I'm trying to show that

if $f=u+iv$ is continuous on the closed unit disk and holomorphic on the open unit disc and $u=v^2$ on the unit circle, then $f$ is constant.

I was thinking to apply maximum and minimum principle for harmonic functions to $v^2-u$ but $v^2-u$ is not necessarily harmonic (it is subharmonic).

I also thought about constructing a conformal map on the image of f that sends the parabola $x=y^2$ to the real line but I'm not sure if one exists.

  • $\begingroup$ Do you know the open mapping theorem? $\endgroup$ – Micah Apr 13 '13 at 4:57
  • $\begingroup$ I think you can use Cauchy's theorem. The integral around the boundary circle must be zero (the theorem works in the case of an integral along the boundary when you have continuous extension -- see the wikipeda page on the theorem). Since the integral must be zero, and the real part is $\ge 0$, we must have $v=0$, from which we see $f$ must be constant by the Cauchy-Riemann equations. $\endgroup$ – Potato Apr 13 '13 at 4:58
  • $\begingroup$ @Potato: The OMT tells us that the interior of the disc is sent to the interior of its image, so the boundary of the image is contained in the parabola $x=y^2$. Now, apply compactness... $\endgroup$ – Micah Apr 13 '13 at 5:09
  • $\begingroup$ @Micah I agree the boundary of the image is in the parabola. I'm not able to extrapolate what you mean by apply compactness ... $\endgroup$ – Digital Gal Apr 13 '13 at 5:17
  • $\begingroup$ @Micah How do we know the boundary of the image is the image of the boundary of the circle? $\endgroup$ – Potato Apr 13 '13 at 5:23

Let $U$ be the open unit disk. The function $$g: \mathbb{C} \rightarrow \mathbb{R}$$ $$g(x + i y) = x - y^2$$ is open. If $f$ is not constant then by the open mapping theorem $g \circ f(U)$ is open. The extreme values of $g \circ f$ on the compact set $\overline{U}$ are therefore attained on the unit circle. Since the minimum is strictly less than the maximum $g \circ f$ is not constant on the unit circle. A contradiction.


Let $D$ and $P$ as above in the solution of Micah.

Claim: $f(D)\subseteq P$.

Suppose there exists a $z_0\in D$ such that $f(z_0)\notin P$. Let's give a unit parametrization $\gamma(t)=e^{2\pi it}$ to unit circle. Then $\sigma(t)=f(\gamma(t))$ is a closed curve contained in P. Note that $n(\sigma,f(z_0))=0$, where $n(,)$ denotes the index of a point with respect to a curve. Now

$n(\sigma,f(z_0))=\Sigma_{j=1}^{n}n(\gamma, z_j)$, where $z_j$'s are the preimages of $f(z_0)$.

The RHS is a positive integer greater than $1$, and hence a contradiction. So the claim is true and hence $f$ has to be constant.

Now the only problem is that the statement $n(\sigma,f(z_0))=\Sigma_{j=1}^{n}n(\gamma, z_j)$ is true when $\gamma$ is a curve contained in a domain $\Omega$ and $f$ is holomorphic on $\Omega$. But as "it turns out for a function $f$ holo. in the interior of $D$ and cont. on D, this property is true."

Remark: Observe that there is nothing special in the Parabola here, one can choose any curve $\sigma$ such that $\mathbb{C}\setminus<\sigma>$ does not have any bounded component.


Let $D$ denote the closed unit disc; let $P=\{x+iy|x=y^2\}$. If $f(D) \subset P$, then $f(D)$ cannot be open. So $f$ is constant by the open mapping theorem.

On the other hand, suppose there is some $z_0 \in D$ with $f(z_0) \not\in P$. Notice that $P$ divides $\Bbb{C}$ into two non-compact path-components. On the other hand, $f(D)$ is compact (since $D$ is). So there must be some $w \not\in f(D)$ contained in the same path-component as $f(z_0)$.

Take a path joining $f(z_0)$ to $w$ which does not intersect $P$. Then it must intersect the boundary of $f(D)$. Since $f(D)$ is compact, it is closed. So there is some $z_1$ such that $f(z_1)$ is contained in the boundary of $f(D)$ and not in $P$.

Since $f(z_1) \not\in P$, $z_1$ is not on the unit circle. Thus we must have $|z_1|<1$, so there is some open $U$ with $z_1 \in U \subset D$. By the open mapping theorem, if $f$ is non-constant, $f(U)$ is a $\Bbb{C}$-open subset of $f(D)$ containing $f(z_1)$. But $f(z_1)$ lies on the boundary of $f(D)$, so no such set exists. Thus $f$ must be constant.

  • $\begingroup$ "Take a path joining $f(z_0)$ to $w$...it must intersect the boundary of $f(D)$." This seems obvious, but how to prove it? $\endgroup$ – Potato Apr 13 '13 at 5:59
  • 1
    $\begingroup$ @Potato: Let $\gamma:[0,1] \to \Bbb{C}$ be such a path, and let $t^*=\sup \{t|\gamma(t) \in f(D)\}$. Then $\gamma(t^*)$ is an element of the boundary of $f(D)$: for any $\epsilon > 0$, $\gamma(t^*+\epsilon) \not\in f(D)$, $\gamma([t^*-\epsilon, t^*))$ intersects $f(D)$, and $\gamma$ is continuous. $\endgroup$ – Micah Apr 13 '13 at 6:04

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