If $e_n=\frac{1}{\sqrt{2 \pi}}e^{inx}$ is a basis for $L^2([0, 2\pi])$, then why $\frac{1}{\sqrt{2 \pi}}e^{-inx}$ as well?

Question

If $\frac{1}{\sqrt{2 \pi}}e^{inx}$ is a basis for $L^2([0, 2\pi])$, then why $\frac{1}{\sqrt{2 \pi}}e^{-inx}$ as well?

This realization allows one to write Fourier coefficients as:

$$\frac{1}{\sqrt{2 \pi}}(f \space | \space e_n)$$

Answer 1

Note the complex conjugate of $\frac{1}{\sqrt{2 \pi}}e^{inx}$ is $\frac{1}{\sqrt{2 \pi}}e^{-inx}$. So, given $f \in L^2([0,2\pi])$, take its complex conjugate and express $\overline{f}$ in terms of the basis of $\frac{1}{\sqrt{2 \pi}}e^{inx}$, then just take the complex conjugate of the whole expression to have $f$ in terms of the required basis functions.