Is $\langle f,g\rangle = ∫(f^*g)\,d\mu$ an inner product? Does the equality $\|f\|_2^2+\|g\|_2^2=\|f+g\|_2^2+\|f−g\|_2^2$ hold? Let $(X,A,\mu)$ be a measure space and $L^2$ the space of all measurable functions with $∫f^2d\mu < \infty$.
Is $⟨f,g⟩ = ∫(f^*g)\,d\mu$ an inner product?
I can show that $⟨f,g⟩=⟨g,f⟩$ and $⟨cf,g⟩=c⟨f,g⟩$
How can I find an example where $⟨f,f⟩=0$, but $f$ is not the zero function?
I'm not sure with the equality $⟨f_1+f_2,g⟩ = ⟨f_1,g⟩+⟨f_2,g⟩$ either.
Does the equality $\|f\|_2^2+\|g\|_2^2=\|f+g\|_2^2+\|f−g\|_2^2$ still hold?
 A: You can't show that $\langle g,f\rangle=\langle f,g\rangle$, because it's false; what's true is that
$$
\langle g,f\rangle=\langle f,g\rangle^*
$$
(where the asterisk denotes complex conjugation). This is obvious from the definition, just observe that
$$
\int h^*\,d\mu=\left(\int h\,d\mu\right)^{\!*}
$$
It is a standard fact that if $f,g\in L^2$, then $fg\in L^1$ and $f^*\in L^2$, which should be proved in advance in order for $\langle f,g\rangle$ to be defined.
Finally it depends on what you mean by $L^2$. If functions that are equal almost everywhere with respect to $\mu$ are identified, then you indeed have an inner product: the linearity properties are easy to verify and
$$
\langle f,f\rangle=\int f^*f\,d\mu=\int|f|^2\,d\mu=0
$$
implies $|f|^2=0\text{ a.e.}[\mu]$, so also $f=0\text{ a.e.}[\mu]$.
If instead you think to $L^2$ without identification of functions equal almost everywhere, a counterexample is any nonzero function which is however zero almost everywhere.
The polarization identity holds for every inner product.
