Existence of a function

Question

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 :)

Answer 1

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 $$ \int_{-\pi}^0g(x)\sin(k\,x)\,dx=-\int_0^\pi \sin x\sin(k\,x)\,dx,\quad k=1,2 $$

Answer 2

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 $b_1=0$.