigtriangleup}(\lambda |u|^{2k}u-\lambda |v|^{2k}v)(s)ds \right \|_{M^{p,1}} \\ &\leq & T C_{T} \sup_{t\in [0, T]} \left \|\lambda |u|^{2k}u-\lambda |v|^{2k}v \right\|_{M^{p,1}}\\ &\leq & T C_{T} \sup_{t\in [0, T]} \left \|\lambda |u|^{2k}(u-v)\right\|_{M^{p,1}}\\ &&+T C_{T} \sup_{t\in [0, T]} \left \|\lambda (|v|^{2k}-|v|^{2k})v \right\|_{M^{p,1} (\mathbb R^{d})}\\ & \lesssim & C_{T} T \left(\|u\|^{2k}_{M^{p,1}}+\|u\|^{2k-2}_{M^{p,1}} \|v\|_{M^{p,1}}+\|v\|_{M^{p,1}}^{2k}\right)\|u-v\|_{M^{p,1}}. \end{eqnarray*} – username Mar 7 '14 at 15:14

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