1

A theorem for differentiability of a function's Fourier series states that:

If f is a piecewise smooth function and if f is also continuous, then the Fourier series of f can be differentiated term by term provided that f(-L) = f(L).

I went on searching for a proof for this theorem but I couldn't find one. Could someone provide a proof as to why this is true?

Bernard
  • 175,478
muhzi
  • 27
  • Using integration by parts you get $(f')^\hat{}(n)=\int_{-\pi}^{\pi}f'e^{-int},dt=in\hat{f}(n)$, hence if you write $f\sim \sum \hat{f}(n)e^{int}$ and differentiate the righthand termwise, you get $\sum in\hat{f}(n)e^{int}$, the Fourier series of $f'$. I don't know if this settle your question, or you are interested in issues of convergence. – user90189 Dec 09 '17 at 03:13
  • @user90189 yes, I wish to know how the condition f(-L) = f(L) satisfies the condition for the convergence of the derivative. – muhzi Dec 09 '17 at 16:53
  • to see what's the condition $f(-L)=f(L)$ about, think of the interval $[-L,L]$ as a circle, that is, identify both extremes of the interval. The condition $f(-L)=f(L)$ means that $f$ is continuous on the circle. For example, take $f(t)=t$ in $[-\pi,\pi]$, then over the circle you see that $f$ jumps from $\pi$ to $-\pi$, hence $f$ is discontinuous. It may help you. – user90189 Dec 09 '17 at 18:27
  • @user90189 I couldn't grasp what you mean by continuous on the circle. Also then, how can I prove the convergence by proving its continuity on a circle interval? – muhzi Dec 09 '17 at 19:36
  • Dear @airomyst, as regards convergence, you can use in this case Dini's or Jordan's test. About the function $f(t)=t$ in $[-\pi,\pi]$, notice that $f\sim \sum_{n\neq 0} \frac{(-1)^n}{in}e^{int}$, and if you naively differentiate you get $1=f'\sim \sum_{n\neq 0} (-1)^ne^{int}$, which is wrong; the problem is that we are not taking into account the jump of $f$ at the extremes of the interval. Think about it :) – user90189 Dec 12 '17 at 03:16

0 Answers0