# The convergemce of the fourier series

I was trying to prove the convergence of the fourier series because I checked wikipedia and found that it is described using knowledge in branches of mathematics that I have/had no clue of,. Here is my attempt, kindly help me out further and check if I am going awry somewhere.

I know that $f(x)=\sum_{k=0}^{\infty}\left\{A_k\cos\left(\frac{2\pi k x}{x_0}\right)+B_ksin\left(\frac{2\pi kx}{x_0}\right)\right\}$ so

$$\left|f(x)\right|\leq \sum_{k=0}^{\infty}\left|A_k\right|\left|\cos\left(\frac{2\pi kx}{x_0}\right)\right|+\left|B_k\right|\left|\sin\left(\frac{2\pi kx}{x_0}\right)\right|\leq\sum_{k=0}^{\infty}\left|A_k\right|+\left|B_k\right|$$. Hence, if could prove that $\sum_{k=0}^{\infty}\left|A_k\right|$ and $\sum_{k=0}^{\infty}\left|B_k\right|$ individually converge, then I could prove that my function converges. So, I attempt to write $$\sum_{k=0}^{\infty}\left|B_k\right|=\sum_{k=0}^{\infty}\left|\int_{x_0}f(\tau)\sin\left(\frac{2\pi k\tau}{x_0}\right)\mathrm{d}\tau\right|\leq\sum_{k=0}^{\infty}\int_{x_0}\left|f(\tau)\right|\left|\sin\left(\frac{2\pi k\tau}{x_0}\right)\right|\mathrm{d}\tau$$. Now, how do I proceed from here, every route that I've tried from here has led me to disappointment. Hints would also be welcome. P.S. I would be really happy if in the hints section, the level of abstraction of the mathematics is restricted to one dimensional simple calculus the way I've proceeded so far, since I'm not knowledgeable enough to be called a mathematician.

• The series of coefficients of a Fourier Series need not converge. – Mark Viola Nov 29 '16 at 21:04
• Then how else do I prove that my Fourier series converges? – ubuntu_noob Nov 29 '16 at 21:06
• You need to specify some condition on $f$ that is strong enough for the series to converge. Indeed there are functions $f$, for which the fourier series will not converge. – iolo Nov 29 '16 at 21:22
• Define convergence. There are different types. If you are speaking of pointwise convergence, then you need to show that $\lim_{N\to \infty}|f(x) -\sum_{n=1}^N c_n e^{inx}|=0$. Are you familiar with the Dirichlet kernel? – Mark Viola Nov 29 '16 at 21:27
• No, I'm not familiar with it, plus I know only one kind of convergence where the sum of partial sum as n tends to infinity tends to a non infinite number – ubuntu_noob Nov 29 '16 at 21:49

The oldest method of proving pointwise convergence for typical situations actually goes back to 1800's. Using the expression for the coefficients, truncating the series at $n=N$, and using trig identities, \begin{align} S_N(f)&=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)dt \\ &+\sum_{n=1}^{N}\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos(nt)dt\cos(nx)\\ &+\sum_{n=1}^{N}\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\sin(nt)dt\sin(nx).\\ &=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)\frac{\sin((N+1/2)(x-t))}{\sin(1/2(x-t))}dt. \end{align} Assuming $f$ is extended periodically with period $2\pi$ to the entire real line, $$S_N(f) =\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x+t)\frac{\sin((N+1/2)t)}{\sin((1/2)t)}dt.$$ Then it can be shown that, regardless of how small you take $\delta$, the following limit equality holds, in the sense that one limit exists iff the other does and, in that case, the two limits are equal: $$\lim_{N\rightarrow\infty}S_N(f)=\lim_{N\rightarrow\infty}\frac{1}{2\pi}\int_{-\delta}^{\delta}f(x+t)\frac{\sin((N+1/2)t)}{\sin((1/2)t)}dt.$$ If you replace $f(t)$ by $f(t)-L$ where $L$ is a candidate limit for the Fourier series, you obtain $S_N(f-L)=S_N(f)-L$ and $$\lim_{N\rightarrow\infty}(S_N(f)-L)=\lim_{N\rightarrow\infty}\frac{1}{2\pi}\int_{-\delta}^{\delta}(f(x+t)-L)\frac{\sin((N+1/2)t)}{\sin((1/2)t)}dt.$$ So, if you have nice enough convergence of $(f(x+t)-L)$ as $t\rightarrow 0$, then the Fourier series at $x$ converges to $L$. For example, convergence will occur if $f$ is differentiable at $x$, or if it has left- and right-hand derivatives at $x$. Or, if, for a fixed $x$, $|f(x+t)-L| \le C|t|^{\alpha}$ holds for some constant $C$, some constant $0 < \alpha < 1$, and for all $t$ near $0$. It doesn't take much to give convergence, but it takes more than just continuity.