Assume that $f\_n$ is a sequence of conformal injective mappings of the unit disk $D$ onto the nested smooth Jordan domains $D\_n\subset D$, such that $\cup\_{n=1}^\infty D\_n=D$ and $D\_n$ are images of $n/(n+1) D$ under a fixed diffeo q.c. mapping of the unit disk onto itself. Assume as well that $f\_n\to id$ uniformly (on D). Is the following sequence $$I\_n=\int\_D\frac{|f\_n'(z)|^2}{|1-z|} dxdy$$ bounded?



Integral and conformal mappings II
The answer is no. All these integrals can be infinite.
Let us fix $a,b$, $0<a<b<1$.


I will first construct a Jordan region $G$
containing $|z|<a$, contained in $|z|<b$, and such that
for $f$ mapping conformally $D$ onto $G$, the integral is infinite.
Consider a smooth (except at the endpoint $b$) simple curve beginning at $a$,
ending at $b$, contained
in $a<|z|<b$ except the ends, and having infinite length.
There is no problem with constructing such a curve.
In addition I will require that the diameter of this curve is less than $b-a$.
Then you take a small "finger" around this curve, and add it to $|z|<a$,
to obtain a Jordan region $G$, whose boundary
is smooth everywhere except the point
$b$, and having the property that every curve in $G$ from the origin to $b$
has infinite length.


Now let $f$ map $D$ onto $G$ conformally and $f(1)=b$.
The $f$-image of every segment in $D$ which ends at $b$ has infinite
length.
This implies that
$$\int\_D\frac{|f'(z)|}{|1-z|}dxdy=\infty.$$
To see this, switch to polar coordinates with the origin at $1$.
In these polar coordinates the part of the integral near $1$ will
be
$$\int\_{0}^{2\pi}d\theta\int\_0^\epsilon\frac{|f'(re^{i\theta})|}{r}rdr.$$
The inner integral is the length of the image of a segment terminating at $1$. So your integral also diverges. Of course my region $G$ is not smooth at one point ($b$),
and for a smooth region the intergal converges. But by approximating my $G$ with a smooth
region $D\_n$ you can make this integral as large as you wish. 


It remains to make sure that this function $f$ is uniformly close to
the identity, but this is because the finger we added has small diameter.


