I currently stuck proving positive definiteness for the following function: $$\left< f,g \right> =\int _{ 0 }^{ 1 }{ f(x)g(x)e^x\;dt }$$ in $C[0,1]$
So far, I have proven $0\leq\left<f,f\right>$ as such:
$\left< f,f \right> =\int _{ 0 }^{ 1 }{ f(x)^2e^xdt }$ . Since $f \in R$, we know that $0\leq f(x)^2$. We also know that $0\leq e^x$, allowing us determine that the slope of $\left< f,f \right>$ ($f(x)^2e^x$) is always positive, meaning that our integral is as well.
I am having trouble proving that $\left<f,f\right>=0 \iff f=0$:
I proved that $f(x)=0\Rightarrow \left< f,f \right>=0$ quite simply by saying that:
$$f(x)=0\Rightarrow \int _{ 0 }^{ 1 }{ (0)^2e^xdt }=0$$
It is proving that $ \left< f,f \right>=0\Rightarrow f(x)=0$ that I am having difficulty with. I am unsure how to continue after the assumption that $\left< f,f \right> =\int _{ 0 }^{ 1 }{ f(x)^2e^xdt } = 0$. It is clear to me that, since $e^x$ cannot be the cause the of result being zero, $f(x)^2$ must be, but I am unsure of how to prove this.
Take $h(x)=f(x)^2e^x$ .Then $h$ is continuous on $[0,1]$ and $\int h(x)dx=0$. First we prove $h(x)=0$. Suppose not, then $\exists x_0 \in [0,1] $ so that $h(x_0)>0$. By continuity of $h$ at $x_0$, $\exists$ $[a,b] \subset [0,1]$ so that $h(x)>0$ for all $x \in [a,b]$. Now $$\int_0^1 h(x) dx>\int_a^bh(x)dx>0$$ which is a contradiction! so $h(x)=0$. But $e^x$ is never zero, so $f^2$ is zero and hence $f=0$