Show that $\beta(f,g)=\int_0^1 f(x)g'(x)dx$ is nondegenerate. We have the bilinear form $\beta:V \times V \to \mathbb{R}, (f,g) \mapsto \int_0^1 f(x)g'(x)dx$ with
$V=\{f\in C^1[0,1] \mid f(0)=f(1)=0 \}$. Show that $\beta$ is nondegenerate.
My attempt: For $0\neq f \in V$ we have $\beta(f,F)=\int_0^1 f(x)^2dx > 0$ where
$F(x)=\int_0^x f(t)dt$. But $F(1)$ is not always $0$. So i tried something like
$G(x)=F(x)-F(1)x$ or $G(x)=F(x)(1-x)$ but i am not able to show $\beta(f,G)\neq 0$.
 A: Let $0 \neq f \in V$.
For $G(x)=F(x)-F(1)x \in V$ with $F(x)=\int_0^xf(t)dt$.
Now $f(x)$ cannot be $F(1)$ for all $x\in [0,1]$ because $f(0)=f(1)=0$ and therefore $f=0$ which would be a contradiction. Therefore we get:
$0<\int_0^1\left(f(x)-F(1)\right)^2dx = \int_0^1f(x)^2dx - 2F(1)\underbrace{\int_0^1f(x)dx}_{=F(1)}+F(1)^2 = \int_0^1f(x)^2dx - F(1)^2$.
This implies:
$\beta(f,G) = \int_0^1 f(x)(f(x)-F(1))dx 
=\int_0^1f(x)^2dx -F(1)\underbrace{\int_0^1f(x)}_{=F(1)}= \int_0^1f(x)^2dx -F(1)^2 \stackrel{}{>}0$.
A: We assume this is known:

Lemma: If $h:(0,1)\to \mathbb R$ is continuous and $h(c)\neq 0$ for some $c\in (0,1),$ then there is an $[a,b]\subset (0,1)$ such that $h(x)$ is either strictly positive or strictly negative on $[a,b].$

A corollary is that we can find a continuously differentiable $g$ which is non-zero and the same sign as $h$ on $(a,b),$ and zero outside $(a,b),$ so we have $g\in V$ and $\int_0^1 gh>0.$
Now, for $f\in V\setminus \{0\},$ let $h=f’.$
$h$ is continuous and not everywhere zero, since otherwise $f$ would be constant and hence $f=0.$
So there a $g\in V$ such that $$\beta(g,f)=\int_0^1 h(t)g(t)\,dt\neq 0.$$
But then we can easily show that: $$\beta(f,g)+\beta(g,f)=\int_0^1 (fg)’(t)=f(1)g(1)-f(0)g(0)=0.$$
So we have $\beta(f,g)\neq 0.$
