Quantitative unique continuation for entire functions Let $\widehat{f}$ be the Fourier transform of $f$ defined by
$$
\widehat{f}(\xi) = \int_{\textbf{R}}e^{2\pi i x \xi}f(x)dx, \quad f\in L^2(\textbf{R}).
$$
Define a set
$$
E:= \{f\in L^2(\textbf{R}): supp \widehat{f} \subset [-1,1]\}.
$$
where supp denotes the support of a function.
Note that if $f\in E$ then $f$ can be extended to an entire function on the plane $\textbf{C}$. Thus if $f=0$ on the interval $[-1,1]$ then $f\equiv 0$ on the line $\textbf{R}$, and of course $f=0$ on $[-2,-2]$. My question arises naturally, namely does there exist a quantitative unique continuation of this kind. Precisely, I want to know:
Question: Is there a constant $C$ independent of $f$ such that
\begin{equation}\label{equ-UCP}
\int_{[-2,2]}|f|^2dx \leq C\int_{[-1,1]}|f|^2dx
\end{equation}
hold for all $f\in E$?
 A: This is not true. To construct an example, consider an auxilliary function $\phi$
whose Fourier transform is supported by $(-1,1)$, and such that $\phi(x)=O(|x|^{-N}),\; x\to\pm\infty$ for all $N>0$. To construct such a function, take an
inverse Fourier transform of an infinitely smooth function supported on $(-1,1)$.
Notice that $\phi$ is entire, of exponential type, by the Wiener-Paley theorem.
Now find a polynomial $P_\epsilon$ such that $|P_\epsilon(x)\phi(x)|<\epsilon$ for $|x|<1$,
but $\int_{-2}^2| P_\epsilon(x)\phi(x)|^2dx\geq 1 $. It is clear that such a polynomial 
exists for every $\epsilon$ (by Runge's theorem, or by Weierstrass theorem, or just write it explicitly).
So the function $P\phi$ violates your inequality. But it is in $L^2(R)$
and and also it entire of the same exponential type as $\phi$, so its Fourier transform is
still supported on $(-1,1)$ by Wiener-Paley.
To make your statement true, you have to restrict $\|\hat{f}\|_2$. Then your class of functions will be a normal family, and you will have your inequality
with $C=C(\|\hat{f}\|_2)$.
