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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.

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  • $\begingroup$ The title seems to be asking exactly the opposite of the question. Are you supposed to show it's NONdegenerate on the space of continuous functions, or that it's degenerate there? Or is it merely that you have the definitions of degenerate/nondegenerate messed up (as the parenthesis suggests)? $\endgroup$ – John Hughes Dec 8 '16 at 5:13
  • $\begingroup$ Anyway, the product $f(x)g(x)$ will always itself be just a function. Can you think of a nonzero function that integrates to zero? Can a nonzero continuous function that integrates to zero? $\endgroup$ – arctic tern Dec 8 '16 at 5:14
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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 $$\langle g,g\rangle=\int_0^1|g|^2=0.$$

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  • $\begingroup$ I see. The fact that the integral of $f^2$ is zero implies that $f$ has to be zero $a.e.$. But if $f$ is continuous and zero $a.e$, then it's identically zero. Am I right? $\endgroup$ – adrija Dec 8 '16 at 5:20
  • $\begingroup$ Exactly! $\ \ \ \ $ $\endgroup$ – Martin Argerami Dec 8 '16 at 5:21
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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 $f^2=0$.

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