In Bushnell and Henniart's The Local Langland's Conjecture for GL(2) they define a right Haar integral on a locally profinite group $G$ as being a non-zero linear functional $$ I: C^{\infty}_{c}(G) \to \mathbb{C}, $$ where the space on the left is the space of all locally constant complex valued functions with compact support on $G$, that satisfies the conditions:
- $I(\rho_{g}f) = I(f)$ where $\rho_{g}$ is right translation by $g \in G$, $f \in C^{\infty}_{c}(G)$.
- $I(f) \geq 0$ for $f \geq 0$.
I assume the ordering in the second condition implies that this condition applies to only functions taking real values (otherwise what could it possibly mean), but does this mean that a Haar integral is required to take real valued functions to real numbers, or is it the case that a linear functional on this space must take real values on real functions (possibly conditional on (1))?
cross-post: MathOverflow
The notation $\geq 0$ makes perfect sense for complex numbers: a complex number $z$ satisfies $z\geq 0$ if $z$ happens to be a nonnegative real number. So condition (2) means that if $f$ happens to take nonnegative real values everywhere, then $I(f)$ is a nonnegative real number.
Nothing in the definition explicitly assumes that the integral of a real-valued function must be real. However, this is a consequence of property (2). Indeed, if $f\in C_c^\infty(G)$ is real-valued, let $g=\max(f,0)$ and $h=\max(-f,0)$. Then $g,h\in C_c^\infty(G)$ and $g,h\geq 0$, so $I(g),I(h)\geq 0$. But $f=g-h$, so $I(f)=I(g)-I(h)$ is real.