How do I show that the bilinear form on functions in $[0,1]$ is degenerate, but becomes nondegenerate when restricted to continuous maps?

Question

For functions $f,g:[0,1]\rightarrow \mathbb R$, the standard bilinear form is given by $\langle f,g\rangle=\int_{0}^{1}f(t)g(t)dt$. I'm trying a problem asking me to show that this form is degenerate (that is, I can find a non-zero continuous function which integrates to zero which every other function), but it becomes nondegenerate on restricting to the space of continuous functions.

I seem to completely lost. I'd appreciate some help. Thanks.

Answer 1

For a continuous function $f$, if you have $0=\langle f,f\rangle=\int|f|^2$, it follows that $f=0$. But if you allow arbitrary functions, you can take for example $$g(t)=\begin{cases}1,&\ t=0\\ 0,&\ t>0\end{cases}$$ Then $$\langle g,g\rangle=\int_0^1|g|^2=0.$$

Answer 2

Consider $f(x)=\begin{cases}1 & x=0\\0 & x \ne 0\end{cases}$. For any function $g$ on $[0,1]$ we have $\langle f,g\rangle=0$.

Suppose $f$ is a continuous function such that $\langle f,g\rangle = 0$ for any continuous $g$. In particular we have $\langle f,f\rangle = \int f^2 =0$. Since $f^2$ is nonnegative and continuous, we must have $f^2=0$.