Proving uniform convergence of a Fourier series Let $f(x):\mathbb{R}\to\mathbb{R}$, $f'(x) > 0$ for $x\in\mathbb{R}$ and $g(x):\mathbb{R}\to\mathbb{R}$ a periodic function with a $2\pi$ period such that $g(x)=f(x)$ if $-\pi < x < \pi$.
I need to prove, or disprove, that the Fourier series of $g(x)$ converge uniformly in $(0, \pi)$.
First, $g$'s Fourier series is
$$g(x)\sim \frac{1}{2}a_0+\sum_{n=1}^{\infty}(a_n\cos nx+b_n\sin nx)$$
I know that for every $x_0\in(0, \pi)$, $g(x_0)=f(x_0)$ is continuous in $x_0$ and both the left and right derivative exists thus for every such $x_0$ the Fourier series of $g(x)$ converge to $f(x_0)$, i.e.
$$\frac{1}{2}a_0+\sum_{n=1}^{\infty}(a_n\cos nx_0+b_n\sin nx_0) = f(x_0)$$
Now all that left is to show that $$\sum_{n=1}^{\infty}(a_n\cos nx+b_n\sin nx)$$ converges to $f(x)$ uniformly for $x\in(0,\pi)$.
 A: $\def\R{\mathbb{R}}\def\N{\mathbb{N}}\def\paren#1{\left(#1\right)}$The counter-example provided by @Conrad is enough to disprove the proposition. A stronger proposition, however, can be proved:

If $f: [-π, π] → \R$ is strictly monotonic and has only a finite number of points of discontinuity, and $g$ is a $2π$-periodic function with $f = g$ on $(-π, π)$, then the Fouries series of $g$ does not converge uniformly on $(0, π)$.

Proof: Denote the partial sums of the Fourier series by $S_N(x) = a_0 + \sum\limits_{n = 1}^N (a_n \cos nx + b_n \sin nx)$ for $N \in \N$. Note that the Dirichlet condition implies that$$
\lim_{N → ∞} S_N(x) = \frac{1}{2} (g(x + 0) + g(x - 0)).\quad \forall x \in \R
$$
Suppose $\{S_N\}$ converges uniformly on $(0, π)$, then\begin{gather*}
\frac{1}{2} (g(π + 0) + g(π - 0)) = \lim_{N → ∞} S_N(π) = \lim_{N → ∞} \lim_{x → π-} S_N(x)\\
\stackrel{(1)}{=} \lim_{x → π-} \lim_{N → ∞} S_N(x) = \lim_{x → π-} \frac{1}{2} (g(x + 0) + g(x - 0)) \stackrel{(2)}{=} g(π - 0),
\end{gather*}
where (1) uses uniform convergence and (2) uses monotonicity of $g$ on $(-π, π)$. But$$
g(π - 0) = f(π - 0) \geqslant f\paren{ \frac{π}{3} } > f\paren{ -\frac{π}{3} } \geqslant f(-π + 0) = g(-π + 0) = g(π + 0),$$
a contradiction.
