Given $\langle f,g \rangle := \frac12\int_{-\pi}^{\pi}f(x)\overline{g(x)}\,dx$ I'm searching an example of $f \neq 0$ such that $\langle f,f\rangle=0$

Question

Given the inner product

$$\langle f,g \rangle := \frac12 \int_{-\pi}^{\pi}f(x)\overline{g(x)}\,dx$$

I'm searching an example of $f \neq 0$ such that $\langle f,f\rangle=0$.

Answer 1

What you want is impossible for one of two reasons. Either

$(1)$ We take it on linear algebra faith that the object is a well defined inner product, so such an $f$ cannot exist because it violates positive definiteness or

$(2)$ We prove it with calculus.

$$\langle f, f \rangle = \frac{1}{2}\int_{-\pi}^{\pi} |f(x)|^2dx$$

The integrand is strictly nonnegative. Thus the only way the integral evaluates to $0$ is if the integrand is $0$:

$$|f(x)|^2 = 0 \: \text{a.e.} \implies f(x) = 0 \: \text{a.e.}$$

Answer 2

Since nothing is said in the question that $f$ must be continuous or that we are considering functions module null sets, then there are examples. Here's one: $$ f(x) = \begin{cases} 0 & (x \neq 0) \\ 1 & (x = 0) \end{cases} $$

Answer 3

If you are not limiting the domain of $f$, you can take $$ f(x)=\begin{cases} |x|-\pi,&\ |x|\geq\pi, \\ \ \\ 0,&\ -\pi<x<\pi \end{cases} $$ Then $f$ is continuous and $$\int_{-\pi}^\pi |f(x)|^2\,dx=0.$$