Is there a compactly supported function that its Fourier transfrom vanishes at given n real points? My question is as follows: Given ${{\lambda }_{1}},\,{{\lambda }_{2}},...,{{\lambda }_{n}}\in \mathbb{R}$ where $\underset{1\le j\le n-1}{\mathop{\min }}\,\left| {{\lambda }_{j+1}}-{{\lambda }_{j}} \right|\ge d>0$. Is there a compactly supported function $f:\mathbb{R}\to \mathbb{R}$ such that its Fourier transform $\hat{f}$, defined by $\hat{f}\left( \lambda  \right)=\int_{\mathbb{R}}{f\left( x \right){{e}^{i\lambda x}}dx}$, satisfies $\hat{f}\left( {{\lambda }_{j}} \right)=0$, $j=1,...,n$ ? Thank you for helping.
 A: The space $L^2(a,b)$ is a Hilbert space of infinite dimension. Therefore there is an
element $f\neq 0$ of this space which is orthogonal to $e^{i\lambda_j x}$. Take this $f$.
You can also find such $f$ is any finite dimensional subspace whose dimension is $>n$.
Just choose a basis and solve a system of linear equations.
A: Consider the polynomial
$
P(\xi)=\prod_{1\le j\le n}(\xi-\lambda_j).
$
The inverse Fourier transform of $(\xi-\lambda_j)$ is 
$$
\int(\xi-\lambda_j) e^{2iπ x\xi} d\xi=(D_x-\lambda_j)(\delta_0)=\frac{\delta'_0}{2iπ}-\lambda_j\delta_0=T_j, \quad\text{support } T_j=\{0\},
$$
Let $F$ be the inverse Fourier transform of $P$:
we have
$$
F=T_1\ast\dots\ast T_n,\quad\text{support } F=\{0\},\quad \hat F= P.
$$
If $\rho$ is smooth compactly supported, the function $F\ast \rho$ is smooth compactly supported
and 
$$
\widehat{F\ast \rho}= P\hat \rho
$$
and thus vanishes at the $\lambda_j$. There are generalizations using the Weierstrass factorization theorem to construct entire functions with prescribed zeroes.
A: The space of $W$ compactly supported continuous functions is infinite dimensional, and the map $f\mapsto (\widehat{f}(\lambda_1),\cdots,{\widehat f}(\lambda _n))$ which is a map from $W$ into ${\mathbb C}^n$ has a nonzero element in the kernel.
