1
$\begingroup$

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 $f\in E$?

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $C=C(\|\hat{f}\|_2)$.

$\endgroup$
2
  • $\begingroup$ Thanks for your answer. But I still don't understand the existence of such a polynomial $P_\varepsilon$. Could you give more details on this? Moreover, the size of $\|\widehat{f}\|_{L^2}$ is not important. Since set $g=f/\|f\|_{L^2}$, the desired conclusion is equivalent to $\int_{[-2,2]}|g|^2dx \leq C\int_{[-1,1]}|g|^2dx$ for all $g\in E$ with $\|g\|_{L^2}=1$. $\endgroup$
    – Wang Ming
    Apr 23, 2017 at 14:58
  • $\begingroup$ Any continuous function on $[-2,2]$ can be uniformly approximated by some polynomial. $\endgroup$ Apr 23, 2017 at 15:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.