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For the sake of concreteness, we consider the linear Schrödinger equation $$ \partial_t u = i\Delta u, \ \ \ \ u(0, x) = u_0(x). $$ The solution is typically obtained by taking the Fourier transform of both sides, giving $\widehat{\partial_t u}(t, \xi) = -i|\xi|^2 \hat{u}(t, \xi)$.

The next step is where I have questions. Assuming that everything is nice enough (for instance, in Tao's book, he assumes $u_0$ is Schwartz), a dominated convergence argument gives $\widehat{\partial_t u}(t, \xi) = \partial_t \hat{u}(t, \xi)$, and then we get an ODE that solves to $$ \hat{u}(t, \xi) =e^{-i|\xi|^2}\hat{u}_0(\xi) \implies u(t, x) = e^{it\Delta}u_0(x). $$ This is then referred to as "the solution to the Schrödinger equation, with initial data $u_0$."

My question: How do we know that there are no other solutions, that may not satisfy the right decay/smoothness criteria to justify pulling the Fourier transform into the time derivative of $u$? I agree that there are no other solutions $u$ that are "nice enough" to justify this. But how do we rule out the existence of solutions $u$ such that $\partial_t \hat{u} \neq \widehat{\partial_t u}$? For instance, I don't understand how just assuming $u_0$ Schwartz is enough to guarantee this.

Any help is much appreciated.

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    $\begingroup$ In my book I only establish uniqueness for solutions that are $C^1_t {\mathcal S}_x$ by this argument. Without some regularity in time and decay in space uniqueness can fail; see Exercise 2.24 of the book. $\endgroup$
    – Terry Tao
    Nov 13, 2019 at 5:54

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In fact, in exactly the same book as you referred to, exercise 2.24 gives a counter example. Tao himself has commented above. Maybe he is too busy to give more details. Roughly speaking, the reason that this uniqueness may fail is that, the constructed counter-example grows too fast near infinity, so that it is out of the scope of the tempered distribution, and the Fourier transform etc fail to make nice sense.

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