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Question about Schwarz Lemma applied to comformal automorphism of a square

1

$begingroup$

Let $f$ be a conformal automorphism of the square $x+iy mid x, y in [-3,3] $ and suppose $f(0) = 2+2i$, show $frac19sqrt 2 < |f’(0)| < frac109sqrt 2$

I’ve tried to consider $f^*$ as a conformal automorphism of a disk containing the square, that coincides with $f$, and got an upper bound of $frac59$. However, I am not sure if such thing exist, and if it does, how to proceed to get the lower bound.

complex-analysis

share|cite|improve this question

edited Mar 24 at 20:19

Martin R

31k33561

asked Mar 24 at 18:57

William AmbroseWilliam Ambrose

213

$endgroup$

add a comment | 

1

$begingroup$

Let $f$ be a conformal automorphism of the square $x+iy mid x, y in [-3,3] $ and suppose $f(0) = 2+2i$, show $frac19sqrt 2 < |f’(0)| < frac109sqrt 2$

I’ve tried to consider $f^*$ as a conformal automorphism of a disk containing the square, that coincides with $f$, and got an upper bound of $frac59$. However, I am not sure if such thing exist, and if it does, how to proceed to get the lower bound.

complex-analysis

share|cite|improve this question

edited Mar 24 at 20:19

Martin R

31k33561

asked Mar 24 at 18:57

William AmbroseWilliam Ambrose

213

$endgroup$

add a comment | 

$begingroup$

Let $f$ be a conformal automorphism of the square $x+iy mid x, y in [-3,3] $ and suppose $f(0) = 2+2i$, show $frac19sqrt 2 < |f’(0)| < frac109sqrt 2$

I’ve tried to consider $f^*$ as a conformal automorphism of a disk containing the square, that coincides with $f$, and got an upper bound of $frac59$. However, I am not sure if such thing exist, and if it does, how to proceed to get the lower bound.

complex-analysis

share|cite|improve this question

edited Mar 24 at 20:19

Martin R

31k33561

asked Mar 24 at 18:57

William AmbroseWilliam Ambrose

213

$endgroup$

share|cite|improve this question

edited Mar 24 at 20:19

Martin R

31k33561

asked Mar 24 at 18:57

William AmbroseWilliam Ambrose

213

share|cite|improve this question

edited Mar 24 at 20:19

Martin R

31k33561

asked Mar 24 at 18:57

William AmbroseWilliam Ambrose

213

edited Mar 24 at 20:19

Martin R

31k33561

edited Mar 24 at 20:19

Martin R

31k33561

asked Mar 24 at 18:57

William AmbroseWilliam Ambrose

213

asked Mar 24 at 18:57

William AmbroseWilliam Ambrose

213

1 Answer
1

active

oldest

votes

3

$begingroup$

$f$ is not necessarily the restriction of an automorphism of a disk containing the square. But you can proceed as follows: Since
$$
B(0, 3) subset x+iy mid x, y in [-3,3] subset B(0, 3sqrt2)
$$
we can define
$$
g: Bbb D to Bbb D ,,quad g(z) = fracf(3z)3 sqrt 2
$$
where $Bbb D$ is the unit disk. Then $g(0) = frac2+2i3 sqrt 2$ and the Schwarz-Pick lemma gives
$$
fracf'(0) sqrt 2 = |g'(0)| le 1 - |g(0)|^2 = 1 - left| frac2+2i3 sqrt 2right|^2 = frac 59 \
implies |f'(0)| le frac5 sqrt 29 , .
$$

---

For the lower bound we use that
$$
(f^-1)' (z) = frac 1f'(f^-1(z)
$$
and in particular
$$
(f^-1)'(2+2i) = frac1f'(0)
$$
so that a lower bound for $|f'(0)|$ is equivalent to an upper bound for $|(f^-1)'(2+2i)|$. And that can be obtained similarly as above: Since
$$
B(2+2i, 1) subset x+iy mid x, y in [-3,3] subset B(0, 3sqrt2)
$$
we can define
$$
h: Bbb D to Bbb D ,,quad h(z) = fracf^-1(z+2+2i)3 sqrt 2
$$
Then $h(0) = 0$ and the Schwarz lemma gives
$$
1 ge |h'(0)| = frac3 sqrt 2 = frac1 \
implies
|f'(0)| ge frac13 sqrt 2 , .
$$

share|cite|improve this answer

edited Mar 24 at 21:08

answered Mar 24 at 20:11

Martin RMartin R

31k33561

$endgroup$

add a comment | 

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3

$begingroup$

$f$ is not necessarily the restriction of an automorphism of a disk containing the square. But you can proceed as follows: Since
$$
B(0, 3) subset x+iy mid x, y in [-3,3] subset B(0, 3sqrt2)
$$
we can define
$$
g: Bbb D to Bbb D ,,quad g(z) = fracf(3z)3 sqrt 2
$$
where $Bbb D$ is the unit disk. Then $g(0) = frac2+2i3 sqrt 2$ and the Schwarz-Pick lemma gives
$$
fracf'(0) sqrt 2 = |g'(0)| le 1 - |g(0)|^2 = 1 - left| frac2+2i3 sqrt 2right|^2 = frac 59 \
implies |f'(0)| le frac5 sqrt 29 , .
$$

---

For the lower bound we use that
$$
(f^-1)' (z) = frac 1f'(f^-1(z)
$$
and in particular
$$
(f^-1)'(2+2i) = frac1f'(0)
$$
so that a lower bound for $|f'(0)|$ is equivalent to an upper bound for $|(f^-1)'(2+2i)|$. And that can be obtained similarly as above: Since
$$
B(2+2i, 1) subset x+iy mid x, y in [-3,3] subset B(0, 3sqrt2)
$$
we can define
$$
h: Bbb D to Bbb D ,,quad h(z) = fracf^-1(z+2+2i)3 sqrt 2
$$
Then $h(0) = 0$ and the Schwarz lemma gives
$$
1 ge |h'(0)| = frac3 sqrt 2 = frac1 \
implies
|f'(0)| ge frac13 sqrt 2 , .
$$

share|cite|improve this answer

edited Mar 24 at 21:08

answered Mar 24 at 20:11

Martin RMartin R

31k33561

$endgroup$

add a comment | 

3

$begingroup$

$f$ is not necessarily the restriction of an automorphism of a disk containing the square. But you can proceed as follows: Since
$$
B(0, 3) subset x+iy mid x, y in [-3,3] subset B(0, 3sqrt2)
$$
we can define
$$
g: Bbb D to Bbb D ,,quad g(z) = fracf(3z)3 sqrt 2
$$
where $Bbb D$ is the unit disk. Then $g(0) = frac2+2i3 sqrt 2$ and the Schwarz-Pick lemma gives
$$
fracf'(0) sqrt 2 = |g'(0)| le 1 - |g(0)|^2 = 1 - left| frac2+2i3 sqrt 2right|^2 = frac 59 \
implies |f'(0)| le frac5 sqrt 29 , .
$$

---

For the lower bound we use that
$$
(f^-1)' (z) = frac 1f'(f^-1(z)
$$
and in particular
$$
(f^-1)'(2+2i) = frac1f'(0)
$$
so that a lower bound for $|f'(0)|$ is equivalent to an upper bound for $|(f^-1)'(2+2i)|$. And that can be obtained similarly as above: Since
$$
B(2+2i, 1) subset x+iy mid x, y in [-3,3] subset B(0, 3sqrt2)
$$
we can define
$$
h: Bbb D to Bbb D ,,quad h(z) = fracf^-1(z+2+2i)3 sqrt 2
$$
Then $h(0) = 0$ and the Schwarz lemma gives
$$
1 ge |h'(0)| = frac3 sqrt 2 = frac1 \
implies
|f'(0)| ge frac13 sqrt 2 , .
$$

share|cite|improve this answer

edited Mar 24 at 21:08

answered Mar 24 at 20:11

Martin RMartin R

31k33561

$endgroup$

add a comment | 

$begingroup$

$f$ is not necessarily the restriction of an automorphism of a disk containing the square. But you can proceed as follows: Since
$$
B(0, 3) subset x+iy mid x, y in [-3,3] subset B(0, 3sqrt2)
$$
we can define
$$
g: Bbb D to Bbb D ,,quad g(z) = fracf(3z)3 sqrt 2
$$
where $Bbb D$ is the unit disk. Then $g(0) = frac2+2i3 sqrt 2$ and the Schwarz-Pick lemma gives
$$
fracf'(0) sqrt 2 = |g'(0)| le 1 - |g(0)|^2 = 1 - left| frac2+2i3 sqrt 2right|^2 = frac 59 \
implies |f'(0)| le frac5 sqrt 29 , .
$$

---

For the lower bound we use that
$$
(f^-1)' (z) = frac 1f'(f^-1(z)
$$
and in particular
$$
(f^-1)'(2+2i) = frac1f'(0)
$$
so that a lower bound for $|f'(0)|$ is equivalent to an upper bound for $|(f^-1)'(2+2i)|$. And that can be obtained similarly as above: Since
$$
B(2+2i, 1) subset x+iy mid x, y in [-3,3] subset B(0, 3sqrt2)
$$
we can define
$$
h: Bbb D to Bbb D ,,quad h(z) = fracf^-1(z+2+2i)3 sqrt 2
$$
Then $h(0) = 0$ and the Schwarz lemma gives
$$
1 ge |h'(0)| = frac3 sqrt 2 = frac1 \
implies
|f'(0)| ge frac13 sqrt 2 , .
$$

share|cite|improve this answer

edited Mar 24 at 21:08

answered Mar 24 at 20:11

Martin RMartin R

31k33561

$endgroup$

share|cite|improve this answer

edited Mar 24 at 21:08

answered Mar 24 at 20:11

Martin RMartin R

31k33561

answered Mar 24 at 20:11

Martin RMartin R

31k33561

answered Mar 24 at 20:11

Martin RMartin R

31k33561