Prove an inner product space

Question

Prove $(L^2[0,1],\|·\|_g)$ is an inner product space, where the norm of $\|·\|_g$ is given by $\|f\|_g=(\int_0^1 g(x)|f(x)|^2dx)^{\frac{1}{2}}$. I know that I need to prove this through proving it satisfies the four properties of inner product space, I just don't know how to express the inner product from the norm. (Consider it in the real space.)

Is $\langle f,g \rangle=\int_0^1 g(x)|f(x)||g(x)|dx$ the corresponding inner product?

Answer 1

The one you have defined is not an inner product because of the absolute values for $f$ and $g$. The absolute value sign destroys linearity.

You have to assume that $g \geq 0$ a.e. for this to be a norm. The inner product is given by $ \langle f, \phi \rangle =\int_0^{1} g(x) f(x) \phi (x) dx$.

[If you are considering complex valued functions change $\phi (x)$ to $\overline {\phi} (x)$].

Answer 2

The inner product is determined by the norm via the polarisation identity.

$\langle f, h \rangle = {1 \over 4} ( \|f+h\|^2 - \|f-h\|^2)$.

Hence $\langle f, h \rangle = {1 \over 4} ( \int g \cdot (f+h)^2 - \int g \cdot (f-h)^2 ) = \int g \cdot f \cdot h $.

Answer 3

You shouldn't use $g$ as the symbol for two functions, one the inner product's weight, the other an argument of the inner product.

Instead we want $\langle f,\,h\rangle=\int_0^1g(x)f^\ast(x)h(x)dx$ in the complex case, or $\langle f,\,h\rangle=\int_0^1g(x)f(x)h(x)dx$ in the real case.

This recovers the desired norm as a special case. Further, it's unique by the polarization identity, which @copperhat already gave in the real case. (In the complex case, it's $4\langle f,\,h\rangle=\sum_{n=0}^3i^{-n}\Vert f+i^nh\Vert_g$.)

You can now verify the inner product axioms require $g>0$ almost everywhere in $[0,\,1]$. (Merely $g\ge0$ isn't enough, because e.g. if $g=0$ on an interval of non-zero width we only have positive semidefiniteness.)