With inner product $\langle f,g\rangle=\int_0^{2\pi}f(t)\overline{g(t)}dt$ on $L_2[0,2\pi]$ how do $||f||$, $||f||_2$ and $\langle f,f\rangle$ relate?

Question

So this is more an issue of understanding syntax. It is given just about anywhere $||f||^{2}=\langle f,f\rangle$ implying $||f||=\sqrt{\langle f,f\rangle}$, but what norm is $||f||$ representing? Which of the following is true: $$||f||^{2}_{2}=\langle f,f\rangle \space \,\text{ or }\, \space ||f||^{2}_{1}=\langle f,f\rangle .$$ Without knowing which of the above (or both) are true, I am unable to have any confidence in computing $||f||_{2}^{2}$.

I hope my confusion makes sense, and I appreciate any help provided.

Answer 1

Firstly, $\|f\|_1$ is not the norm of $L_2[0,2\pi]$ by definition; $\|f\|_2$ is. Secondly, $\|f\|_1$ does not satisfy the Parallelogram law (see the section “Parallelogram law” and “Real and complex parts of inner products” of this wiki page), so it cannot define a inner product. Thus, for your question, the norm is definitely $\|f\|_2$.