$u$ satisfies Schrodinger equation implies $\mathcal{F}^{-1} \left(\chi_{2}(\xi) \hat{u} \right) $ also?

Question

Consider Schr\"odinger equation (SE):

$i \frac{\partial }{\partial t}u (x,t )+ \Delta u(x,t) =0, (x, t)\in \mathbb R^{N}\times \mathbb R.$ $u(0,x)=\phi(x).$

Then, formally, the solution of (SE) can be written as

$u(x,t)= \mathcal{F}^{-1}\left( e^{-4\pi^2 |\xi|^2 it} \hat{\phi} \right) (x).$

See Section 1.1 and equation (1.2) for details.

Define $\chi_{2}(D)u:= \mathcal{F}^{-1} \left(\chi_{2}(\xi) \hat{u} \right) $ where $\chi_{2}(\xi) = \chi_{[2, 4]}(\xi)$ (Characteristic function) (and $\mathcal{F}^{-1}$ denotes inverse Fourier transform and $\hat{\cdot}$ denotes the Fourier transform.)

My Question: If $u$ satisfies (SE), then can we say $\chi_2(D)u$ also satisfies (SE)?

Answer 1

Yes, it will still satisfy the Schrödinger-equation.

Though now it will solve it for the initial data $\mathcal{F}^{-1} \chi_2(\xi)\phi(\xi)$ instead of $\phi$ as it did before.