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Spring 2009 - Problem 12

change of variables, conformal mappings

Let $Q$ be the closed unit square in the complex plane $\C$ and let $R$ be the closed rectangle in $\C$ with vertices $\set{0, 2, i, 2+i}$ . Prove that there does not exist a surjective homeomorphism $\func{f}{Q}{R}$ that is conformal on the interior $Q^\itr$ and that maps corners to corners.

Solution.

Suppose $\func{f}{Q}{R}$ is such a mapping. Parametrize $\partial{Q}$ counter-clockwise by $\gamma$ . Since $f$ is continuous and injective, this means that $f \circ \gamma$ is a simple closed curve. Suppose $\p{f \circ \gamma} \cap R^{\itr} \neq \emptyset$ . Then by the Jordan curve theorem, this implies the existence of $w_1$ in the interior region of $f \circ \gamma$ and $w_2$ in the exterior region. But the interior and exterior regions are disconnected, which implies that $f\p{Q^{\itr}}$ is disconnected, which is impossible as it is the continuous image of a connected set. Thus, $f \circ \gamma \subseteq \partial{R}$ , and because $f$ maps corners to corners in an injective way, it follows that $f \circ \gamma = \partial{R}$ .

Since $f$ is injective, it traverses the vertices of $R$ in a clockwise or counter-clockwise way. By reversing $\gamma$ if necessary, we may assume that $f\p{0} = 0$ , $f\p{1} = 2$ , $f\p{1 + i} = 2 + i$ , and $f\p{i} = i$ . For $y \in \br{0, 1}$ consider the segment $C_y = \br{iy, 1 + iy}$ parameterized by $x \mapsto x + iy$ . Because of how $f \circ \gamma$ traverses $\partial{R}$ , it follows $f\p{iy}$ and $f\p{1 + iy}$ are on the sides of length $1$ of $R$ , and it follows that $\diam{f\p{C_\gamma}} \geq 2$ . Thus,

\diam{f\p{C_\gamma}} = \int_0^1 \abs{f'\p{x + iy}} \,\diff{x} \geq 2.

On the other hand, by Cauchy-Schwarz,

\begin{aligned} \int_0^1 \int_0^1 \abs{f'\p{x + iy}} \,\diff{x} \,\diff{y} &= \int_Q \abs{f'\p{x + iy}} \,\diff{A} \\ &\leq \p{\int_Q \abs{f'\p{x + iy}}^2 \,\diff{A}}^{1/2} \p{\int_Q \diff{A}}^{1/2} \\ &= \p{\int_Q \abs{f'\p{x + iy}}^2 \,\diff{A}}^{1/2} \p{\int_Q \diff{A}}^{1/2}. \end{aligned}

The Jacobian of $f = u + iv$ as a function $\R^2 \to \R^2$ is given by

\diff{f} = \begin{pmatrix} u_x & u_y \\ v_x & v_y \end{pmatrix}.

Combined with the Cauchy-Riemann equations, this has determinant

u_xv_y - v_xu_y = \p{u_x}^2 + \p{v_x^2} = \abs{\pder{f}{x}}^2 = \abs{f'}^2.

Thus, by the change of variables formula,

\int_Q \abs{f'\p{x + iy}}^2 \,\diff{A} = \int_{f\p{Q}} \,\diff{A} = \area{R}.

Thus,

\begin{aligned} 2 &\leq \int_0^1 \int_0^1 \abs{f'\p{x + iy}} \,\diff{x} \,\diff{y} \\ &\leq \p{\area{R}}^{1/2} \p{\area{Q}}^{1/2} \\ &= \sqrt{2}, \end{aligned}

which is impossible. Thus, no such $f$ exists, which completes the proof.