Show that $\beta(f,g) = \int_{0}^{1} f(x)g‘(x) dx$ defines a nonsingular bilinear map on the vectorspace $V=\lbrace f \in C^1([0,1]) \mid f(0)=0=f(1) \rbrace$.
(Where $f \in C^1$ means that f has a continous derivative)
The problem for me is that $V$ is not finite-dimensional. I guess I have to show that there isnt any $0 \neq h \in V$ such that $\beta (f,h) = 0 \quad \forall f \in V$.
I think it should be $f(0)=f(1)=0$ in the definition of $V$. Suppose $h'(x)>0$ for some $x \in (0,1)$. You can easily construct a function $f \in V$ such that $\int_0^{1} fh'>0$ (ask me if you need details). Similarly $h'(x)<0$ leads to a contradiction. Thus $h'=0$ on $(0,1)$. By continuity $h'=0$ on $[0,1]$. Thus $h$ is a constant. Since $h(0)=0$ it follows that $h \equiv 0$.