Let $b$ be a non-extreme point in the unit ball of $H^\infty$. Let $(a,b)$ be a [corona pair](http://www.theta.ro/jot/archive/1986-016-001/1986-016-001-003.pdf), that is $|a|+|b|$ is bounded away from zero in the unit disc. Also let $|a|^2$ satisfy Hunt-Muckenhoupt condition that is 


$$\sup\_{I}|I|^{-2}\int\_I|a|^2d\theta\int\_I|a|^{-2}d\theta<\infty,$$
where $I$ ranges over all subarcs of $\partial \mathbb{D}.$ 


The question is: do we have some particular examples of such pairs?


Thanks in advance



An corona pair (a,b) satisfies (A_2)
You are probably already aware of the connection between these conditions and the deBranges-Rovnyak spaces $\mathcal H(b)$, in particular Chapter IX of Sarason's "Sub-Hardy Hilbert Spaces..." book. To construct explicit examples, one can use a theorem of Davis and McCarthy (Corollary 2.5 of "Multipliers of de Branges spaces", Michigan Math J. 38 (1991), no. 2, 225–240, MR1098860). Assume $b$ is non extreme and $a$ is the outer function with $|a|^2+|b|^2=1$ a.e. on the circle. Their result shows that the pair of conditions you give is equivalent to the single condition that $|a/(1-b)|^2$ is an $A\_2$ weight. By the posited relation between $a$ and $b$, this amounts to saying that $\frac{1-|b|^2}{|1-b|^2}$ is $A\_2$, and this can be used to construct examples: in particular let $w>0$ be any $A\_2$ weight on the circle satisfying 
$$\int\_{\mathbb T}\log w(\zeta)\, dm(\zeta)>-\infty
$$
and define $b$ by the Herglotz integral
$$
\frac{1+b(z)}{1-b(z)}=\int\_{\mathbb T}\frac{1+z\overline{\zeta}}{1-z\overline{\zeta}}w(\zeta)\, dm(\zeta).
$$
Then by Fatou's theorem we have $w=\frac{1-|b|^2}{|1-b|^2}$ a.e. on $\mathbb T$. By the log-integrability of $w$, $b$ will be non-extreme. Taking $a$ as above gives such a pair, and every pair with $a$ outer and $|a|^2+|b|^2=1$ arises this way. (Any $b$ with $\|b\|\_\infty <1$ gives a trivial example.)


*EDIT*: Actually, it is easy to see that an $A\_2$ weight $w$ must be log integrable, since if $w, 1/w$ are both in $L^1(\mathbb T)$ then so is $\log w$. Thus any choice of $w$ will work.


