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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.


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Weierstrass' $M$ test. –  Pedro Tamaroff Oct 20 '13 at 19:39
    
Hello and thanks for your quick reply! From what I gather, the M-test just tells you if your series converges uniformly. How do I use it to show that the series uniformly converges to original function with the boundaries included? Apologies if my question is vague. –  Christopher Nguyen Oct 20 '13 at 19:58
1  
You already know what your series converges to. With the $M$ test, you know convergence is uniform. –  Pedro Tamaroff Oct 20 '13 at 19:59
    
to close the argument there are various ways, but most of them will express the truncated Fourier series by a convolution of the function with a sinc function . then perhaps show $L^2$ convergence –  Evan Oct 20 '13 at 20:36
    
oh wait it is automatic by the fact that the complex exponentials form an orthonormal basis –  Evan Oct 20 '13 at 20:39

2 Answers 2

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.

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If you have a fourier series with coeffiencts that go like $\frac{1}{k^n}$, then in general the series will conform uniformly if $n>1$. This follows from the Weierstrass M test and recalling that $\sum_k \frac{1}{k^n}$ converges for $n>1$.

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