Existence of a function I need some help:
I am thinking about this problem. Any advice would be appreciated.
Let's fix $\epsilon>0$. Does there exists some $f\in C^0([0,\pi])$ such that: 


*

*$f\mid_{[\epsilon,\pi-\epsilon]}>0$

*$f=\sum_{k=3}^\infty{a_k\cos(kx)+b_k\sin(kx)}$?
Thanks :)
 A: Yes, there exist such functions. Let for instance
$$
f(x)=\begin{cases}\sin x & 0\le x\le\pi,\\g(x) & -\pi\le x<0,\end{cases}
$$
where $g\colon[-\pi,0]\to\mathbb{R}$ is to be chosen in such a way thet $f$ is continuous and piecewise $C^1$ on $[-\pi,\pi]$ and
$$
\int_{-\pi}^0g(x)\cos(k\,x)\,dx=-\int_0^\pi \sin x\cos(k\,x)\,dx,\quad k=0,1,2
$$
and
$$
\int_{-\pi}^0g(x)\sin(k\,x)\,dx=-\int_0^\pi \sin x\sin(k\,x)\,dx,\quad k=1,2
$$
A: To get a concrete simple example, think of a function $f$ expanded in $[0,\pi]$ in a sine-only Fourier series (it is enough to think of $f$ extended to $[-\pi,0]$ as an odd function). Coefficients $b_k$ of this Fourier series are given by $b_k=\int_0^\pi f(x)\sin(kx)\,dx$ (apart a normalization factor). It is then enough to choose $f$ so that $b_1=b_2=0$. One easy example is the following:
$$
f(x)=\cases{
qx^2+(2/\epsilon-q\epsilon)x-1& if $0\le x<\epsilon$\cr
1& if $\epsilon\le x\le\pi-\epsilon$\cr
q(\pi-x)^2+(2/\epsilon-q\epsilon)(\pi-x)-1& if $\pi-\epsilon< x\le\pi$\cr
}
$$
where $q=(1-2\sin\epsilon/\epsilon)/(2\cos\epsilon+\epsilon\sin\epsilon-2)$.
This is $1$ on $[\epsilon,\pi-\epsilon]$, while $f(0)=f(\pi)=-1$ and the graph of $f$ in an arc of parabola on the outer small intervals. All $b_k$ vanish for even $k$, while $q$ is chosen so that $b_1=0$.
