We consider the one dimensional cubic nonlinear Shr\"odinger equation (NLS): $$i\partial_{t}\phi (x,t) +\Delta \phi (x,t)= \pm |\phi (x,t)|^{2} \phi(x,t), \ (x, t\in \mathbb R),$$ $$\phi (x,0) = \phi_{0}(x)\in H^{s}(\mathbb R);$$ where $H^{s}(\mathbb R)$ is usual Sobolev space.
In 1978 Ginibre-Velo have shown that the above NLS is globally wellposed: that is, for the initial data in $\phi_{0}\in H^{1}(\mathbb R)$, the NLS has a unique solution in $C(\mathbb R, H^{1}(\mathbb R)).$ In 1987 Yoshio Tsutsumi have shown that: for the initial data in $\phi_{0}\in L^{2}(\mathbb R)$, the NLS has a unique global solution in $C(\mathbb R, L^{2}(\mathbb R)).$
We note that, $H^{1}(\mathbb R)\subset H^{s}(\mathbb R) \subset H^{0}(\mathbb R)=L^{2}(\mathbb R), (0<s<1).$
My Question: Let $\phi_{0}\in H^{s}(\mathbb R), (0<s<1).$
Is it true that for the given initial data $\phi_{0}\in H^{s}(\mathbb R),$ the cubic NLS has a unique global solution in $C(\mathbb R, H^{s}(\mathbb R))$ ? If yes, how to prove it ? (or proper reference ?)
[See also this question on MO]
Thanks,
Edit: Vague Ideas:
Fact: Suppose we have a local existence for cubic NLS in $H^{s}(0<s<1):$ that is, for the given initial time $u(t_{0}, x)=u_{0}(x)\in H^{s}(0<s<1), $ the cubic NLS has a unique local solution in $C(\mathbb [t_{0}, t_{0}+T_{1}^{*}]: H^{s}(\mathbb R)).$ (I guess this true)
Now one may apply the above result for the initial time $t_{0}+T_{1}^{\ast}$: more specifically, for the given initial data, $u(t_{0}+T^{\ast})=u_{0}(x)\in H^{s}(\mathbb R) \ (0<s<1);$ the NLS has unique local solution in $C([t_{0}+T^{\ast},t_{0}+T^{\ast}+T^{\ast}_{1}], H^{s}(\mathbb R)).$
If we start with $I_{1}= [t_{0}, t_{0}+T_{1}^{*}]$, $I_{2}=[t_{0}+T_{1}^{\ast}, t_{0}+T_{1}^{\ast}+T_{2}^{\ast}]$, and so on... .; and it will cover $[t_{0}, \infty)$). And on each individual intervals (as explained here ) $I_{j}$ we have a solution of cubic NLS in $C(I_{j}: H^{s}(\mathbb R)).$
My difficulties: But does it mean that the cubic NLS has a global solution in $C([t_{0}, \infty): H^{s}(\mathbb R))$ ? (Or I have to do bit more work to conclude this fact ?)