# Integral and conformal mappings II

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?

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.

• If $D_n$ is smooth (for example $C^2$), then $|f'(z)|\le C_n$, so why the integral diverges? I am assuming that $D_n$ are images of $n/(n+1) D$ under a $C^2-$$K$ q.c. diffeomorphic mapping of the unit disk onto itself. – user36162 Nov 8 '13 at 18:36
• I think user36162 is right. I don't see why $I_n$ would diverge since $f_n'$ must be bounded in $D$. – Malik Younsi Nov 8 '13 at 19:15
• If one wants $D_n$ to be smooth, it is sufficient to consider Alexandre's sequence a bit rescaled: $f(r_nz)$ with $r_n<1$ but so close to $1$ that the corresponding integral is e.g. greater than $n$. – Pietro Majer Nov 8 '13 at 19:57
• Probably it is a correct construction. – user36162 Nov 8 '13 at 20:17
• In my example, the boundary of $G=D_n$ is not smooth at one point, namely $b$. For smooth boundary, the integral will of course converge. But can be made arbitrarily large by approximating my $G$ with smooth regions. – Alexandre Eremenko Nov 8 '13 at 20:28