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Define [tex] (f|g) = \int_0^2 (1+x^2)f(x)g(x)dx[/tex] on [tex] V = C([0,2],\mathbb{R})[/tex].

Then this is an inner product space because it obeys the four axioms

1. [tex](f|f) = \int_0^2 (1+x^2)f^2(x)dx \geq 0[/tex] since [tex] f^2(x) \geq 0 [/tex] for all [tex] x \in [0,2][/tex]

2. [tex](f|g) = \int_0^2 (1+x^2)f(x)g(x)dx = \int_0^2 (1+x^2)g(x)f(x)dx = (g|f)[/tex]

3. [tex](f+g|s) = \int_0^2 (1+x^2)((f(x)+g(x))s(x)dx = \int_0^2(1+x^2)f(x)s(x)dx + \int_0^2(1+x^2)g(x)s(x)dx = (f|s) + (g|s)[/tex]

4. [tex](\lambda f|g) = \int_0^2 \lambda(1+x^2)f(x)g(x)dx = \lambda\int_0^2(1+x^2)f(x)g(x)dx = \lambda(f|g)[/tex]

Further, because [tex] f^2(x) \geq 0[/tex] and [tex]f(x)[/tex] is continuous on [tex][0,2][/tex], it follows that

[tex]\int_0^2f^2(x)dx = 0 \Leftrightarrow f(x) = 0 \forall x \in [0,2][/tex]

This further proves that axiom 1 holds.

My query is, I have only ever had to prove things like

[tex] (f|g) = \int_a^b f(x)g(x)dx [/tex]

I don't exactly know what do do with the [tex](1+x^2)[/tex] bit in there.

Can anyone tell me where this comes into the calculations?

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# Homework Help: Inner Product Spaces

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