Uniqueness for tempered distributional Cauchy problems 
Question. Assume that $U\in C^1(\,[0, \infty)\to \mathcal{S}'(\mathbb{R}^n)\,)$ is a solution to the following tempered distributional Cauchy problem 
  $$\tag{CP}\begin{cases}
\frac{ d U}{dt} = f \cdot U(t), & t>0 \\
U(0) = 0
\end{cases}
$$ 
  where $f\in C^\infty(\mathbb{R}^n)$ is a smooth function not depending on $t$. Is it true that $U(t)=0$ at all times $t>0$?

The background for this question comes from a passage in the book on PDEs by Michael E. Taylor.  The author claims that the problem 
\begin{equation}
\begin{cases}
\frac{\partial u}{\partial t} -\Delta u =0, & t>0,\ x\in \mathbb{R}^n \\
u(0, x)=u_0
\end{cases}
\end{equation}
has a unique tempered distributional solution if $u_0$ is a tempered distribution. (This is not entirely trivial, as uniqueness fails if one imposes no growth conditions on $u$). The proof via Fourier transform is essentially reduced to the statement that the Cauchy problem 
$$
\begin{cases}
\frac{\partial \hat{u}}{\partial t}(t, \xi) + \lvert \xi\rvert^2 \hat{u}(t, \xi)=0, & t > 0 \\
\hat{u}(0, \xi)=0\\
\hat{u}\in C^1(\, [0, \infty)_t \to \mathcal{S}'(\mathbb{R}^n_\xi)\, )
\end{cases}
$$
has the unique solution $\hat{u}(t, \xi)\equiv 0$. This does not seem obvious to me and that's why I am posing this question. 
 A: I think this is a great question.  This is the direction I would try.  So I presume that you are able to prove this if $f$ is compactly supported (since multiplication by $f$ would then be a continuous operation from $\mathcal S \to \mathcal S$).  Secondly, I presume that there is a theorem that says if $\mu, \nu \in \mathcal S'$, and $\mu(\phi) = \nu(\phi)$ for all $\phi \in \mathcal S$ which are compactly supported, then $\mu = \nu$.  I think that whatever proof is used to show this would have to be an important part of my argument.
So pick a bump function $\rho \in \mathcal S$ such that $\rho(x) = 1$ if $|x| \le 1$, and $\rho(x) = 0$ if $|x| \ge 2$, and $\rho(x) \in [0,1]$ for all $x \in \mathbb R^n$.
Then for each positive integer $m$, let $U_m$ be the solution to the equation
$$ \frac d{dt} U_m = \rho(\cdot/m) f \cdot U_m ,\quad U_m(0) = 0$$
and show that $U_m \equiv 0$.  Then show that $U(t,\cdot) = U_m(t,\cdot)$ on $B(0,m)$.  And then conclude $U = \lim_{m\to\infty} U_m = 0$.
A: Here's a partial solution suggested to me by my thesis director. It is partial because it needs some additional assumption on $f$. Either $f$ must be real valued and each of its derivatives must be bounded above, or $f$ must be complex valued and each of its derivatives must be bounded. $^{[1]}$ In particular, this method of proof works when $f$ is the symbol of the Laplacian and also when $f$ has compact support. 
This partial result could then be plugged into Stephen Montgomery-Smith's great answer to eliminate the additional assumption on $f$. I haven't still checked the details, though.
Let us consider our Cauchy problem together with a suitable dual problem: 
$$\tag{CP}
\begin{cases}
\partial_t U - f\cdot U = 0, & U \in C^1(\, [0, +\infty) \to \mathcal{S}'\, ) \\
U(0) = 0 
\end{cases}
$$
and
$$
\tag{CP'}
\begin{cases}
\partial_t \phi + f\cdot \phi = 0, &  t\in(0, T) \\
\phi(T)=\psi \in \mathcal{S} 
\end{cases}
$$
Here $T > 0$ and $\psi\in \mathcal{S}$ are arbitrary. Our task is to prove that 
$$
\langle U(T), \psi\rangle=0.
$$
Now the dual problem (CP') has the unique classical solution 
$$
\phi(t, \xi) = e^{( T - t )f(\xi)} \psi(\xi). 
$$
Our assumptions on $f$ imply that $\phi \in C^1(\,[0, +\infty)\to \mathcal{S}\, )$. We may therefore use $\phi(t)$ as a test function for $U(t)$. The resulting pairing is constant, because 
$$
\begin{split}
\frac{d}{dt} \langle U(t), \phi(t) \rangle & = \langle f U (t), \phi(t) \rangle \\ 
& = \langle U(t), f\phi(t)\rangle \\
& = -\frac{d}{dt} \langle U(t), \phi(t) \rangle.
\end{split}
$$
We infer that 
$$
\langle U(T), \psi\rangle = \langle U(T), \phi(T)\rangle=\langle U(0), \phi(0)\rangle = 0.
$$
Since $T$ and $\psi$ are arbitrary, we conclude that $U\equiv 0$. 

$^{[1]}$ Actually, we need the slightly less stringent requirement that $e^{\lambda f}$ is a slowly increasing function for any $\lambda>0$. For more information one may look into this French Wikipedia page. 
