Defining an inner product on C[-1,1]

1. Jul 24, 2016

icesalmon

1. The problem statement, all variables and given/known data
Consider the vector space of all continuous functions on the interval C[-1,1]. That is V = C[-1,1]
show that <f(x),g(x)> = ∫(-1,1) x2f(x)g(x)dx defines an inner product on C[-1,1].
I have shown <f,g> = <g,f>, <kf,g> = k<f,g>, <f+g,h> = <f,h> + <g,h> and I am trying to show that given any continuous function over [-1,1] that <f,f> ≥ 0 and <f,f> = 0 iff f(x) = 0

3. The attempt at a solution
<f,f> = ∫(-1,1)x2[f(x)]2dx = ∫(-1,1)[xf(x)]2dx. I am stuck here, this f(x) could be anything that is continuous on [-1,1] over a symmetric interval and x2 is an even function while f(x)2 could be even or odd. Maybe I can say something about f(x)2 being even or odd, if x2f2(x) is odd then integrating it over a symmetric interval will be zero but if x2f2(x) is even then integrating it over a symmetric interval will just be twice the value of ∫[0,1](xf(x))2. I'm not sure how to proceed because I feel I need to force the integrand to be an even function and even then how would I know that the value of the integral is positive? Any help to push me in the right direction would be much appreciated, thank you.

2. Jul 24, 2016

Ray Vickson

Can the function $(x f(x))^2$ ever be $< 0$ anywhwere in $(-1,1)$? Can you think of a real number whose square is negative?

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