Oxymoron
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Question)
Define (f|g) = \int_0^2 (1+x^2)f(x)g(x)dx on V = C([0,2],\mathbb{R}).
Then this is an inner product space because it obeys the four axioms
1. (f|f) = \int_0^2 (1+x^2)f^2(x)dx \geq 0 since f^2(x) \geq 0 for all x \in [0,2]
2. (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)
3. (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)
4. (\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)
Further, because f^2(x) \geq 0 and f(x) is continuous on [0,2], it follows that
\int_0^2f^2(x)dx = 0 \Leftrightarrow f(x) = 0 \forall x \in [0,2]
This further proves that axiom 1 holds.
My query is, I have only ever had to prove things like
(f|g) = \int_a^b f(x)g(x)dx
I don't exactly know what do do with the (1+x^2) bit in there.
Can anyone tell me where this comes into the calculations?
Define (f|g) = \int_0^2 (1+x^2)f(x)g(x)dx on V = C([0,2],\mathbb{R}).
Then this is an inner product space because it obeys the four axioms
1. (f|f) = \int_0^2 (1+x^2)f^2(x)dx \geq 0 since f^2(x) \geq 0 for all x \in [0,2]
2. (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)
3. (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)
4. (\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)
Further, because f^2(x) \geq 0 and f(x) is continuous on [0,2], it follows that
\int_0^2f^2(x)dx = 0 \Leftrightarrow f(x) = 0 \forall x \in [0,2]
This further proves that axiom 1 holds.
My query is, I have only ever had to prove things like
(f|g) = \int_a^b f(x)g(x)dx
I don't exactly know what do do with the (1+x^2) bit in there.
Can anyone tell me where this comes into the calculations?
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