1. ## Inner Product

Let V be the space of continuous functions. Show that <f(x), g(x)> = the integral from 0 to 1 of f(x)g(x)dx is an inner product.

I know all of the axioms needed to prove it's an inner product. I'm confused on the definiteness axiom though. For example if it's <v,v>, then it's easy to prove definiteness because it's the same variable; ==> v=0. But in this case, they're two different functions, and I'm not sure how to go about that. And also, is the method for proving the integral as an inner product the same way as proving <f(x), g(x)> is an inner product? Thanks for the help.

2. $\displaystyle <f(x), f(x)>=\int_0^1 f^2(x) dx$. As $\displaystyle f^2(x)>0$ for $\displaystyle f(x)\neq 0$, the integral is greater than $\displaystyle 0$. For $\displaystyle f(x)=0$ the integral is worth 0 so this condition of inner product is satisfied.
Yes, you have to show the 4 properties of inner product to conclude that it is one.

3. For bi-linearity:

Say that you're given <v,w> and in order to prove the bi-linearity axiom it would be <v1+v2, w> = <v1, w> + <v2, w>. What would you do in this case? Could you do <f(x) + 2f(x) , g(x)> = <f(x), g(x)> + <2f(x), g(x)>? Thanks.

4. Originally Posted by Fel
For bi-linearity:

Say that you're given <v,w> and in order to prove the bi-linearity axiom it would be <v1+v2, w> = <v1, w> + <v2, w>. What would you do in this case? Could you do <f(x) + 2f(x) , g(x)> = <f(x), g(x)> + <2f(x), g(x)>? Thanks.
In this case you have to take an arbitrary $\displaystyle v_2$, not $\displaystyle v_2=2_v1$ as you did.
So I'd pick up $\displaystyle <f(x)+g(x), h(x)>$. And I want to see that it's worth $\displaystyle <f(x),h(x)>+<g(x),h(x)>$ using the given function (inner product that is. But as we do not know if it is one until we prove it, I just called it function).
Can you show the relation?