We are asked to show that the set $\mathcal{B} = \{e_n\}_{n\in\mathbb{Z}}$, where $e_n(x) = e^{2\pi inx}$, is an orthonormal basis for $\mathcal{H} = L^2[-1/2,1/2]$. I have shown that:
- each of the functions is an element of $\mathcal{H}$;
- the set $\mathcal{B}$ is orthonormal.
It remains to show that $\mathcal{B}$ spans $\mathcal{H}$. For this, I am using a proposition which states that an orthonormal set $\{a_k\}_{k \in \mathbb{Z}}$ spans a Hilbert space if and only if:
$$\left<f,a_k\right> = 0 \text{ for all } k \implies f = 0.$$
I have assumed the "only if" direction above and am seeking clarification on the following line:
$$0 = \left<f,e_n\right> = \int_{-1/2}^{1/2}f(x)\cdot e^{-2\pi inx}\,\mathrm{d}x.$$
Since $e^{-2\pi inx} \ne 0$ for all $n$ and $x$, it must be the case that $f(x)=0$. Can someone provide a rigorous argument for this please!
Well, after showing that the fourier transform is one-to-one, i.e. the mapping $$\mathcal{F}:L^2(-\frac12,\frac12)\to c_0(\mathbb{Z}) $$ with $$f\mapsto\mathcal{F}(f):=(\hat{f}(n))_{n=-\infty}^\infty $$
Then $\langle f, e_n\rangle=0$ for all $n$ means precisely that $\mathcal{F}(f)=(0)$, the zero element of $c_0(\mathbb{Z})$. But $\mathcal{F}$ is linear, so since it is one-to-one this means that $\ker(\mathcal{F})=0$, so $f=0$ in $L^2$, i.e. $f=0$ almost everywhere.