How do the inner products on $L^2$ look like?

Question

I was wondering whether all scalar products in $(L^2[0,1],\lambda)$ are given by $\langle f,g \rangle := \int f(x)g(x) \cdot w(x) d\lambda(x)$? If this is true, what are the exact conditions that we have on $w$? Probably Frèchet-Riesz could help here.

Answer 1

No, this is not true. There are plenty more scalar products. Think of $\mathbb{R}^n$. Then, any symmetric, positive definite matrix defines a scalar product. Your scalar products only corresponds to the diagonal matrices.

On $(0,1)$ with the Lebesgue measure, also $$(f,g) = \int f(x) \, g(x) dx + \int f(x) \, dx \, \int g(x) \, dx$$ is a scalar product, which is not of your form. More generally, take a Hilbert space $X$ and a bounded linear operator $A : L^2(0,1) \to X$ and define $$(f,g) = \int f(x) \, g(x) \, dx + (Af, Ag)_X.$$