I need to show that the Fourier Series of |x| in the interval $(-\pi, \pi)$ converges uniformly to |x| in $[-\pi, \pi]$.
I know that
|x| = $\frac{\pi}{2}$ + $\frac{2}{\pi}$$\sum\limits_{k=1}^{\infty} \frac{(-1)^n-1}{k^2}cos(kx)$
I know that to show that this Fourier series converges uniformly, I have show that
$max$ | |x| – $\frac{\pi}{2}$ + $\frac{2}{\pi}$$\sum\limits_{k=1}^{\infty} \frac{(-1)^k-1}{k^2}cos(kx)$ | $\rightarrow$ 0
I've tried separating and looking at just the even and the odd terms of the Fourier Series. Any ideas? Thanks for the help.
Best Answer
If $f$ is an $L^2$ function on $(-\pi,\pi)$ and the Fourier series of $f$ converges uniformly to some function $g$, then $f=g$ almost everywhere. (As others said)
Indeed, we know (from the fact that the exponentials form a basis) that the Fourier series converges to $f$ in $L^2$. On the other hand, it also converges to $g$ in $L^2$, since uniform convergence implies $L^2$ convergence. Thus $f$ and $g$ are the same element of $L^2$. As functions, they may be different on a null set. The precise statement is: $f$ has a continuous representative, and that representative is $g$.
In this problem, $f$ is given as a continuous function, and since the convergence is uniform by Weierstrass, the conclusion follows.