I encountered the following problem but I have no idea how to proceed.. Basically, I have a class of periodic function satisfying certain properties, and since I do not have explicit form of them I use their Fourier expansions, and hope to arrive the minimum value of $g(x)+g(\alpha-x)$ over interval $x\in(0,\alpha]$, where $\alpha\in(0,\pi/2]$.
Problem setting
Consider a $2\pi$-periodic function $g(x)$, $x\in\mathbb{E}$. By Fourier theorem, it can be represented as $$g(x)=a_0+\sum_{n=1}^\infty a_n\sin nx+\sum_{n=1}^\infty b_n\cos nx$$
The odd function $g(x)$ $$g(x)=\sum_{n=1}^\infty b_n\sin nx$$ where $b_n=\frac{1}{\pi} \int_{-\pi}^{\pi}g(x)\sin nx\ dx,\forall n=1,2,\cdots$. satisfying the following constraints
(1) $g(\pi/2)=1$
(2) $g'(x)>0,x\in(0,\frac{\pi}{2}]$
(3) $g(x)$ is symmetric to $x=\pi/2$.
As an example, $g(x)=\sin x$ is one of candidates.
By direct plug-in, from (1) we have $\sum_{n=1}^\infty b_n\sin\frac{n\pi}{2}=\pi$.
From (2) we have $\sum_{n=1}^\infty \Big\{\Big(\int_{-\pi}^{\pi}g'(x)\sin nx+ng(x)\cos nx\ dx\Big)\sin nx+ \frac{n}{\pi}\cos nx \Big(\int_{-\pi}^{\pi}g(x)\sin nx\ dx\Big)\Big\}>0$
My question
Is it possible to obtain the lower bound of $\min_{x\in(0,\alpha]}g(x)+g(\alpha-x)>0$?
By direct plug-in, it is equivalent to $\min_{x\in(0,\alpha]}\sum_{n=1}^\infty \big[\frac{1}{\pi} \int_{-\pi}^{\pi}g(x)\sin nx\ dx\big]\big(\cos nx+\cos(n(\alpha-x))\big)>0 $
Swap $\min$ and $\sum$, we obtain $\sum_{n=1}^\infty \min_{x\in(0,\alpha]}\big[\int_{-\pi}^{\pi}g(x)\sin nx\ dx\big]\sin\frac{n\alpha}{2}\cos(nx-\frac{n\alpha}{2})>0 $
But then, how should I proceed with this? I feel this might be an ill-posed problem, so maybe should add more constraints on $g$ to make it well-posed?