I was told a heuristic for thinking about the $A_p$ characteristic of a weight is that the closer the characteristic is to $1$, the closer the weight is to being flat, i.e. constant.
To gain an intuition for this heuristic, I went through a few examples. I was able to show that for any simple function/weight $w$ which is nonzero a.e., $[w]_{A_p} \geq 1$ with equality if and only if $w$ is constant a.e. However I'm having trouble generalizing this to arbitrary weights, in particular the "equality if and only if" part.
I've been able to reduce this to the case of when $p=2$. So how can I show that $[w]_{A_2} = 1$ implies $w$ is constant a.e.? Is it possible to bootstrap from the case of simple functions?
Quibble: The terminology "$f$ is constant a.e" doesn't quite make sense. It says that the set of $x$ such that $f(x)$ is not constant has measure zero, but for a given $x$ saying "$f(x)$ is constant" is problematic.
Of course it's clear that you mean "there exists $c$ such that $f=c$ a.e.". The same issue arises with the definition of $L^\infty$: by definition $f\in L^\infty$ if there exists $c$ such that $|f|\le c$ a.e. Saying "$f$ is bounded a.e." doesn't quite make sense; this is why people say "$f$ is essentially bounded" instead.
So it seems to me we should say
OK, now for the math:
Evidently $w>0$, and $$[w]_{A_2}=\sup_Q\left(\frac1{|Q|}\int_Q w\right) \left(\frac1{|Q|}\int_Q \frac1w\right).$$
Of course Cauchy-Schwarz shows that $[w]_{A_2}\ge1$. So if $[w]_{A_2}=1$, we have equality in Cauchy-Schwarz for every $Q$, so for every $Q$ the condition for equality in Cauchy-Schwarz shows there exists $c_Q$ such that $$\frac1w=c_Qw$$almost everywhere on $Q$. It follows easily that in fact $\frac1w=cw$ almost everywhere, so $w$ is essentially constant.
Edit You say you've reduced the general case to $p=2$. It's not clear to me how that reduction would go, but in fact the same argument must work for $1<p<\infty$, starting with an application of Holder's inequality to $\int fg$< where $f=w^{1/p}$ and $g=w^{-1/p}$.