Fourier series and uniform convergence for a given function.

Question

The Fourier series of $x$ is $$\sum_{n=1}^{\infty} 2(-1)^n\frac{\sin(nx)}{n}$$ I can see that the series does not converge pointwise to x in $x=\pi$ My question is: what happen in $[0,a]$ when $a<\pi$ ,my intution says that we should get uniform convergence, but i cant prove that is true.

For general, what can we say about uniform convergence of Fourier series of a function which is continuous, the derivative continuous but $f(\pi) \neq f(-\pi)$ in $[0,a]$ when $a<\pi$ ?

What about this specific case? I did not succeed to determine if there is a uniform convergence.

I am interested in this question because I tried to calculate the Gibs phenomenon in the function $-1+x$ at $[0,\pi] $and $1+x$ at $ [-\pi,0]$ We know that the Gibs phenomenon in the function 1 occur at $\pi /2n $ so because $x$ is a nice function, I think that the Gibs phenomenon of the given function above occur in $\pi /2n $. So, if the Fourier series of x uniform convergence, then that should be true.

Answer 1

Let $a\in \mathbb C$, $|a|\le 1$ and $a\ne -1$. Then (integration below is complex along the interval $[0,a]\subset\mathbb C$) $$ \sum_{k=1}^n (-1)^k\frac{a^k}{k}=\int_0^a \left(\sum_{k=1}^n (-t)^{k-1}\right) dt =-\int_0^a \frac{1-(-t)^n}{1+t} \,dt\\ =-\log (1+a)+(-1)^n\int_0^a \frac{t^n}{1+t} \,dt= -\log (1+a)+\delta_n $$ where $\log z$ is the principal branch defined in $\mathbb C\setminus(-\infty,0]$ and $$ |\delta_n|\le\int_0^a \frac{|t^n|}{|1+t|} \,|dt| \le \frac{1}{n(|1+a|)}. $$ Hence, for every $x\ne (2k+1)\pi i$ $$ \left|\sum_{k=1}^n (-1)^k\frac{\mathrm{e}^{ikx}}{k}+\log(1+\mathrm{e}^{ix})\right|\le \frac{1}{n(|1+\mathrm{e}^{ix}|)}, $$ and in particular, the series $\sum_{n=1}^\infty (-1)^n\frac{\mathrm{e}^{inx}}{n}$ converges uniformly in any closed interval of $x$ not containing a $(2k+1)\pi$, and so does its imaginary part $$ \sum_{n=1}^\infty (-1)^n\frac{\sin (nx)}{n} $$