Question:
$f(x) =\begin{cases} \frac{\sin{x}}{|x|} \ \ \ \text{for} \ x \neq 0 \\ 0 \ \ \ \ \ \ \ \ \text{for} \ x = 0 \end{cases}$ where $f(x) $ is defined on $[-\pi,\pi]$. Does the Fourier series converge to the function $f(x)$ on the given interval? If so give proof.
Relevant Theorems
Theorem 1:
Let $f \in L^1[-\pi,\pi] $, and let $x \in [-\pi,\pi] \ $ such that $f(x)$ is differentiable everywhere then $S_N (x) \rightarrow f(x) \ \ \text{as} \ \ N \rightarrow \infty $
Theorem 2:
If $\int^{\pi}_{0} \frac{|f(x + \tau) - f(x^+)+f(x-\tau)-f(x^-)|}{\tau}d\tau < \infty$
Then $S_N (f)(x) \rightarrow \frac{f(x^+)+f(x^-)}{2} \ \ \text{as} \ \ N \rightarrow \infty$
Attempted solution:
I had several questions on solving this question. Clearly $f$ is differentiable everywhere and $f \in L^1[-\pi,\pi] \ $ that is $f$ is integrable. From theorem this implies that $S_N (x) \rightarrow f(x) \ \ \text{as} \ \ N \rightarrow \infty $ for all $x$ on the interval $x \neq 0 $. Now to address the discontinuity I understand that the right hand limit is $1$ and the left hand limit is $-1$ implying by theorem 2 that if the condition is satisfied the series should converge to $\frac{-1+1}{2} = 0$ which is the value of $f(0)$. Therefore the series converges everywhere on the interval. The trouble for me is understanding how I know that the condition for theorem 2 is satisfied, how do I even go about proving that the condition is satisfied in a general sense?
Let $x=0$. Then $f(x^+)=\lim_{x\to0^+}\dfrac{\sin x}{|x|}=1$. We have $$ |f(t)-1|=\frac{|\sin t-t|}{t}\le C\,t^2,\quad t>0, $$ and $$ \int_0^\pi\frac{|f(t)-1|}{t}\,dt<\infty. $$ Similarly, $f(x^+)=-1$ and $$ \int_0^\pi\frac{|f(-t)+1|}{t}\,dt<\infty. $$