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What exactly is Fourier Transformation? For functions on the Schwartz Space $S(\Bbb R^n)$, we may define,

$$ \hat{u}(\xi) := \int e^{-ix\xi} u(x) \, dx $$ This formula seems to come out of no where for me.


It satisfies so many amazing properties that seems nonintuitive, to name a few:

(1) $$ ||f||_2 =||\hat{f}||_2$$

(2) $$ \hat{\hat{u}} = u^{-}$$


Is there a general/intuitive interpretation for how (i) One comes up with the formula, and (ii) why we should expect results as (1) and (2)?

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    $\begingroup$ One way to see it is to think of a function on the line by cutting it off on a large interval and then periodically extending it. The operation of applying the FT and then the inverse FT is what you get when you write out the Fourier series of this periodically extended function and then make the cutoff interval arbitrarily large. The inner integral of the Fourier coefficients was already there (from the $L^2$ inner product) while the outer sum over Fourier modes becomes an integral as the cutoff interval becomes large. $\endgroup$ – Ian Dec 22 '18 at 4:58
  • $\begingroup$ Do you know the Fourier series ? And the discrete Fourier transform ? $\endgroup$ – reuns Dec 22 '18 at 5:05
  • $\begingroup$ It is basically a change of bases. As mentioned previously, for a better understanding, go to the origins and look at Fourier series for periodical functions $\endgroup$ – Damien Dec 22 '18 at 7:40
  • $\begingroup$ Ok, thanks, I do not know any and will look into what all comments and posts have recommended. $\endgroup$ – Bryan Shih Dec 22 '18 at 9:33
  • $\begingroup$ Also, what I said above only actually makes sense for a function that is at least locally $L^2$. Fourier transform theory also makes sense for a function that is $L^1$ and it is possible for a function to have a nasty enough singularity that it is even globally $L^1$ without being locally $L^2$. My proposed intuition falls apart in this case. $\endgroup$ – Ian Dec 22 '18 at 17:09
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The basic idea was to decompose a function into sine and cosine waves with varying amplitudes. Complex numbers make the math quite a bit neater, but the basic idea is still the same even though technically we are now decomposing the function into "complex exponentials". The equation then comes from the fact that you can represent sinusoids with complex exponentials.

As for the properties, doing the math reveals the properties. But at least for the second property, Fourier Transform and its inverse are not the same so I wonder if that even holds.

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Fourier originally came up with the Fourier transform, but did so using $\sin$ and $\cos$ functions. The exponential series and transforms weren't used at that time. Fourier used a limit of the Fourier series as the period grew longer and longer in order to come up with integral expansions. He argued that the sums turned to integrals as the period grew without bound. His arguments were not rigorous, but they resulted in correct expressions. The exponential series on $[-L,L]$ is $$ f(x) = \sum_{n=-\infty}^{\infty}\frac{1}{2\pi L}\int_{-L\pi}^{L\pi}f(y)e^{-iny/L}dy\cdot e^{inx/L} $$

You can imagine--though probably not correctly justify--that the limit as $L\rightarrow\infty$ looks like a Riemann sum for $$ f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}f(y)e^{-isy}dy\cdot e^{isx}ds $$ I've never seen a correct argument where this is made rigorous; but it is good motivation for the continuous case, and the final result is correct.

The discrete expansion of $f$ is a series $$ \sum_{n=-\infty}^{\infty}\langle f,e^{-iny/L}\rangle e^{inx/L} $$ where $\langle f,e^{-iny/L}\rangle = \frac{1}{2\pi L}\int_{-L\pi}^{L\pi}f(y)e^{-iny/L}dy$ is the amplitude of $e^{inx/L}$. The limit as $L\rightarrow\infty$ gives way to an integral, which may be interpreted as a "continuous" sum $\int$ given by $$ f(x) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \langle f,e^{isy}\rangle e^{isx}ds, $$

where $ \langle f,e^{isy}\rangle = \int_{-\infty}^{\infty}f(y)e^{-isy}ds. $

The discrete Parseval relation $$ \|f\|^2 = \frac{1}{2\pi}\sum_{n=-\infty}^{\infty} |\langle f,e^{iny/L}\rangle|^2 $$ becomes an integral relation $$ \|f\|^2 = \frac{1}{2\pi}\int_{-\infty}^{\infty} |\langle f,e^{ins}\rangle|^2 ds $$ I think it's good to see this analogy between the discrete and continuous. The intuition is great, but it is not a substitute for a rigorous development which, unfortunately, cannot follow this line of argument in any simple way.

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  • $\begingroup$ Are you sure the $L^2$ theory can't be sorted out along these lines? The $L^1$ theory obviously cannot, but I don't see the problem with the $L^2$ theory. $\endgroup$ – Ian Dec 23 '18 at 22:23

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