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Let $w(x)$ be a strictly positive continuous function on [a,b]. Define a form on $C[a,b]$ by the formula $\langle f,g \rangle _w = \int^b_a f(x)g(x)dx$ for $f,g \in C[a,b]$. Show that it is an inner product.

Ive never seen the inner product expressed as a integral before so I'm fairly stumped.

Any guidance would help,

Cheers

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Shouldn't the inner product depend on $w$ in some way? –  Chris Eagle Jan 31 '13 at 18:58
2  
Shouldn't your integral have a $w(x)$ in it, somewhere? –  David Wheeler Jan 31 '13 at 18:58
    
Once you've sorted that out: do you know what "inner product" means? Which parts of the definition are you having trouble proving? –  Chris Eagle Jan 31 '13 at 18:58
    
Maybe you mean $$\int_a^b fg\space dw(x)?$$ Also, I agree with @ChrisEagle. If what you have written is really what you mean, proving it via the definition is very straightforward. –  anon271828 Jan 31 '13 at 19:06
    
What properties were you told that must be satisfied in order for something to be an inner product? –  JohnD Jan 31 '13 at 19:19
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1 Answer

up vote 1 down vote accepted

I am guessing that you meant $\langle f ,g \rangle_w = \int_a^b w(t)f(t)g(t) dt$.

You need (you are working in the reals here) to show symmetry (ie $\langle f ,g \rangle_w = \langle g, f \rangle_w $), linearity in the first argument (ie, $\langle \lambda f ,g \rangle_w = \lambda \langle f ,g \rangle_w $ and $\langle f+h ,g \rangle_w = \langle f ,g \rangle_w + \langle h ,g \rangle_w$) and positive definiteness (ie, $\langle f ,f \rangle_w \geq 0$, and $\langle f ,f \rangle_w = 0$ iff $f=0$).

The first two properties follow immediately from the commutativity of multiplication and linearity of the integral. The latter follows from properties of the integral when integrating non-negative functions.

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