$\textbf{The Problem:}$ Suppose that for any smooth function $F:[-L,L]\to\mathbb C$ satisfying $F(L)=F(-L)$ we can write for some $c_n\in\mathbb C$ $$F(x)=\sum^{\infty}_{n=-\infty}c_ne^{in\pi x/L}.$$ Show that if $f:(0,L)\to\mathbb R$ is smooth and satisfies $f'(0)=f'(L)=0$ then we can write for some $b_n\in\mathbb R$ $$f(x)=\sum^{\infty}_{n=0}b_n\cos\left(\frac{n\pi x}{L}\right).$$
$\textbf{My Attempt:}$ Let $F:[-L,L]\to\mathbb R$ be defined by $F(x)=f(x)$ if $x\geq0$ and $F(x)=f(-x)$ if $x<0.$ Then using the hypothesis we get $$\begin{align*}F(x)&=\sum^{\infty}_{n=-\infty}c_ne^{in\pi x/L}\\ &=c_0+\sum^{\infty}_{n=1}c_n\left[\cos\left(\frac{n\pi x}{L}\right)+i\sin\left(\frac{n\pi x}{L}\right)\right]+\sum^{-1}_{n=-\infty}c_n\left[\cos\left(\frac{n\pi x}{L}\right)+i\sin\left(\frac{n\pi x}{L}\right)\right]\\ &=c_0+\sum^{\infty}_{n=1}c_n\left[\cos\left(\frac{n\pi x}{L}\right)+i\sin\left(\frac{n\pi x}{L}\right)\right]+\sum^{\infty}_{n=1}c_{-n}\left[\cos\left(\frac{n\pi x}{L}\right)-i\sin\left(\frac{n\pi x}{L}\right)\right]\\ &=c_0+\sum^{\infty}_{n=1}(c_n+c_{-n})\cos\left(\frac{n\pi x}{L}\right)+\sum^{\infty}_{n=1}i(c_n-c_{-n})\sin\left(\frac{n\pi x}{L}\right).\\ \end{align*}$$ Now since $F(x)=F(-x),$ we see that $$\sum^{\infty}_{n=1}i(c_n-c_{-n})\sin\left(\frac{n\pi x}{L}\right)=0.$$
By Kavi Rama Murthy's remark in his answer here: https://math.stackexchange.com/a/3183135/595519, we see that the above implies that it must be case that $c_n=c_{-n}$ for all $n\in\mathbb N_{>0}.$ Finally, since $\overline{F(x)}=F(x)$, it follows that $2c_n=2\overline{c_n}.$ Therefore we get that for some real numbers $b_n$ with $b_n=2c_n$ for $n\geq1$ and $b_0=c_0$ we have $$f(x)=\sum^{\infty}_{n=0}b_n\cos\left(\frac{n\pi x}{L}\right).$$
Is my reasoning above correct?
Any feedback is much appreciated, and thanks to Kavi Rama Murthy for his insight in the previous problem I linked above, which is crucial to solve this one as well.