[SOLVED] bilinear mapping

Assume a<b. $\displaystyle C[a,b]$ denotes the set of all continuous functions $\displaystyle [a,b] \rightarrow \mathbb{R}$ and that all continuous functions $\displaystyle [a,b] \rightarrow \mathbb{R}$ are integrable. So we have a function $\displaystyle I: C[a,b] \times C[a,b] \rightarrow \mathbb{R}$ given by $\displaystyle I(f,g)= \int^b_a f(t)g(t)dt$.

(I have already proved and concluded that I is a bilinear form).

Here's the question:

**Is this bilinear form non-degenerate? Justify your answer** **by stating explicitly any results you use from calculus.**

My Attempt

The bilinear form I is non-degenerate

1. If I(f,g)=0 $\displaystyle \forall f \in C[a,b]$ then g=0

2. If I(f,g)=0 $\displaystyle \forall g \in C[a,b] $then f=0

I personally think this bilinear form is non-degenerate. I will try to give a proof by contradiction. For a contradiction suppose $\displaystyle I(f,g) = \int_a^b f(t)g(t)\, dt = 0$ but $\displaystyle g \neq 0$.

I'm not sure how to carry on and finish this proof from this point. I don't know how to use calculus here... I know that integrals represent areas under the graphs. And in this case we know know $\displaystyle g(t) \neq 0$.

And f is not defined, if we define f = sin(t) and a=-1, b=1 we get

$\displaystyle

\int^1_{-1} sin(t) =0

$

But this is a counter example to what we are trying to prove!

I have to submit this tomorrow, so I appreciate it if anyone could please help me complete this proof...urgently