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Is there any way to prove the Fourier Theorem ?

Any single valued periodic function can be represented by a summation of simple harmonic terms having frequencies which are the integral multiples of the frequency of the periodic function.

$f(t) = a_0 + \displaystyle \sum_{n=1}^{\infty} (a_n \cos n\omega t + b_n \sin n \omega t)$

  • Of course! Every theorem needs a proof, otherwise it is no theorem. A quick search gave me Carleson's theorem. If this is what you are looking for then you can certainly google for a proof. Otherwise you can find futher information on Wikipedia here and especially here. – M. Winter Dec 09 '17 at 14:34
  • @M.Winter Quite interesting that proof is from 1960s and Fourier lived and worked 150-200 years earlier. – mathreadler Dec 09 '17 at 14:41
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    There isn't one "Fourier theorem". There are lot of ways a Fourier series can converge (pointwise, uniform, $L^2$ etc.) and a lot of different conditions that may be put on $f$ to ensure each form convergence. The resulting theorems span a very wide range of difficulty.... – Angina Seng Dec 09 '17 at 15:07
  • In applied maths, for most applications, this convergence theorem is enough : for $f \in L^2([-\pi,\pi])$, the Fourier series converges to $f$ in the $L^2$ sense, that is $\lim_{N \to \infty}\int_{-\pi}^\pi |f(t)-f_N(t)|^2 dt = 0$, where $f_N(t) = a_0 + \sum_{n=1}^{N} a_n \cos n\omega t + b_n \sin n \omega t$. Note it doesn't mean $\lim_{N \to \infty} f_N(t) = f(t)$ for every $t$ ! – reuns Dec 09 '17 at 17:12

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Depends on what you mean by "represented by". If you mean the series actually converges to $f(t)$ for every $t$ then this can't be proved because it's false, even for continuous $f$.

But $f$ can nonetheless be recovered from its Fourier series. For example if $f$ is continuous and we define $$A_r(t)=a_0+\sum_{n=1}^{\infty}r^n (a_n \cos n\omega t + b_n \sin n \omega t)$$for $0<r<1$ then $A_r\to f$ uniformly as $r\to1$; one certainly might regard this as a sense in which $f$ is "represented by" its Fourier series. (The proof proceeds by showing that $A_r$ is the convolution $f*P_r$, where $P_r$ is the Poisson kernel.)

David C. Ullrich
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  • This theorem is from my wave book. I wrote it as written in the book. In physics, that theorem is used to analyze the graphs of different periodic functions. –  Dec 09 '17 at 16:41
  • I'm actually a physics undergraduate. Asked the question out of curiosity. –  Dec 09 '17 at 16:48
  • @SkyWalker I'm certain you wrote it exactly as in that book - this non-theorem appears in a lot of places. The authors there didn't want to worry about technical details - the theorem is true under stronger conditions on $f$, for example $f$ continuously differentiable is more than enough. – David C. Ullrich Dec 09 '17 at 17:08