Non-compactness of support of linear KdV equation solution The last question in Linares and Ponce's 'Introduction to Nonlinear Dispersive Equations's first chapter asks the reader to prove that, if the following IVP is given:
$$\begin{cases} 
\partial_t u + \partial_x^3 u = 0 & \cr
u(x,0) = u_0(x)
\end{cases}$$
Then, if $u_0 \in C^{\infty}_c(\mathbb{R})$, we cannot have $u(\cdot,t) \in C^{\infty}_c(\mathbb{R})$ for any $t\in \mathbb{R}^{*}$.
My attempt
I was trying to take Fourier transforms on both sides, so that we would have
$$e^{8\pi^3 i t \xi^3} \widehat{u_0}(\xi) = \widehat{u}(\xi,t).$$
Along with that, I wanted to use Paley-Wiener's theorem, which states that a function $f \in C^{\infty}_c(\mathbb{R}) \iff \widehat{f}$ has an analytic continuation satisfying 
$$ |\widehat{f} (x+iy)| \le c_k \frac{e^{2\pi M |y|}}{(1+|x+iy|)^k},$$
For any $k\in \mathbb{N}$, where $c_k$ is a constant, and $M$ is the least radius such that the closure of $B(0,M)$ contains the support of $f$. 
My idea was to pick specific values of $x$ and $y$ - depending on the initial data - and then, plugging in the expression for $\widehat{u} (\cdot,t)$ on the formula above arriving at a contradiction. Still, this is not completely clear for me, and I would like to know if anyone could help me complete this answer.
P.S.: If it is of any help, professor Linares has sent me an e-mail 1 hour ago stating that this should be the right direction. 
 A: Let me sketch a solution by contradiction: We may assume $t>0$, as for the other case we simply invert the roles of the functions involved. 
STEP 1: We are going to use the inversion formula
$$ u(x,t) = \int_{\mathbb{R}} e^{8\pi^3 it \xi^3} \widehat{u_0}(\xi) e^{2\pi i x\xi} d\xi.$$
Instead of integrating on $\mathbb{R}$,  we may, by a contour integration and some little techinicalities justified by the Paley-Wiener theorem and the positivity of $t$, change it to
$$ u(x,t) = \int_{\mathbb{R}} e^{8\pi^3 it (\xi+i\eta)^3} \widehat{u_0}(\xi+i\eta)e^{2\pi i x(\xi+i\eta)} d\xi,$$
Where $\eta >0$ is fixed. 
STEP 2: We simply change $x$ by $x+iy$ on this last integral. It is easy to see that for little $|y|$ it is convergent, and then we may define, for this set of $y$, the function 
$$u(z;t) = u(x+iy,t).$$
By Morera's Theorem, this is analytic inside this region. 
STEP 3: We conclude, by noting that an analytic function whose restriction to the real line is of compact support is, by analytic continuation, identically zero. 
Of course, we have purposely missed out some technical details, which are easily filled by interested readers :) 
