Show that the formula defines an inner product on X

Question

Let $X=C[-1,1]$ be the space of continuous functions $f:[-1,1]\rightarrow \mathbb{R}$. For $f,g\in X$ define: $$\langle f,g\rangle_2=\int_{-1}^{1}|t|f(t)g(t)dt$$ The property i'm struggling with is that $$\langle f,f\rangle_2 \geq 0 \mbox{ and } \langle f,f\rangle_2=0 \iff f=0$$ How would I go about proving this? thanks.

Answer 1

$$\langle f,f\rangle_2=0 \iff \int_{-1}^1|t|f^2 dt=0$$

Claim: $g(t)=|t|f^2=0, t\in[-1,1]$.

First note $g(t)\geq 0$ and continuous. If $g(t_0)>0$ for some $t_0$, then you can find a neighborhood $I$ of $t_0$ s.t. $g(t)>\frac{1}{2}g(t_0)>0$ by continuity, hence $$\int_{-1}^1g(t) dt\ge \int_{I}g(t) dt>0$$

Then you can show $f=0, t\in[-1,1]$ by similar argument.

Conversely, if $f=0, t\in[-1,1]$, clearly you get $\langle f,f\rangle_2=0$.