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

[icesalmon]

Homework Statement

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

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.

 

[Ray Vickson]

Homework Statement

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

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.

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?