Schwarz's lemma application on a composition of functions
$begingroup$
Let $Delta={z in mathbb{C} | |z| < 1}$. Suppose that $F:Deltato mathbb{C}$ is holomorphic and injective. Assume that $g: Delta to F(Delta)$ is holomorphic and $g(0)=F(0)$. Prove that $|g'(0)|leq |F'(0)|$ and that
$$g(D(0,r)) subseteq F(D(0,r)), 0<r<1.$$
I proved the first part, I cannot prove the second part.
The function $F:Delta to F(Delta)$ is bijective so it has an inverse, define $h(z) : Delta to Delta : z mapsto F^{-1}circ g(z)$. $h$ is holomorphic and $h(0) = 0$, so we can apply Schwarz's lemma to $h$. This gives us that $|h'(0)|leq 1$ and $|h(z)|le|z|$ for $|z| < 1$. $|h'(0)| leq 1 implies |g'(0)|leq |F'(0)|$.
I assume you would use $|h(z)|<|z|$ for $|z| < 1$ to prove the second part, but I'm unable to work it out.
complex-analysis
edited Jan 15 at 14:45
Martin R
30.5k33558
asked Jan 15 at 14:13
pureundergradpureundergrad
547211
$endgroup$
add a comment |
$begingroup$
Let $Delta={z in mathbb{C} | |z| < 1}$. Suppose that $F:Deltato mathbb{C}$ is holomorphic and injective. Assume that $g: Delta to F(Delta)$ is holomorphic and $g(0)=F(0)$. Prove that $|g'(0)|leq |F'(0)|$ and that
$$g(D(0,r)) subseteq F(D(0,r)), 0<r<1.$$
I proved the first part, I cannot prove the second part.
The function $F:Delta to F(Delta)$ is bijective so it has an inverse, define $h(z) : Delta to Delta : z mapsto F^{-1}circ g(z)$. $h$ is holomorphic and $h(0) = 0$, so we can apply Schwarz's lemma to $h$. This gives us that $|h'(0)|leq 1$ and $|h(z)|le|z|$ for $|z| < 1$. $|h'(0)| leq 1 implies |g'(0)|leq |F'(0)|$.
I assume you would use $|h(z)|<|z|$ for $|z| < 1$ to prove the second part, but I'm unable to work it out.
complex-analysis
edited Jan 15 at 14:45
Martin R
30.5k33558
asked Jan 15 at 14:13
pureundergradpureundergrad
547211
$endgroup$
add a comment |
$begingroup$
Let $Delta={z in mathbb{C} | |z| < 1}$. Suppose that $F:Deltato mathbb{C}$ is holomorphic and injective. Assume that $g: Delta to F(Delta)$ is holomorphic and $g(0)=F(0)$. Prove that $|g'(0)|leq |F'(0)|$ and that
$$g(D(0,r)) subseteq F(D(0,r)), 0<r<1.$$
I proved the first part, I cannot prove the second part.
The function $F:Delta to F(Delta)$ is bijective so it has an inverse, define $h(z) : Delta to Delta : z mapsto F^{-1}circ g(z)$. $h$ is holomorphic and $h(0) = 0$, so we can apply Schwarz's lemma to $h$. This gives us that $|h'(0)|leq 1$ and $|h(z)|le|z|$ for $|z| < 1$. $|h'(0)| leq 1 implies |g'(0)|leq |F'(0)|$.
I assume you would use $|h(z)|<|z|$ for $|z| < 1$ to prove the second part, but I'm unable to work it out.
complex-analysis
edited Jan 15 at 14:45
Martin R
30.5k33558
asked Jan 15 at 14:13
pureundergradpureundergrad
547211
$endgroup$
Let $Delta={z in mathbb{C} | |z| < 1}$. Suppose that $F:Deltato mathbb{C}$ is holomorphic and injective. Assume that $g: Delta to F(Delta)$ is holomorphic and $g(0)=F(0)$. Prove that $|g'(0)|leq |F'(0)|$ and that
$$g(D(0,r)) subseteq F(D(0,r)), 0<r<1.$$
I proved the first part, I cannot prove the second part.
The function $F:Delta to F(Delta)$ is bijective so it has an inverse, define $h(z) : Delta to Delta : z mapsto F^{-1}circ g(z)$. $h$ is holomorphic and $h(0) = 0$, so we can apply Schwarz's lemma to $h$. This gives us that $|h'(0)|leq 1$ and $|h(z)|le|z|$ for $|z| < 1$. $|h'(0)| leq 1 implies |g'(0)|leq |F'(0)|$.
I assume you would use $|h(z)|<|z|$ for $|z| < 1$ to prove the second part, but I'm unable to work it out.
complex-analysis
complex-analysis
edited Jan 15 at 14:45
Martin R
30.5k33558
asked Jan 15 at 14:13
pureundergradpureundergrad
547211
edited Jan 15 at 14:45
Martin R
30.5k33558
asked Jan 15 at 14:13
pureundergradpureundergrad
547211
edited Jan 15 at 14:45
Martin R
30.5k33558
edited Jan 15 at 14:45
Martin R
30.5k33558
edited Jan 15 at 14:45
Martin R
30.5k33558
30.5k33558
asked Jan 15 at 14:13
pureundergradpureundergrad
547211
asked Jan 15 at 14:13
pureundergradpureundergrad
547211
asked Jan 15 at 14:13
pureundergradpureundergrad
547211
547211
add a comment |
add a comment |
1 Answer
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$begingroup$
Assume that $w in g(D(0,r))$. Then $w = g(z)$ for some $z in D(0,r)$ and
$$
|F^{-1}(w)| = |F^{-1}(g(z))| = |h(z)| le |z| < r
$$
which means that $F^{-1}(w) in D(0, r)$ or equivalently, $w in F(D(0, r))$.
answered Jan 15 at 14:34
Martin RMartin R
30.5k33558
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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$begingroup$
Assume that $w in g(D(0,r))$. Then $w = g(z)$ for some $z in D(0,r)$ and
$$
|F^{-1}(w)| = |F^{-1}(g(z))| = |h(z)| le |z| < r
$$
which means that $F^{-1}(w) in D(0, r)$ or equivalently, $w in F(D(0, r))$.
answered Jan 15 at 14:34
Martin RMartin R
30.5k33558
$endgroup$
add a comment |
$begingroup$
Assume that $w in g(D(0,r))$. Then $w = g(z)$ for some $z in D(0,r)$ and
$$
|F^{-1}(w)| = |F^{-1}(g(z))| = |h(z)| le |z| < r
$$
which means that $F^{-1}(w) in D(0, r)$ or equivalently, $w in F(D(0, r))$.
answered Jan 15 at 14:34
Martin RMartin R
30.5k33558
$endgroup$
add a comment |
$begingroup$
Assume that $w in g(D(0,r))$. Then $w = g(z)$ for some $z in D(0,r)$ and
$$
|F^{-1}(w)| = |F^{-1}(g(z))| = |h(z)| le |z| < r
$$
which means that $F^{-1}(w) in D(0, r)$ or equivalently, $w in F(D(0, r))$.
answered Jan 15 at 14:34
Martin RMartin R
30.5k33558
$endgroup$
Assume that $w in g(D(0,r))$. Then $w = g(z)$ for some $z in D(0,r)$ and
$$
|F^{-1}(w)| = |F^{-1}(g(z))| = |h(z)| le |z| < r
$$
which means that $F^{-1}(w) in D(0, r)$ or equivalently, $w in F(D(0, r))$.
answered Jan 15 at 14:34
Martin RMartin R
30.5k33558
answered Jan 15 at 14:34
Martin RMartin R
30.5k33558
answered Jan 15 at 14:34
Martin RMartin R
30.5k33558
answered Jan 15 at 14:34
Martin RMartin R
30.5k33558
30.5k33558
add a comment |
add a comment |
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