Prove/Disprove$ \langle f,f\rangle = 2\int_0^{0.5} \left | f(t) \right |^2 \, dt -\int_{0.5}^1 \left | f(t) \right |^2 \, dt = 0 \Rightarrow f = 0$

Question

It's a part of proving an inner product space, I can't find a way to prove that nor disprove that $$ \langle f,f\rangle = 2\int_0^{0.5} \left| f(t) \right|^2 \, dt -\int_{0.5}^1 \left | f(t) \right |^2 \, dt = 0 \Rightarrow f = 0$$

We know $C^1[0,1]$ are all the continuous functions $f[0,1] \rightarrow \mathbb{C}$ with continous deravatives on $[0,1],$ the inner product that needs to be shown is over $C^1[0,1]$.

I can't see for example why this can't be a possibility that will result in the fact that $f\neq0$ but still $\langle f,f\rangle = 0$

Answer 1

The statement is wrong. As a result, the bilinear form is not a inner product.

A counter example is given by $$ f(t) = 2^t\,. $$

Answer 2

Try $f(x)=x^k$, where $k=\frac{\log_23-1}{2}.$