Lusin area integral formula

[Tolaso J Kos]

Let $f:\mathbb{C} \rightarrow \mathbb{C}$ be an analytic and $1-1$ function and let $\mathbb{D}$ be the open unitary disk. Prove that:

$$\iint \limits_{\mathbb{D}} \left |f'(z) \right| \, {\rm d}z = \text{area}(f \left(\mathbb{D} \right))$$

[Papapetros Vaggelis]

Hello Tolaso.

Here is a solution. I am not so sure about the given solution, so, what's your opinion ?

For the last equation \(\displaystyle{\left(\ast\right)}\), check change of variables

We consider that \(\displaystyle{\mathbb{C}\simeq \mathbb{R}^2}\) and we can write :

\(\displaystyle{f:\mathbb{R}^2\longrightarrow \mathbb{R}^2\,,(x,y)\mapsto \left(u(x,y),v(x,y)\right)}\) .

The functions \(\displaystyle{u\,,v:\mathbb{R}^2\longrightarrow \mathbb{R}}\) are infinitely many times

differentiable since the function \(\displaystyle{f}\) is analytic.

Also, the functions \(\displaystyle{u\,,v}\) satisfy the \(\displaystyle{\rm{Cauchy-Riemann}}\) equations, that is:
$$u_{x}(x,y)=v_{y}(x,y)\,\,,u_{y}(x,y)=-v_{x}(x,y)\,,\left(x,y\right)\in\mathbb{R}^2$$

Now, \(\displaystyle{f^\prime(x,y)=\begin{pmatrix}
u_{x}(x,y) & u_{y}(x,y)\\
v_{x}(x,y) & v_{y}(x,y)
\end{pmatrix}\,,\left(x,y\right)\in\mathbb{R}^2}\) .

with $$\det(f^\prime(x,y))=(u_{x}\,v_{y}-v_{x}\,u_{y})(x,y)=u_{x}^2(x,y)+u_{y}^2(x,y)\neq 0\,,\forall\,\left(x,y\right)\in\mathbb{R}^2$$

Indeed, suppose that \(\displaystyle{u_{x}(x_0,y_0)=v_{y}(x_0,y_0)=u_{y}(x_0,y_0)=v_{y}(x_0,y_0)}\)

for some \(\displaystyle{\left(x_0,y_0\right)\in\mathbb{R}^2}\) . Then, \(\displaystyle{f^\prime(x_0,y_0)=\mathbb{O}}\)

and the function \(\displaystyle{f}\) is constant on an open ball which contains \(\displaystyle{\left(x_0,y_0\right)}\),

a contradiction, since the function \(\displaystyle{f}\) is one to one.

Now,

$$\begin{aligned} \rm{area}f(D)&=\int_{f(D)}\mathrm{d}u\land \mathrm{d}v\\&=\int_{D}\left|\rm{det}(f^\prime(x,y))\right|\,\mathrm{d}x\,\land \mathrm{d}y\\&=\int_{D}\left|f^\prime(z)\right|\,\mathrm{d}z\end{aligned}\,\,(\ast)$$

Edit by T: Updated link!