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I want to calculate $f(y)$ such that $$\int_{-\infty}^\infty \frac{f(y)}{1+(x-y)^2}\mathrm dy =0 \quad \forall x$$

Can we prove that the solution to this problem is $f(y)=0?$

Thank you so much

jimjim
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Mamal
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1 Answers1

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$\newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $\widetilde{\vphantom{\large A}\cdots}$ ( tilde's ) are Fourier Transforms. \begin{align} 0 &= \int_{-\infty}^{\infty}\dd x\,\expo{-\ic kx} \int_{-\infty}^\infty {{\rm f}\pars{y} \over 1 + \pars{x - y}^{2}}\,\dd y = \int_{-\infty}^{\infty}\dd x\,{\rm f}\pars{y}\expo{-\ic ky} \int_{-\infty}^\infty {\expo{-\ic kx} \over \pars{x - \ic}\pars{x + \ic}}\,\dd x \\[3mm]&= \tilde{\rm f}\pars{k}\pi\expo{-\verts{k}} \quad\imp\quad \tilde{\rm f}\pars{k} = 0 \quad\imp\quad {\rm f}\pars{x} = 0 \end{align}

Felix Marin
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    Thank you so much. However this solution is valid for the $f(y)$s that their fourier transform converge. A function like $f(y)$ may exist that satisfy our equation and have diverging Fourier transform. – Mamal Oct 19 '13 at 07:23
  • @Mamal Yes. It's true. I didn't mention that but it was obvious that is valid whenever the Fourier transform exists. – Felix Marin Oct 19 '13 at 07:39
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    Can we prove that the $f(y)$s satisfying this equation have to convergne in terms of Fourier transform? – Mamal Oct 19 '13 at 08:52
  • @Mamal It's an interesting question. I don't know the answer. – Felix Marin Oct 19 '13 at 09:09