The uniform slice estimate

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

Let \(F:\mathbb D\to D\) and its inverse \(g\) be the maps of A planar conformal map. For an integer \(k\ge2\) and \(v+it\in D\), put \[ \begin{aligned} J_k(v,t)=k^2\int_0^v&|g(h+it)|^{2k-2}\\ &\cdot|g'(h+it)|^2\,dh , \end{aligned} \] the integral along the horizontal segment from \(it\) to \(v+it\), which lies in \(D\) by Theorem 5.2(d). These signed integrals will be momentum coordinates in the last lesson. This lesson proves that \(|J_k|/k\) is at most \(\frac\pi4|g|^{2k}\) plus an error that tends to \(0\) uniformly on \(D\) (Proposition 6.1). The constant \(\pi/4\) comes from the bound \(A<2/\pi\) of Theorem 5.2(b); the difficulty is near the tips \(\pm i\) of \(D\), where the slices are short and the height may approach \(\pm1\) as \(k\) grows. The estimate is OpenAI's [OpenAI-SB, Proposition 3.2].

1. The statement

Proposition 6.1. Each \(J_k\) is smooth on \(D\), odd in \(v\), and \[ \begin{aligned} &\partial_vJ_k(v,t)\\ &\quad=k^2|g(v+it)|^{2k-2}|g'(v+it)|^2\ge0. \end{aligned} \] There are numbers \(\varepsilon_k\ge0\), depending only on \(F\), with \(\varepsilon_k\to0\), such that for every \(k\ge2\) and every \(v+it\in D\), \[ \frac{|J_k(v,t)|}k\le\frac\pi4\,|g(v+it)|^{2k}+\varepsilon_k . \]

2. Two elementary bounds

Lemma 6.2. For every \(X>0\), \(\dfrac1{\arctan X}-\dfrac2\pi\le\dfrac2X\).

Proof. For \(0<X\le1\), concavity of \(\arctan\) on \([0,1]\) gives \(\arctan X\ge\frac\pi4X\ge\frac X2\), so \(1/\arctan X\le2/X\). For \(X\ge1\), \(\frac\pi2-\arctan X=\arctan\frac1X\le\frac1X\) and \(\arctan X\ge\frac\pi4\), so \[ \begin{aligned} \frac1{\arctan X}-\frac2\pi&=\frac{\frac\pi2-\arctan X}{\frac\pi2\arctan X}\\ &\le\frac{8}{\pi^2X}\le\frac2X.\qquad\square \end{aligned} \]

Lemma 6.3. For integers \(k\ge1\) and \(0\le r_a<1\), \[ \begin{aligned} I(k,r_a)&=k\int_{r_a}^1r^{2k-1}\sqrt{\frac{1-r}{r-r_a}}\,dr\\ &\le1+\frac\pi e . \end{aligned} \]

Proof. Substitute \(y=k(1-r)\) and put \(H=k(1-r_a)\); then \[ \begin{aligned} &I(k,r_a)\\ &\quad=\int_0^H\Bigl(1-\frac yk\Bigr)^{2k-1}\sqrt{\frac y{H-y}}\,dy\\ &\quad\le\int_0^He^{-y}\sqrt{\frac y{H-y}}\,dy, \end{aligned} \] because \(0\le1-y/k\le1\), \(2k-1\ge k\) and \(\log(1-x)\le-x\). On \([0,H/2]\) the square root is at most \(1\), which contributes at most \(\int_0^\infty e^{-y}dy=1\). On \([H/2,H]\) bound \(e^{-y}\) by \(e^{-H/2}\) and use \(\int_0^H\sqrt{y/(H-y)}\,dy=\frac{\pi H}2\) (substitute \(y=H\sin^2\phi\)); this contributes at most \(\frac{\pi H}2e^{-H/2}\le\frac\pi e\), since \(xe^{-x}\le e^{-1}\). \(\square\)

3. The proof

Step 1: smoothness and the radial form. For fixed \(v+it\in D\), the segment from \(it\) to \(v+it\) is a compact subset of the open set \(D\), and so are the segments for all nearby endpoints. Writing \[ \begin{aligned} J_k(v,t)=k^2v\int_0^1&|g(av+it)|^{2k-2}\\ &\cdot|g'(av+it)|^2\,da, \end{aligned} \] the integrand is smooth in \((a,v,t)\), because \(|g|^{2k-2}=(g\bar g)^{k-1}\) with \(k\) an integer; differentiation under the integral sign (core course Measure and Integration; Fremlin, Measure Theory, Volume 1, 123D) shows that \(J_k\) is smooth, and the fundamental theorem of calculus gives \(\partial_vJ_k\). By Theorem 5.2(d), \(g(-h+it)=-\overline{g(h+it)}\); differentiating \(F(-\bar w)=-\overline{F(w)}\) gives \(|F'(-\bar w)|=|F'(w)|\), hence \(|g'(-h+it)|=|g'(h+it)|\) by \(g'=1/F'(g)\). So the integrand is even in \(h\), and \(J_k\) is odd in \(v\). It suffices to treat \(v>0\).

Fix \(v>0\) with \(v+it\in D\), and put \(R=|g(v+it)|\) and \(r_0=T^{-1}(|t|)\). By Theorem 5.2(e), the positive half of the slice at height \(t\) is the image of the arc \(r\mapsto re^{i\theta(r,t)}\), \(r_0<r<1\), with \(h=v(r,t)\); along it \(|g|=r\), \(|g'|=1/|F'(w)|\), and \(dh=\frac{|wF'(w)|^2}{rA}\,dr\) with \(A=A(r,\theta(r,t))\). Since \(|wF'(w)|^2=r^2|F'(w)|^2\), \[ |g|^{2k-2}|g'|^2\,dh=\frac{r^{2k-1}}A\,dr, \] \[ \begin{aligned} &\frac{J_k(v,t)}k\\ &\quad=k\int_{r_0}^R\frac{r^{2k-1}}{A(r,\theta(r,t))}\,dr. \end{aligned} \tag{6.1} \] The change of variables is made on segments \([h_1,v]\) with \(h_1>0\) and then \(h_1\downarrow0\); both integrands are nonnegative and the limit is the same. Since \(0<A<2/\pi\), write \(\frac1A=\frac\pi2+\bigl(\frac1A-\frac\pi2\bigr)\) with a nonnegative second term; the constant part of (6.1) contributes \(\frac\pi4(R^{2k}-r_0^{2k})\le\frac\pi4R^{2k}\). It remains to bound the contribution of the excess uniformly.

Step 2: the excess near the tips. Let \(\frac12\le r<1\) and \(w=re^{i\theta}\) with \(|\theta|<\pi/2\); put \(s=\frac\pi2-|\theta|\) and \(X=\frac{2r\sin s}{1-r^2}\), so that \(A=\frac4{\pi^2}\arctan X\). By Lemma 6.2, \(\sin s\ge2s/\pi\) and \((1+r)/r\le3\), \[ \begin{aligned} \frac1A&\le\frac{\pi^2}4\Bigl(\frac2\pi+\frac2X\Bigr)\\ &=\frac\pi2+\frac{\pi^2(1-r)(1+r)}{4r\sin s}, \end{aligned} \] so that \[ \frac1A\le\frac\pi2+\frac{3\pi^3}8\,\frac{1-r}s. \tag{6.2} \] On the other hand, the gap between \(|\operatorname{Im}F(w)|\) and the endpoint value \(T(r)\) is, by (5.1) and \(\cos(\frac\pi2-\xi)=\sin\xi\), \[ \begin{aligned} \Delta&=T(r)-|\operatorname{Im}F(w)|\\ &=\int_0^s\frac4{\pi^2}\arctan\frac{2r\sin\xi}{1-r^2}\,d\xi\\ &\le\frac{4r\,s^2}{\pi^2(1-r^2)}\le\frac2{\pi^2}\,\frac{s^2}{1-r}, \end{aligned} \] using \(\arctan y\le y\), \(\sin\xi\le\xi\) and \(2r\le1+r\). Hence \(s\ge\frac\pi{\sqrt2}\sqrt{\Delta(1-r)}\), and (6.2) becomes, with \(C_0=3\sqrt2\pi^2/8\), \[ \frac1A\le\frac\pi2+C_0\sqrt{\frac{1-r}{\Delta}}. \tag{6.3} \] On the slice at height \(t\), \(w=re^{i\theta(r,t)}\) and \(\Delta=T(r)-|t|\).

Step 3: a compact part and a tip part. Fix \(r_*\in(\frac12,1)\). Along the positive slice the radius \(|g|\) increases, so the points with \(|g|\le r_*\) form an initial segment of \([0,v]\), of length at most the diameter of \(D\). On it \(|g'|=1/|F'(g)|\le\max_{|w|\le r_*}|F'(w)|^{-1}\), so its contribution to \(J_k/k\) is at most \[ \begin{gathered} C_*\,k\,r_*^{2k-2},\\ C_*=\operatorname{diam}(D)\max_{|w|\le r_*}|F'(w)|^{-2}. \end{gathered} \] If \(R>r_a=\max(r_0,r_*)\), the remaining contribution is the part of (6.1) with \(r_a<r<R\). Let \(L_*=\inf_{r_*\le r<1}T'(r)>0\). Since \(T(r_a)\ge T(r_0)=|t|\), the mean value theorem gives \(T(r)-|t|\ge L_*(r-r_a)\) for \(r>r_a\). By (6.3), the remaining contribution is at most \[ \begin{aligned} &\frac\pi4\bigl(R^{2k}-r_a^{2k}\bigr)+\frac{C_0}{\sqrt{L_*}}\,I(k,r_a)\\ &\quad\le\frac\pi4R^{2k}+\frac{C_0}{\sqrt{L_*}}\Bigl(1+\frac\pi e\Bigr), \end{aligned} \] where the integral up to \(R\) was enlarged to the integral up to \(1\) and Lemma 6.3 was used. In all cases, \[ \begin{aligned} &\frac{|J_k(v,t)|}k-\frac\pi4|g(v+it)|^{2k}\\ &\quad\le C_*k\,r_*^{2k-2}\\ &\qquad+\frac{C_0(1+\pi/e)}{\sqrt{L_*}}. \end{aligned} \tag{6.4} \]

Step 4: uniformity. Define \[ \begin{aligned} &\varepsilon_k=\max\Bigl\{0,\ \sup_{v+it\in D}\Bigl(\frac{|J_k(v,t)|}k\\ &\qquad\qquad-\frac\pi4|g(v+it)|^{2k}\Bigr)\Bigr\}, \end{aligned} \] finite by (6.4). By Theorem 5.2(c), \(T'(r)\ge\frac8{\pi^2}\operatorname{arctanh}r\), so \(L_*\ge\frac8{\pi^2}\operatorname{arctanh}r_*\to\infty\) as \(r_*\to1\). Given \(\varepsilon>0\), first choose \(r_*\) with \(C_0(1+\pi/e)/\sqrt{L_*}<\varepsilon/2\), and then \(N\) with \(C_*k\,r_*^{2k-2}<\varepsilon/2\) for \(k\ge N\). Then \(\varepsilon_k<\varepsilon\) for \(k\ge N\). \(\square\)

The height \(t\), the lower radius \(r_0\) and the upper radius \(R\) may all depend on \(k\); the bound (6.4) does not depend on them, which is why the error is uniform up to the tips.

4. Exercises

6.1 (easy). Show that \(J_k(v,t)/k\ge\frac\pi4\bigl(|g(v+it)|^{2k}-r_0^{2k}\bigr)\) for \(v>0\), so the constant \(\pi/4\) in Proposition 6.1 cannot be lowered along slices with \(r_0\) small.

6.2 (medium). For \(t=0\), show that \(J_k(v,0)/k=k\int_0^R r^{2k-1}/A(r,0)\,dr\) with \(A(r,0)=\frac4{\pi^2}\arctan\frac{2r}{1-r^2}\), and that \(J_k(v,0)/k-\frac\pi4R^{2k}\to0\) as \(k\to\infty\), uniformly in \(R\le1\).

6.3 (easy). Verify \(\int_0^H\sqrt{y/(H-y)}\,dy=\pi H/2\).

5. Solutions

6.1. In (6.1), \(1/A>\pi/2\).

6.2. At height \(0\), \(\theta(r,0)=0\) and \(r_0=0\). For \(r<1\), \(\frac{2r}{1-r^2}=\tan(2\arctan r)\), so \(A(r,0)=\frac8{\pi^2}\arctan r\) and \(\frac1A-\frac\pi2=\frac{\pi^2}8\bigl(\frac1{\arctan r}-\frac4\pi\bigr)\ge0\), which is bounded near \(r=1\) and at most \(c/r\) near \(0\). The excess \(k\int_0^Rr^{2k-1}(\frac1A-\frac\pi2)\,dr\) is at most \(k\int_0^1r^{2k-1}\cdot c'(1-r+r^{-1}\mathbf 1_{r<1/2})\,dr\), which tends to \(0\): the term \(k\int_0^1r^{2k-1}(1-r)\,dr=\frac k{2k(2k+1)}\) and the term \(k\int_0^{1/2}r^{2k-2}dr\) both tend to \(0\). Here \(\frac1{\arctan r}-\frac4\pi\le c''(1-r)\) on \([\frac12,1]\) because the function vanishes at \(r=1\) and has bounded derivative there.

6.3. With \(y=H\sin^2\phi\), \(dy=2H\sin\phi\cos\phi\,d\phi\) and \(\sqrt{y/(H-y)}=\tan\phi\); the integral becomes \(\int_0^{\pi/2}2H\sin^2\phi\,d\phi=\frac{\pi H}2\).

References