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Sorry I don't know how to give an appropriate title.

In the complex plane, suppose there is a graph $x+if(x)$ separating the plane into two unbounded components, where $f(x)$ is smooth and bounded, with all derivatives bounded. $f(x)$ could oscillate at $\infty$. Let $\Omega$ denote the region under the graph.

By Riemann mapping theorem, there is a Riemann mapping $\Phi: \Omega\rightarrow \mathbb{P}_-$, where $\mathbb{P}_-$ is the lower half plane.

My question is, is it possible to find a Riemann mapping $\Phi$ such that $\Phi$ grows almost like the identity, in the sense that $\Phi(z)-z$ is bounded?

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This is not a complete answer, just as suggestion: move things over to the unit disk, with the Cayley transform $$ \varphi(z)=\frac{z+i}{z-i} . $$ Then $\Phi$ will be as desired precisely if $F=\varphi\Phi\varphi^{-1}$ satisfies $F(w)=1+O(w-1)$ near $w=1$.

The boundary behavior of Riemann maps has of course been studied extensively, though the standard results (see for example the second volume of Conway's book for this) don't seem to answer your question immediately. The boundary curve of your region is $C^{\infty}$, except at $w=1$. It is given by $$ \gamma(s) = 1 - \frac{2is}{1+is(f(1/s)+1)} ; $$ I've changed the parameter to $s=1/x$. This is still differentiable at $s=0$ also, but the derivative may be discontinuous. We do obtain a continuous (up to the boundary) Riemann map, but of course that is less than what you are asking for.

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  • $\begingroup$ Thanks. I tried but not get the desired result. Maybe the functions I am considering is two general. $\endgroup$
    – qingtang
    Commented Feb 19, 2016 at 18:04
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You can always find a map $\Phi$ (unique up to shifts) such that $\Im\mathfrak{m}(\Phi(z)-z)$ is bounded, but, in general, $\Re\mathfrak{e}(\Phi(z)-z)$ may be unbounded.

Let $h$ be the bounded harmonic function in $\Omega$ satisfying the boundary conditions $h(z)=\Im\mathfrak{m} z$, $z\in\partial\Omega$ (such a function can be constructed, e. g., as the exit expectation of the Brownian motion). The function $g(z)=\Im\mathfrak{m}z-h(z)$ is zero on the boundary, and, by maximum principle, is negative. Therefore, if $\varphi$ is a conformal map from the unit disc to $\Omega$, then $g(\varphi(z))$ is a multiple of the Poisson kernel in the disc. From this, you can see that if $\tilde{g}$ denotes the harmonic conjugate of $g$, then $\tilde{g}|_{\partial\Omega}$ is increasing and tends to $\pm \infty$ as $\Re\mathfrak{e}z\to\pm\infty$. The argument principle shows that $\tilde{g}+ig$ is a conformal map.

Observe, however, that if $f$ is a smoothed step function, then the normal derivative of $h$ (which is also the tangential derivative of $\tilde{h}$) behaves as $\Re\mathfrak{e}(const/z)$ as $\Re\mathfrak{e}z$ tends to infinity, so $\tilde{h}$ grows logarithmically. This is the worst possible behaviour.

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  • $\begingroup$ @Kostya_l Thanks. The domain is unbounded, the maximum principle might not be true in general, so can we really guarantee that $g(z)$ negative? $\endgroup$
    – qingtang
    Commented Feb 19, 2016 at 18:06
  • $\begingroup$ yep. If $\psi(z)$ is a conformal mapping from the upper half-plane to $\Omega$ that maps $\mathbb{R}$ to $\partial\Omega$, then $g(\psi(z))$ is a harmonic function in the upper half-plane which is bounded from above and vanishes on the real line. The only such function, up to scaling, is $-\mathfrak{Im}z$ $\endgroup$
    – Kostya_I
    Commented Feb 19, 2016 at 22:35
  • $\begingroup$ @qingtang Alternatively, you can give yourself an "epsilon of room" and apply the maximum principle to $g-\epsilon\log|z-z_0|$ for some $g_0\notin\Omega$ and send $\epsilon\to0$. $\endgroup$
    – Fan Zheng
    Commented May 28, 2018 at 19:06

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