Inner product space of continuous function

[Lily@pie]

Homework Statement

C[a,b] is a vector space of continuos real valued functions. for f,g in C[a,b]
<f,g>=∫f(x)g(x)dx, [a,b]

Give a completely rigorous proof that if <f,f>=0, then f=0

2. The attempt at a solution
I tried to prove this by contrapositive, "f≠0 implies that <f,f>≠0

When f(x)≠0, f(x)<0 or f(x)>0. However, [f(x)]²>0 in both cases. Therefore, there exist a<c<b and a real number d such that [f(c)]²=d > 0

Since f is a continuous function, f² is also a continuos function.
By definition of continuity,
For all ε>0, exist δ>0 such that
for all x : |x-c|<δ implies that |f²(x)-f²(c)|<ε
We choose ε<d/2 and get that
-d/2<f²(x)-f²(c)<d/2
0<d/2<f²(x)<3d/2
So, for all x that is in the interval [c-δ,c+δ], f²(x)>0
(can we conclude anything about the x not in this interval? like they'll be 0?)

The following is the part where I'm not sure...
To compute the integral ∫$^{a}_{b}$f²(x)
we divide [a,b] into n-subinterval
a=x₁<x₂<...<x_i-1<c-δ<c+δ<x_i+1<...<x_n=b
Let x$^{i}_{*}$= midpoint of each interval [x$_{i}$,x$_{i+1}$] where we denote c-δ<x<c+δ
∫$^{a}_{b}$f²(x)≈f$^{2}$(x$_{1}$)(x₂-x₁)+...+f$^{2}$(x)δ+...+f$^{2}$(x$_{n-1}$)(x_n-x_n-1)
>f$^{2}$(x)δ>0 I'm really not sure about this step

So since f≠0 implies <f,f>=∫$^{a}_{b}$f²(x)≠0.<f,f>=0 implies f=0

Thanks

 

[HallsofIvy]

If f is not identically 0, then there exist some $x_0$ such that $f(x_0)$ is not 0. Since f is continuous, there exist some interval around $x_0$ such that f(x) is not 0 for any x in that interval. Also use the fact that $f^2(x)\ge 0$ for any real valued function, f, and all x.

 

[Lily@pie]

so did i do it correctly?