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For given function, $$f(x) = \begin{cases} 1, & |x|\leq\frac{\pi}{2} \\ 0, & \pi\geq|x|>\frac{\pi}{2} \\ \end{cases}$$

The calculated Fourier series is:

$$\begin{align} a_0 &= \frac{1}{\pi}\int\limits_{-\pi/2}^{\pi/2}1 \ dx=1 \\ a_n &= \frac{1}{\pi}\int\limits_{-\pi/2}^{\pi/2}1\cdot \cos(nx) \ dx=\frac{2\sin(\frac{\pi n}{2})}{n\pi} \\ b_n &= \frac{1}{\pi}\int\limits_{-\pi/2}^{\pi/2}1\cdot \sin(nx) \ dx=0 \\ f(x) &= \frac{1}{2}+\frac{1}{\pi}\sum\limits_{n=1}^{\infty}\frac{2\cdot (-1)^n}{n}\cos(nx) \end{align}$$

And how to determine the type of convergence, to be $L^2$-convergence, pointwise convergence or uniform convergence?

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The $\approx$ above is incorrect; it would be correct for a partial sum. –  Ron Gordon Mar 20 '13 at 23:20
    
It is uniform convergence type, see the answer. –  Mhenni Benghorbal Mar 21 '13 at 1:54
    
@algebra: See my answer and do not worry about the downvote. –  Mhenni Benghorbal Mar 24 '13 at 3:01
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4 Answers

Your original $f(x)$ has jump discontinuities so the Fourier series converges in $L^2$ but not pointwise. And since it doesn't converge pointwise, it doesn't converge uniformly.

Pointwise is easy to see because at $x=\pi/2$ your $f(x)$ takes value 1. But at $x=\pi/2$ the fourier series converges to 1/2, the mean of the left and right values.

To answer your question in general, first there are many type of converges. Since you only ask for three types,

  1. For $L^2$ convergence, if your original $f(x)$ is square summable, then its fourier series will converge to $f(x)$ in $L^2$.

  2. For pointwise convergence, if your $f(x)$ WITH its periodic extension is continuous on all of $\mathbb{R}$, then this is sufficient for its fourier series to converge to $f(x)$ pointwise. The periodic extension being continuous means the function on one period is continuous AND the left and the right values are equal so the periodic extension is continuous.

  3. For uniform convergence, pointwise convergence implies uniform convergence.

These are just some of the implications. Here are a lot more details than perhaps you ever wanted.

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3. does not hold in general, we need extra regularity on f (e.g. absolute continuity, Hoelder, etc.) A counter-example can be found here –  AndrewG Mar 21 '13 at 3:29
    
Of course you are right...was just being a little careless. –  Fixed Point Mar 21 '13 at 5:02
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A Fourier series converges to the function it represents according to an $L^2$ mean error over a period.

$$\lim_{n \rightarrow \infty} \int_{-\pi/2}^{\pi/2} dx \left | f(x)-\sum_{k=0}^n a_k \cos{k x} \right|^2 =0$$

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Could we say since $f(x)$ is not continuous, then it cannot be uniformly convergence ? –  algebra Mar 20 '13 at 23:36
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The series you've written indeed converges uniformly -- but it also isn't the Fourier series of your original function, which (like Fixed Point said) has jump discontinuities. Whenever there are jump discontinuities, you get the Gibb's phenomenon, which ruins uniform convergence.

You calculated $a_n$ correctly, but $\sin(\frac{n \pi}{2}) = 0$ when $n$ is even, so your series should instead be $$ f(x) = \frac{1}{2} + \frac{1}{\pi} \sum_{n=1}^{\infty} \frac{2}{2n-1} (-1)^{n+1} \cos\left[(2n-1)x\right]$$

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A related problem. You can use Abel's uniform convergence test. See here. Or you can use the following theorem:

Theorem: The Fourier series of a $2\pi$-periodic continuous and piecewise smooth function converges uniformly.

Note that, uniform convergence implies $L_{2}$ convergence.

Added: The series converges uniformly on the open sub-intervals, since the function is piecewise smooth.

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@downvoter: what's the downvote for? –  Mhenni Benghorbal Mar 24 '13 at 0:38
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