Mixing
$alph$-Holder continuous holomorphic function on the unit disk
$begingroup$
Suppose $f colon mathbb{D} to mathbb{C}$ is a holomorphic function on the unit disk such that for some $C > 0$ and $alpha in (0,1)$ we have
$$vert f(z) - f(w) vert leq C vert z - w vert^{alpha}$$
for all $z,w in mathbb{C}$.
Show that
$$vert f'(z) vert leq Avert 1 - vert z vert vert^{alpha - 1}$$
for some constant $A > 0$ depending only on $C$.
Dividing both sides of the given inequality by $vert z - w vert$ we get
$$frac{vert f(z) - f(w) vert}{vert z - wvert} leq Cvert z - w vert^{alpha - 1} leq C vert vert z vert - vert w vert vert^{alpha - 1}.$$
It seems like we want to replace the $w$ on the right-hand side by $1$ somehow and replace the left-hand side by some constant multiple of $vert f'(z) vert$, but I'm not sure how to do this.
I would appreciate hints over full solutions.
complex-analysis
asked Jan 24 at 5:18
Ethan AlwaiseEthan Alwaise
6,411717
$endgroup$
add a comment |
$begingroup$
Suppose $f colon mathbb{D} to mathbb{C}$ is a holomorphic function on the unit disk such that for some $C > 0$ and $alpha in (0,1)$ we have
$$vert f(z) - f(w) vert leq C vert z - w vert^{alpha}$$
for all $z,w in mathbb{C}$.
Show that
$$vert f'(z) vert leq Avert 1 - vert z vert vert^{alpha - 1}$$
for some constant $A > 0$ depending only on $C$.
Dividing both sides of the given inequality by $vert z - w vert$ we get
$$frac{vert f(z) - f(w) vert}{vert z - wvert} leq Cvert z - w vert^{alpha - 1} leq C vert vert z vert - vert w vert vert^{alpha - 1}.$$
It seems like we want to replace the $w$ on the right-hand side by $1$ somehow and replace the left-hand side by some constant multiple of $vert f'(z) vert$, but I'm not sure how to do this.
I would appreciate hints over full solutions.
complex-analysis
asked Jan 24 at 5:18
Ethan AlwaiseEthan Alwaise
6,411717
$endgroup$
add a comment |
$begingroup$
Suppose $f colon mathbb{D} to mathbb{C}$ is a holomorphic function on the unit disk such that for some $C > 0$ and $alpha in (0,1)$ we have
$$vert f(z) - f(w) vert leq C vert z - w vert^{alpha}$$
for all $z,w in mathbb{C}$.
Show that
$$vert f'(z) vert leq Avert 1 - vert z vert vert^{alpha - 1}$$
for some constant $A > 0$ depending only on $C$.
Dividing both sides of the given inequality by $vert z - w vert$ we get
$$frac{vert f(z) - f(w) vert}{vert z - wvert} leq Cvert z - w vert^{alpha - 1} leq C vert vert z vert - vert w vert vert^{alpha - 1}.$$
It seems like we want to replace the $w$ on the right-hand side by $1$ somehow and replace the left-hand side by some constant multiple of $vert f'(z) vert$, but I'm not sure how to do this.
I would appreciate hints over full solutions.
complex-analysis
asked Jan 24 at 5:18
Ethan AlwaiseEthan Alwaise
6,411717
$endgroup$
Suppose $f colon mathbb{D} to mathbb{C}$ is a holomorphic function on the unit disk such that for some $C > 0$ and $alpha in (0,1)$ we have
$$vert f(z) - f(w) vert leq C vert z - w vert^{alpha}$$
for all $z,w in mathbb{C}$.
Show that
$$vert f'(z) vert leq Avert 1 - vert z vert vert^{alpha - 1}$$
for some constant $A > 0$ depending only on $C$.
Dividing both sides of the given inequality by $vert z - w vert$ we get
$$frac{vert f(z) - f(w) vert}{vert z - wvert} leq Cvert z - w vert^{alpha - 1} leq C vert vert z vert - vert w vert vert^{alpha - 1}.$$
It seems like we want to replace the $w$ on the right-hand side by $1$ somehow and replace the left-hand side by some constant multiple of $vert f'(z) vert$, but I'm not sure how to do this.
I would appreciate hints over full solutions.
complex-analysis
complex-analysis
asked Jan 24 at 5:18
Ethan AlwaiseEthan Alwaise
6,411717
asked Jan 24 at 5:18
Ethan AlwaiseEthan Alwaise
6,411717
asked Jan 24 at 5:18
Ethan AlwaiseEthan Alwaise
6,411717
asked Jan 24 at 5:18
Ethan AlwaiseEthan Alwaise
6,411717
asked Jan 24 at 5:18
Ethan AlwaiseEthan Alwaise
6,411717
6,411717
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I think $A=C$ works. Fix $z$. Let $g(w)=frac {f(w)-f(z)} {w-z}$ for $w neq z$ and $g(z)=f'(z)$. Then $g$ is holomorphic. We have $|g(w)| leq C|w-z|^{alpha -1} leq C(1- epsilon -|z|)^{alpha -1}$ for $|w|=1-epsilon$ (because $|w-z| geq |w|-|z|$ and $alpha -1 <0$). By Maximum Modulus Principle it follows that the same true for $|w| leq 1-epsilon$. In particular we get $|f'(z)| leq C(1- epsilon -|z|)^{alpha -1}$. Letting $epsilon to 0$ completes the proof.
answered Jan 24 at 5:56
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I think $A=C$ works. Fix $z$. Let $g(w)=frac {f(w)-f(z)} {w-z}$ for $w neq z$ and $g(z)=f'(z)$. Then $g$ is holomorphic. We have $|g(w)| leq C|w-z|^{alpha -1} leq C(1- epsilon -|z|)^{alpha -1}$ for $|w|=1-epsilon$ (because $|w-z| geq |w|-|z|$ and $alpha -1 <0$). By Maximum Modulus Principle it follows that the same true for $|w| leq 1-epsilon$. In particular we get $|f'(z)| leq C(1- epsilon -|z|)^{alpha -1}$. Letting $epsilon to 0$ completes the proof.
answered Jan 24 at 5:56
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
$endgroup$
add a comment |
$begingroup$
I think $A=C$ works. Fix $z$. Let $g(w)=frac {f(w)-f(z)} {w-z}$ for $w neq z$ and $g(z)=f'(z)$. Then $g$ is holomorphic. We have $|g(w)| leq C|w-z|^{alpha -1} leq C(1- epsilon -|z|)^{alpha -1}$ for $|w|=1-epsilon$ (because $|w-z| geq |w|-|z|$ and $alpha -1 <0$). By Maximum Modulus Principle it follows that the same true for $|w| leq 1-epsilon$. In particular we get $|f'(z)| leq C(1- epsilon -|z|)^{alpha -1}$. Letting $epsilon to 0$ completes the proof.
answered Jan 24 at 5:56
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
$endgroup$
add a comment |
$begingroup$
I think $A=C$ works. Fix $z$. Let $g(w)=frac {f(w)-f(z)} {w-z}$ for $w neq z$ and $g(z)=f'(z)$. Then $g$ is holomorphic. We have $|g(w)| leq C|w-z|^{alpha -1} leq C(1- epsilon -|z|)^{alpha -1}$ for $|w|=1-epsilon$ (because $|w-z| geq |w|-|z|$ and $alpha -1 <0$). By Maximum Modulus Principle it follows that the same true for $|w| leq 1-epsilon$. In particular we get $|f'(z)| leq C(1- epsilon -|z|)^{alpha -1}$. Letting $epsilon to 0$ completes the proof.
answered Jan 24 at 5:56
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
$endgroup$
I think $A=C$ works. Fix $z$. Let $g(w)=frac {f(w)-f(z)} {w-z}$ for $w neq z$ and $g(z)=f'(z)$. Then $g$ is holomorphic. We have $|g(w)| leq C|w-z|^{alpha -1} leq C(1- epsilon -|z|)^{alpha -1}$ for $|w|=1-epsilon$ (because $|w-z| geq |w|-|z|$ and $alpha -1 <0$). By Maximum Modulus Principle it follows that the same true for $|w| leq 1-epsilon$. In particular we get $|f'(z)| leq C(1- epsilon -|z|)^{alpha -1}$. Letting $epsilon to 0$ completes the proof.
answered Jan 24 at 5:56
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
answered Jan 24 at 5:56
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
answered Jan 24 at 5:56
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
answered Jan 24 at 5:56
Kavi Rama MurthyKavi Rama Murthy
67.5k53067
67.5k53067
add a comment |
add a comment |
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