Prove an inner product space 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?
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 $.
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 $\overline {\phi} (x)$].