$\textbf{The Problem:}$ Suppose that for any smooth function $F:[-L,L]\to\mathbb C$ satisfying $F(L)=F(-L)$ we can write for some $c_n\in\mathbb C$ $$F(x)=\sum^{\infty}_{n=-\infty}c_ne^{in\pi x/L}.$$ a) Show that if $f:(0,L)\to\mathbb R$ is smooth and satisfies $f(0)=f(L)=0$ then we can write for some $a_n\in\mathbb R$ $$f(x)=\sum^{\infty}_{n=1}a_n\sin\left(\frac{n\pi x}{L}\right).$$

$\textbf{My Attempt:}$ I start by defining the function $F:[-L,L]\to\mathbb R$ by setting $F(x)=f(x)$ if $x\geq0$ and $F(x)=-f(-x)$ if $x<0.$ Then we can use said representation to obtain $$\begin{align*}F(x)&=\sum^{\infty}_{n=-\infty}c_ne^{in\pi x/L}\\ &=c_0+\sum^{\infty}_{n=1}c_n\left[\cos\left(\frac{n\pi x}{L}\right)+i\sin\left(\frac{n\pi x}{L}\right)\right]+\sum^{-1}_{n=-\infty}c_n\left[\cos\left(\frac{n\pi x}{L}\right)+i\sin\left(\frac{n\pi x}{L}\right)\right]\\ &=c_0+\sum^{\infty}_{n=1}c_n\left[\cos\left(\frac{n\pi x}{L}\right)+i\sin\left(\frac{n\pi x}{L}\right)\right]+\sum^{\infty}_{n=1}c_{-n}\left[\cos\left(\frac{n\pi x}{L}\right)-i\sin\left(\frac{n\pi x}{L}\right)\right]\\ &=c_0+\sum^{\infty}_{n=1}(c_n+c_{-n})\cos\left(\frac{n\pi x}{L}\right)+\sum^{\infty}_{n=1}i(c_n-c_{-n})\sin\left(\frac{n\pi x}{L}\right).\\ \end{align*}$$ Now since $F(x)=-F(-x),$ we see that $$c_0+\sum^{\infty}_{n=1}(c_n+c_{-n})\cos\left(\frac{n\pi x}{L}\right)=-c_0-\sum^{\infty}_{n=1}(c_n+c_{-n})\cos\left(\frac{n\pi x}{L}\right),$$ and hence $$c_0+\sum^{\infty}_{n=1}(c_n+c_{-n})\cos\left(\frac{n\pi x}{L}\right)=0.$$ $\color{blue}{\text{Since the above holds for all $x\in[-L,L],$ we must have that $c_0=0$ and $c_n=-c_{-n}$ for all $n\in\mathbb N$.}}$

Finally, since $\overline{F(x)}=F(x)$, it follows that $ic_n=\overline{ic_n}.$

Is my reasoning above correct? I think I was not rigorous enough in the deduction in $\color{blue}{\text{blue}}.$


Your argument is fine. The blue part requires some information about the functions $\cos(\frac {n\pi x} L)$. These functions are orthogonal in $L^{2} (-L,L)$ and hence $\sum \alpha_n \cos(\frac {n\pi x} L)=0$ implies $\alpha_n=0$ for all $n$.


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