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Consider a function $f$ defined on the real line. Consider the restriction of the function to the interval $[-L,L]$ and periodically extend the function using a Fourier series $$f_L(x) = \sum_{n=-\infty}^\infty \hat{f}(k_n)e^{ik_nx}$$ where $k_n=\pi n/L$ and where $\hat{f}(k_n)$ denotes the Fourier coefficient. For sufficiently large $L$, it seems plausible that we may approximate the Fourier series with an integral. In particular, let us extend $\hat{f}(k_n)$ to non-integral values through $$\hat{f}(k)=\frac{1}{2L}\int_{-L}^L f_L(x)\,e^{-ikx}\ dx $$ and write the approximation as $$\tilde{f}_L(x)=\frac{2L}{(2\pi)}\int_{-\infty}^\infty\hat{f}(k)\,e^{ikx}\ dk$$ To provide a bit of context, I first encountered such integral approximations when studying cosmology. It seems common in physics to take such approximations where we formally make the replacement $$\sum_{n=-\infty}^\infty \longrightarrow \frac{2L}{(2\pi)}\int_{-\infty}^\infty$$ in the limit of large $L$, which is called the continuum limit.

My question is under what conditions is such an approximation valid? What kind of smoothness and regularity conditions need to be placed on $f$? How well does $\tilde{f}$ approximate $f$ on $[-L,L]$ (for sufficiently large $L$)? Are there any error bounds?

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  • $\begingroup$ How are the Fourier coefficients $\hat f(k_n)$ initially calculated (in the first displayed formula)? In the second displayed formula, wouldn't $f_L$ be equal to the original function $f$? $\endgroup$ – timur Sep 30 '13 at 23:57
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    $\begingroup$ @timur The only difference between $\hat{f}(k_n)$ and $\hat{f}(k)$ is that I allow non-integral values for the latter, i.e. $k_n$ is restricted to be $\frac{\pi n}{L}$ for integer values of $n$ but there is no such restriction on $k$. They are given by the same integral formula (the second formula). In the second formula, $f_L$ would indeed be the same as $L$ since the integral is restricted to $[-L,L]$ and $f_L$ is just the periodic extension of $f$ (and hence equal on $[-L,L]$. It was just a choice on my part to write $f_L$. $\endgroup$ – EuYu Oct 1 '13 at 0:46
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I claim that $\tilde f_L$ is in fact equal to $f$ on $(-L,L)$, and equal to $0$ outside $[-L,L]$.

To see this, we write $$ \hat{f}(k) =\frac{1}{2L}\int_{-L}^L f_L(x)\,e^{-ikx}\ dx =\frac{1}{2L}\int_{-\infty}^\infty \chi_L(x)f(x)\,e^{-ikx}\ dx, $$ where $$ \chi_L(x) = \begin{cases} 1,&|x|<L,\\ 0,&\textrm{otherwise}, \end{cases} $$ is the characteristic function of the interval $(-L,L)$. We substitute it into the formula for $\tilde f_L$, to get $$ \tilde f_L(x) =\frac{1}{2\pi}\int_{-\infty}^\infty \int_{-\infty}^\infty \chi_L(y)f(y)\,e^{-iky}e^{ikx}\ dy \ dx. $$ This is just the Fourier inversion formula for the function $\chi_Lf$, which means that $\tilde f_L=\chi_Lf$. Possible assumptions that make this true are very mild. For instance, it is true if $f$ is locally integrable, i.e., $f\in L^1_{\mathrm{loc}}(\mathbb{R})$. Under this assumption, of course, the equality $\tilde f_L=\chi_Lf$ must be treated in the $L^1$ sense.

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    $\begingroup$ Ooh, that is a lot nicer than I expected. Thank you. $\endgroup$ – EuYu Oct 1 '13 at 2:10
  • $\begingroup$ Interesting this will actually be useful the problem that i'm encountering withen my Fourier Analysis book $\endgroup$ – Zophikel Mar 20 '17 at 2:31

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