If you work in real analysis see the formulation B).
By $\mathbb{D} $ and $\mathbb{T} $ we denote the unit disk and the unit circle respectively.
A)For $p\in \mathbb{D} $, let $H(p)$ denote the family of all harmonics maps
$f$ from $\mathbb{D}$ into itself with $f(0)=p$ which are conformal at
$0$.
B) Set $L(p)= \sup \{ |f'(0)|: f\in H(p) \}$ and $K(p)=\frac{L(p)}{1-|p|^2}$.
Find $K(p)$. It is known that $1 \leq K(p)\leq \frac{4}{\pi}$ and $K(0)= 1$.
We write $f(t)$ instead of $f(e^{it})$.
Let $Q(p)$ denote the family of all homeomorphisms (more generally weak)
of $\mathbb{T}$ onto itself such that $a_0(f)=\frac{1}{2\pi}\int_0
^{2\pi}f(t) dt =p$
and $b_1(f)= \int_0 ^{2\pi}f(t) e^{it} dt =0$. Set $a_1(f)= \frac{1}{2\pi} \int_0
^{2\pi}f(t) e^{-it} dt$, $\bar{L}(p)= \sup \{ |a_1(f)|: f \in Q(p) \}$
and $\bar{K}(p)=\frac{L(p)}{1-|p|^2}$.
It seems that using the RCK-theorem we can show $\bar{K}(p)=K(p)$ and that $\bar{K}(p)>1$ in general.
Let $P_n$ be the polygonal line defined by points $(0,1), (\frac{\pi}{2}-1/n,1), (\frac{\pi}{2},0),(\pi - 1/n,1/n) $ and $(\pi, - 1)$,
and $x_n$ even function which equals $P_n$ on $[0,\pi]$ and $y_n$ odd function such that $y_n=\sqrt{1-x_n^2}$ on $[0,\pi/2]$ and $y_n=-\sqrt{1-x_n^2}$ on $[\pi/2,\pi]$.
Set $z_n=x_n +i y_n$. Check that $a_0(z_n)=1/2 +o(1)$ and
$a_1(z_n)= 2/\pi$.
Since $y_n$ is odd, $a_1(z_n)= a_1(x_n)$ and $\pi a_1(x_n)= 2
\int_0 ^{\pi/2} \cos t + o(1)= 2+ o(1) $ and therefore $a_1(z_n)= 2/\pi$.
For $p=1/2$ we have $1-p^2=1-1/4 = 3/4$ and $\bar{K}(1/2)\geq
c:= \frac{2}{\pi} \frac{4}{3}$.
But $c1$?
@Miodrag
Follow the work of J.H.He in PRL ,Int.J.Mod.Phys B or mine in proc.Ind.Nat.Sci.Aca -A
You can also matlab.
Prof B.Rath
@ Miodrag Mateljević
1. BAD NEWS: Without changes your example is inappropriate: every one of the functions z_n possesses non-removable discontinuity.
2. GOOD NEWS: The following is true:
If x_n is the even extension over [-\pi, \pi] of your function from [0, \pi] and if y_n is the odd extension of your function, i.e. y_n = \sign y_n \cdot \sqrt{ 1 - {x_n}^2 }, then z_n = x_n + i y_n is a orientation-preserving homeomorphic self-mapping of the circle \mathbb{T}. Moreover
a_{-1} (z_n) = 0, a_0{z_n} \to 0.5, a_1{z_n} \to 2/\pi.
Hence, in particular in the limit we have: a_1 / a_0 = 4/\pi.
3. CONCLUSION: If I have not got wrong, and if your knowledge that K \in [1, 4/\pi] is correct, then your example solves your problem for p=0.5. Congratulations!
Best wishes, Joachim
Dear
@Biswanath Rath · 27.11 · North Orissa University
I can not find the paper you mentioned.
Dear
@Joachim Domsta
Thanks. We need to check again.
Best wishes,Miodrag
@Miodrag
Give your e.mail , I will send the scan copy of J.H.He and Mine
Biswanath
My email is [email protected]
In connection with the posed extremal problem we outline a result.
Let $\mathbb{S}_0= \{w\in\C : |\Re w|< 1\}$ and $\rho_0$ hyperbolic density on $\mathbb{S}_0$.
\begin{equation}\label{eq:dhyp-strip0}
\rho_{0}(w)= \frac{\pi}{2} \frac{1}{\cos (\frac{\pi}{2} u)}.
\end{equation}
If $F$ is holomorphic map from $\D$ into $\mathbb{S}_0$, then by Ahlfors-Schwarz lemma
\begin{equation}\label{eq:Ah-Sch1a}
\rho_0 (F(z)) |F'(z)|\leq 2 (1- |z|^2)^{-1},\quad z\in \D.
\end{equation}
Theorem.
If $f$ is harmonic map from $\D$ into $(-1,1)$ with $f(0)=0$, then
\[
|\nabla f(z)| \leq \frac{4}{\pi} \cos (\frac{\pi}{2}f(z)) (1- |z|^2)^{-1}, \quad z\in \D.
\]
See also D. Kalaj and M. Vuorinen,
On harmonic functions and the Schwarz lemma, Proc. Amer. Math. Soc. 140 (2012), no. 1, 161-165.
IMPORTANT. By this note I am disclaiming my statement, that the proposed correction of Miodrag's example gives a solution to the problem for K(0.5). This due to wrong division by 0.5 instead of 1 - 0.5^2 = 0.75 which gives for the ratio value less than 1 :(
Joachim Domsta
@Miodrag Mateljevi\'c
After analysis of couple of examples I came to a problem of deriviation of the given by you bounds for K. Could you please present any source of the upper estimate by [ 4/ \pi ]?
Best regards
Dear @Joachim Domsta, ·
See Sections 3 and 4 in
Hyperbolic geometry and Schwarz lemma.
Please, can you elaborate your statement that
: every one of the functions z_n possesses non-removable discontinuity.
Best regards
Dear @Miodrag Mateljevi\'c
1. Thanks for the reference.
2. I have drawn your odd extension of $P_n$ over $ [-\pi, \pi] $ and extended it periodically with period $2\pi$. The resulting function possesses two jumps from -1 to 1 at 0 and and $\pi$ (+ periodicity). The corresponding selfmapping of $\mathbb T$ determined by $ z_n = x_n + y_n$ has jumps between the poles 1 and -1. Thus it is not homeomorphic.
Perhaps you have meant even extention for x, as in my proposal, haven't you?
3. Another questions:
3.1 Probably, the fourth point should be $(\pi - 1/n,-1/n)$ instead of $ (\pi - 1/n,1/n)$ to get (stricly) decreasing $P_n$ over $ [{\pi/2},\pi]$; but then why you admitted it to be constant in $ [0,{\pi / 2)]$?
3.2. Why are you deviding by $1 -|p|^2$ instead multiplying by $1 - 0^2$. In the formulas presented by you we have $1 - |z|^2$, and not $ 1 - |f(z)|^2 $. Probably the problem is with division by $\cos( \frac{\pi}{2} p $.
3.3. How did you get $\frac{2}{\sqrt{3}}$
Best regards
Dear @Joachim Domsta,
Thanks very much.
The idea is to construct x_n and y_n such that x_n is a rough model of cos and y_n is a rough model of sin. I used the notion of odd as even and vice versa.
Let $P_n$ be the polygonal line defined by points $(0,1), (\frac{\pi}{2}-1/n,1), (\frac{\pi}{2},0),(\pi - 1/n,1/n) $ and $(\pi, - 1)$,
and $x_n$ even function which equals $P_n$ on $[0,\pi]$ and $y_n$ odd function such that $y_n=\sqrt{1-x_n^2}$ on $[0,\pi/2]$ and $y_n=-\sqrt{1-x_n^2}$ on $[\pi/2,\pi]$.
Set $z_n=x_n +i y_n$.
I will try to answer 3.2. and 3.3 as soon as possible.
Best regards
If $f=u+iv$, $u$ is even and $v$ is odd, then (i) is equivalent
with (i') $\int_0 ^{2\pi} u \cos t dt= \int_0 ^{2\pi} v \sin
t dt$.
Since $a_1(f)=a_1(f) + b_1(f)$, we have (I) $\pi
a_1(f)= \int_0 ^{2\pi}f(t) \cos t$.
Since $y_n$ is odd, $a_1(z_n)= a_1(x_n)$ and $\pi a_1(x_n)= 2
\int_0 ^{\pi/2} \cos t + o(1)= 2+ o(1) $ and therefore $
a_1(z_n)= 2/\pi$.
For $p=1/2$ we have $1-p^2=1-1/4 = 3/4$ and $\bar{K}(1/2)\geq
c:= \frac{2}{\pi} \frac{4}{3}$.
But $c1$?
Dear Miodrag Mateljevi\'c
I am glad, that you are confirming the calculation I have post in my first answer, which was:
"If x_n is the even extension over [-\pi, \pi] of your function from [0, \pi] and if y_n is the odd extension of your function, i.e. y_n = \sign y_n \cdot \sqrt{ 1 - {x_n}^2 }, then z_n = x_n + i y_n is a orientation-preserving homeomorphic self-mapping of the circle \mathbb{T}. Moreover . . . a_0{z_n} \to 0.5, a_1{z_n} \to 2/\pi."
Thus we do not differ now about these two values. BUT, STILL
Q161219JD: Why should we multiply $a_1 $$ by $ (1-a_0^{2} )^{-1} $
instead of $ \frac{ (1-0^2)^1 }{ \cos( \frac{\pi}{2} a_0) }$ ?
If I do not misunderstand, according to the "Result" posted by you we should deal with
z=0 and f(z) = f(0) = a_0 and f'(z) = f'(0) = a_1.
Thus, we should look rather for
K*(0.5} := \sup\{ \frac{ a_1(f) }{ \cos ( \frac{\pi}{2} \cdot a_0(f) ): a_0 = 0.5 \}
which is in ${1; \frac{4}{\pi}$. Shouldn't we?
PS: The corrected example and corrected ratio still lead to
c* = \frac{ 2}{\pi} \cdot \cos \frac{\pi}{4} } < 1.
Up to now, I do not know any example with c* >1, the more, the extremal value K(0.5).
Best regards
@Miodrag Mateljevi\'c
Dear sir, I decided to wait for answers leading us closer to your problem.
Sincerely, Joachim Domsta
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