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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.

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    $\begingroup$ Here is how to do it for one point $\lambda$. Take any two (nice) compactly supported functions $A,B$. Then consider a linear combination with $y\hat A(\lambda)+z\hat B(\lambda)=0$. Does this help? Another useful fact is that $\widehat{(f\star g)}=\hat f\cdot\hat g$. $\endgroup$ Dec 6, 2014 at 12:34
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    $\begingroup$ user555, the method of @NAME_IN_CAPS allows one to construct a compactly supported $f_1$ whose FT vanishes at a prescribed $\lambda_1$. Do this for each $\lambda_j$. Then $f=f_1* f_2 *\dots*f_n$ has the properties you require. $\endgroup$
    – Yemon Choi
    Dec 6, 2014 at 13:55
  • $\begingroup$ Another suggestion: since the FT of the indicator function of the interval [-1,1] is something like $2\sin(\lambda)/\lambda$ you can easily tweak this to get a compactly supported function whose FT vanishes at a prescribed point. $\endgroup$
    – Yemon Choi
    Dec 6, 2014 at 14:01
  • $\begingroup$ Yes I understand your hint. These help me very much. Thank you very much. $\endgroup$
    – Baily
    Dec 6, 2014 at 14:26
  • $\begingroup$ In my "solution", one needs $A\neq cB$, if the linear combination is to be nonzero. Then again, perhaps a simpler solution to the originally stated exercise is just the zero function. I'm still trying to understand what the $\min$ condition is doing. $\endgroup$ Dec 6, 2014 at 14:41

3 Answers 3

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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.

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  • $\begingroup$ I agree with your answer, but it could not be generalized to an infinite number of zeroes, a natural extension of the question. The Weierstrass factorization theorem allows to construct entire functions with prescribed zeroes and in some case these functions could be of exponential type, thus with compactly supported Fourier transforms, e.g $\sin z$. $\endgroup$
    – Bazin
    Dec 8, 2014 at 11:51
  • $\begingroup$ You did not ask about infinitely many zeros. If you ask, I will answer. $\endgroup$ Dec 8, 2014 at 15:29
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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.

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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.

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