Find the Fourier series of a function , determining whether it converges pointwise/uniformly

Question

I've been trying to solve an exercise from a test. I could use your help :)

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Question

Consider the function:

$g(x) = \frac{\pi}{4} $ for $x\in [0,\pi ]$ and $ -\frac{\pi}{4}$ for $x\in (-\pi,0 ) $. Let $f(n)$ be $g(x)$'s periodic continuation.

$S_N f(x) = A_0 +\sum_{n=1}^{N}A_n \cos{nx}+i\sum_{n=1}^{N}B_n \sin{nx}$ where $A_0 = \hat{f}(0), A_n = \hat{f}(n)+\hat{f}(-n), B_n = \hat{f}(n)-\hat{f}(-n)$.

Note: The Fourier series of $f$ is $\lim_{N \to \infty } S_Nf$

1. Find $S_Nf$.
2. Determine if $S_Nf(x)$ pointwise converges for $x\in \left[-\pi,\pi \right]$.
3. Determine if $S_Nf(x)$ uniformly converges in $ \left[-\pi,\pi \right]$.
4. Prove $\lim_{N \to \infty} S_Nf \left( \frac{1}{\pi N}\right) = \int_{0}^{1} \frac{\sin{\left( \frac{t}{\pi}\right)}}{2t} \ dt $

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My attempt

Sol for 1:

$\hat{f}(n)= \frac{1}{2in}$ for $n\in \mathbb{Z}_{odd} $ , else it's $0$.

Hence, $S_N f (x) = \sum_{k=0}^{K}\frac{1}{(2k+1)} \sin{((2k+1)x)}$ where $K=\left\lfloor \frac{N}{2}\right\rfloor$

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Sol for 3:

$S_N f (x)$ is a continuous function, therefore if $S_Nf$ uniformly converges we know it has to be to $f$. But $f$ isn't continuous. Therefore we will conclude $S_Nf$ does not uniformly converges.

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I would appreciate your help in 2 and 4 (if you do please be formal so I can understand better). Thank you!

Answer 1

Solution to (2)

There is a Theorem by Dirichlet that states:

"Given a $2\pi$-periodic function $f$ which is generally continuous and whose derivative is generally continuous in $[-\pi,+\pi]$ (by a generally continuous function I mean that it has finitely many dicontinuities in its domain and that each on of them is a jump dicontinuity), its Fourier Series converges uniformly to $f$ in every sub-interval in which $f$ is continuos and converges to the mean value of the left and right limits of $f$ at each point of discontinuity of $f$"

From this theorem it follows immediately that $S_N(f)$ converges pointwise to some function in $[-\pi,+\pi]$ but that function is not exactly $f$ (e.g, $\lim_{N \to +\infty} S_N(f)(0)=0$). You could also use this theorem to answer point (3).

Solution to (4)

First of all notice that we can compute $\lim_{N\to+\infty}S_N(f)(\frac{1}{\pi N})$ by computing $\lim_{N\to+\infty}S_{2N+1}(f)(\frac{1}{\pi (2N+1)})$, since it is an injective change of variable. Then, having written the expression for this limit explicitly:

$$ \lim_{N\to+\infty} \sum_{n=0}^{N} \frac{1}{2n+1} \sin(\frac{2n+1}{\pi (2N+1)}) $$

we let $x_{n}=\frac{2n+1}{\pi (2N+1)}, \hspace{3mm} n=0,...,N$ and $\Delta x = \frac{2}{\pi (2N+1)}$ in order to obtain:

$$ \lim_{N\to+\infty} \sum_{n=0}^{N} \frac{\sin(x_n)}{2x_n} \Delta x $$

but this Riemann Sum converges exactly to the desired integral.