Showing that $\int_{0}^{1} f\, \overline{g} \,dx$ isn't an inner product on $C[0,2]$.

Question

My reasoning is just that the inner products on $C[0,2]$ are defined to be the integrals from $0$ to $2$… and the one in the problem is from $0$ to $1$, so it can’t be an inner product based on that. Is this reasoning too simplistic or even incorrect? I don’t see any other way to show that it’s not an inner product on $C[0,2]$.

Answer 1

Your reasoning boils down to the following argument: "The inner products on function spaces I've seen are of the form X. This is not of the form X, so it is not an inner product on a function space."

Generalizing, you are essentially saying: "I'm used to <type of object> looking like X. So all <type of object>s must look like X".

This kind of reasoning is called inductive, and it is essential for the sciences, where the rules of our universe are carved in stone, but we don't know them and we try to find or at least approximate those rules by making many observations and then assuming that what was true for all of our observations until now will probably be true in general.

This is not how reasoning in math works. Mathematical reasoning is generally deductive, that is, we know the rules because we set them ourselves, and now we want to find out what we can do with those rules.

That is, in maths you have to work with the rules that we set ourselves. In this case the rules are: Inner products are sesquilinear, symmetric and positive definite. You have to appeal to these rules, not your previous experiences. Your previous experiences may inform you about how to work with the rules, but you can't appeal to them when making your final argument.

Here specifically, it is a basic exercise to prove that the given operation is not positive definite: Try to find a nonzero functions $f$ such that $\int_0^1f(x)\bar f(x)\mathrm dx=0$. Use the fact that the integral from 0 to 1 doesn't depend on how the function behaves on $(1,2)$.