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I have the following expression: $$\sum_{k}\left(\int_{-\infty}^{\infty}e^{-ikx}\, f(k')dk'-\int e^{ikx}h(x)dx\right)\left(\int_{-\infty}^{\infty}e^{ikx}\,\bar{f(k')}dk'-\int_{\mathbb{R}}e^{-ikx}h(x)dx\right).$$

I want to choose the appropriate $f$ in terms of $h$ to make this above expression go to zero. I want to use Fourier series. Also, a little clarification on notation: in the integral, $k$ is fixed for now, since we are summing over all the $k$, as seen on the far left. I am trying to approach this by bring $e^{\pm ikx}$ out of the respective integrals where we are integrating with respect to the variable $k'$ and then proceeding. But that leads to some pathologies when trying to make this term go to zero.

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With $C:=\int_{-\infty}^{\infty}f(k')dk'$ you get $\int_{-\infty}^{\infty}e^{-ikx}\, f(k')dk'=e^{-ikx}C$ and $\int_{-\infty}^{\infty}e^{ikx}\,\bar{f(k')}dk'=e^{ikx}\,\bar{C}$, so that $f$ can be completely replaced by $C$ in your expression. Are you sure this is the correct expression? Another question: Is $h(x)$ a reel function, such that $h(x)=\bar{h}(x)$? –  Thomas Klimpel Dec 30 '11 at 22:13
    
@Thomas- yes h(x) is indeed a real function. But what does that do here? –  john Dec 30 '11 at 22:15
    
It allows to rewrite your expression as $\sum_{k}\left|\int_{-\infty}^{\infty}e^{-ikx}\, f(k')dk'-\int e^{ikx}h(x)dx\right|^2$. Now since each term of the sum is positive, we know that we must make each term identical to zero, if we want that the entire expression goes to zero. –  Thomas Klimpel Dec 30 '11 at 22:21
    
Thanks Thomas, but this is actually where I am having the trouble. The reason I posted this is because I am unable to show each term is equal to 0 for each $k$. –  john Dec 30 '11 at 22:40
    
Well, unless $h(x)$ is very special, not all terms will be equal to zero. You probably have to go back one step, and check that you have the correct expression. You might also tell us where this expression came from, and why you expect that there is an appropriate $f$ to make this expression go to zero. –  Thomas Klimpel Dec 30 '11 at 22:53

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