MATHEMATICS

P. P. BELINSKII

ON THE AREA MEASURE UNDER A QUASICONFORMAL MAPPING

(Presented by Academician M. A. Lavrent'ev on 6 III 1958)

It is known that under a \(q\)-quasiconformal mapping with small \(1 - 1/q = \varepsilon\), the mapping is close to a conformal one, and, under the corresponding normalization, close to the identity mapping \((^1,^2)\). However, up to now estimates of the change of area under such a mapping have been unknown. In the present note an estimate is given for the change of area of the form \(|\Delta \sigma| \leq \sqrt{\varepsilon} + O(\varepsilon)\) for \(q\)-quasiconformal mappings. Along the way, the fact is once again proved that under quasiconformal mappings, sets of measure zero go into sets of measure zero \((^3)\).

Let the rectangle \(K: 0 < x < \lambda,\ 0 < y < 1/\lambda\) of the plane \(z = x + iy\) be mapped \(q\)-quasiconformally onto the rectangle \(0 < u < \mu,\ 0 < v < 1/\mu\) of the plane \(w = u + iv\), in such a way that the vertices go into the corresponding vertices. Let us compute \(\operatorname{Im} \int_{\Gamma} w(z)\,dz\), where \(\Gamma\) is the boundary of the rectangle \(K\):

\[ \operatorname{Im}\int_{\Gamma} w(z)\,dz = \int_{\Gamma} u\,dy + v\,dx = \int_0^{1/\lambda} \mu\,dy - \int_0^\lambda \frac{1}{\mu}\,dx = \frac{\mu}{\lambda} - \frac{\lambda}{\mu}. \]

On the other hand:

\[ \left|\int_{\Gamma} w(z)\,dz\right| \leq \iint_K |-(u_y + v_x) + i(u_x - v_y)|\,dx\,dy \leq \left(\sqrt{q} - \frac{1}{\sqrt{q}}\right) \iint_K \sqrt{\frac{D(u,v)}{D(x,y)}}\,dx\,dy . \]

Comparing with the preceding, we obtain

\[ \left|\frac{\mu}{\lambda} - \frac{\lambda}{\mu}\right| \leq \left(\sqrt{q} - \frac{1}{\sqrt{q}}\right) \iint_K \sqrt{\frac{D(u,v)}{D(x,y)}}\,dx\,dy . \tag{1} \]

Now let, for definiteness, \(\mu \geq \lambda\). Subject the plane \(w\) to a compression in the direction of the \(v\)-axis by \(t \geq 1\) times and to a stretching along the \(u\)-axis by the same factor (in the case \(\mu < \lambda\) we interchange \(u\) and \(v\)). In this case there will be obtained a mapping of the rectangle \(K\) onto a rectangle with sides \(t\mu\) and \(1/t\mu\). The characteristic \(p(z)\), obviously, will not exceed \(t^2 q\), and the Jacobian will not change. Then, according to (1):

\[ \left|\frac{t\mu}{\lambda} - \frac{\lambda}{t\mu}\right| \leq \left(t\sqrt{q} - \frac{1}{t\sqrt{q}}\right) \iint_K \sqrt{\frac{D(u,v)}{D(x,y)}}\,dx\,dy . \]

Dividing both sides of the obtained inequality by \(t\), passing to the limit as \(t \to \infty\), and taking into account that \(\mu \geq \lambda\), we obtain

\[ \frac{1}{\sqrt{q}} \leq \iint_K \sqrt{\frac{D(u,v)}{D(x,y)}}\,dx\,dy . \tag{2} \]

We divide the rectangle \(K\) into two subsets \(S_1\) and \(S_2\). Applying Bunyakovsky’s inequality to (2), we obtain

\[ \frac{1}{\sqrt{q}} \leq \left[ \operatorname{mes} S_1 \cdot \iint_{S_1} \frac{D(u,v)}{D(x,y)}\, dx\, dy \right]^{1/2} + \left[ \operatorname{mes} S_2 \cdot \iint_{S_2} \frac{D(u,v)}{D(x,y)}\, dx\, dy \right]^{1/2}. \tag{3} \]

From the inequality obtained one can derive yet another proof of the fact that a mapping from the closure of the class of continuously differentiable \(q\)-quasiconformal mappings carries a set of measure zero into a set of measure zero.

To prove this, let us first note that inequality (3) remains valid, since it is known that Green’s formula is legitimately applicable in the case of a mapping belonging to the closure of the class of \(q\)-quasiconformal mappings.

Denote the images of the sets \(S_1\) and \(S_2\) by \(S'_1\) and \(S'_2\). It is clear that

\[ \operatorname{mes} S'_2 \geq \iint_{S_2} \frac{D(u,v)}{D(x,y)}\, dx\, dy . \]

If \(\operatorname{mes} S_1 = 0\), then from (3) it follows that

\[ \operatorname{mes} S'_1 = 1-\operatorname{mes} S'_2 \leq 1-\iint_{S_2} \frac{D(u,v)}{D(x,y)}\, dx\, dy \leq 1-\frac{1}{q}. \tag{4} \]

If we assume that \(\operatorname{mes} S'_1 > 0\), then, considering the mapping in a neighborhood of some point of density of \(S'_1\), it is possible, by means of auxiliary conformal transformations, to obtain a mapping carrying a set of measure zero into a set whose measure is arbitrarily close to one, which contradicts (4).

Denote \(\operatorname{mes} S_1\) by \(\sigma\), and \(\operatorname{mes} S'_1\) by \(\sigma+\Delta\sigma\). Then (3) gives

\[ \frac{1}{\sqrt{q}} \leq \sqrt{\sigma(\sigma+\Delta\sigma)} + \sqrt{(1-\sigma)(1-\sigma-\Delta\sigma)} . \tag{3'} \]

Solving the last inequality with respect to \(\Delta\sigma\) and denoting \(1-1/q=\varepsilon\), we obtain

\[ |\Delta\sigma| \leq \varepsilon |1-2\sigma| + 2\sqrt{\sigma(1-\sigma)}\sqrt{\varepsilon+\varepsilon^2}, \tag{5} \]

or, for small \(\varepsilon\),

\[ |\Delta\sigma| \leq \sqrt{\varepsilon}+O(\varepsilon). \tag{5'} \]

Inequality (5′) is easily transferred to the case of a mapping of the disk onto itself with the normalization \(w(0)=0\).

Mathematical Institute
Siberian Branch of the Academy of Sciences of the USSR

Received
15 II 1958

CITED LITERATURE

\(^{1}\) M. A. Lavrent’ev, Matem. sborn., 42, issue 2 (1935).
\(^{2}\) P. P. Belinskii, DAN, 91, No. 2 (1953).
\(^{3}\) I. P. Pesin, DAN, 102, No. 2 (1955).