Show that $f(z)$ has a zero in $mathbb{D}$ and that $|z_{0}|>frac{1}{M}$
$begingroup$
Suppose that $f(z)$ is analytic on the closed unit disk $overline{mathbb{D}}$ and $1<|f(z)|<M$ for $|z|=1$, while $f(0)=1$. Show that $f(z)$ has a zero in $mathbb{D}$ and that such zero $z_{0}$ satisfies $|z_{0}|>frac{1}{M}$.
My attempt:
Suppose $f$ has no zero in the unit disk. Then, by minimum modulus principle, $f$ attains its minimum on the boundary. But maximum modulus principle suggests that $f$ attains its maximum on the boundary. This implies that $|f(z)|=1$ for all $z$, and this is a contradiction to our initial assumption that $1<|f(z)|<M$ for $|z|=1$. Hence,$f(z)$ has a zero in $mathbb{D}$.
Now consider the mapping $$phi_{M^{-1}}=frac{M^{-1}-z}{1-overline{M^{-1}}z}$$. Then $phi_{M^{-1}}left(frac{f(z)}{M}right):mathbb{D}rightarrowmathbb{D}$ and maps $0$ to $0$.
Then, by Schwarz Lemma, $$left|phi_{M^{-1}}left(frac{f(z_0)}{M}right)right|=left|frac{1}{M}right|leq|z_0|$$
Any clarification is appreciated.
complex-analysis
asked Jan 13 at 18:19
Ya GYa G
541211
$endgroup$
add a comment |
$begingroup$
Suppose that $f(z)$ is analytic on the closed unit disk $overline{mathbb{D}}$ and $1<|f(z)|<M$ for $|z|=1$, while $f(0)=1$. Show that $f(z)$ has a zero in $mathbb{D}$ and that such zero $z_{0}$ satisfies $|z_{0}|>frac{1}{M}$.
My attempt:
Suppose $f$ has no zero in the unit disk. Then, by minimum modulus principle, $f$ attains its minimum on the boundary. But maximum modulus principle suggests that $f$ attains its maximum on the boundary. This implies that $|f(z)|=1$ for all $z$, and this is a contradiction to our initial assumption that $1<|f(z)|<M$ for $|z|=1$. Hence,$f(z)$ has a zero in $mathbb{D}$.
Now consider the mapping $$phi_{M^{-1}}=frac{M^{-1}-z}{1-overline{M^{-1}}z}$$. Then $phi_{M^{-1}}left(frac{f(z)}{M}right):mathbb{D}rightarrowmathbb{D}$ and maps $0$ to $0$.
Then, by Schwarz Lemma, $$left|phi_{M^{-1}}left(frac{f(z_0)}{M}right)right|=left|frac{1}{M}right|leq|z_0|$$
Any clarification is appreciated.
complex-analysis
asked Jan 13 at 18:19
Ya GYa G
541211
$endgroup$
add a comment |
$begingroup$
Suppose that $f(z)$ is analytic on the closed unit disk $overline{mathbb{D}}$ and $1<|f(z)|<M$ for $|z|=1$, while $f(0)=1$. Show that $f(z)$ has a zero in $mathbb{D}$ and that such zero $z_{0}$ satisfies $|z_{0}|>frac{1}{M}$.
My attempt:
Suppose $f$ has no zero in the unit disk. Then, by minimum modulus principle, $f$ attains its minimum on the boundary. But maximum modulus principle suggests that $f$ attains its maximum on the boundary. This implies that $|f(z)|=1$ for all $z$, and this is a contradiction to our initial assumption that $1<|f(z)|<M$ for $|z|=1$. Hence,$f(z)$ has a zero in $mathbb{D}$.
Now consider the mapping $$phi_{M^{-1}}=frac{M^{-1}-z}{1-overline{M^{-1}}z}$$. Then $phi_{M^{-1}}left(frac{f(z)}{M}right):mathbb{D}rightarrowmathbb{D}$ and maps $0$ to $0$.
Then, by Schwarz Lemma, $$left|phi_{M^{-1}}left(frac{f(z_0)}{M}right)right|=left|frac{1}{M}right|leq|z_0|$$
Any clarification is appreciated.
complex-analysis
asked Jan 13 at 18:19
Ya GYa G
541211
$endgroup$
Suppose that $f(z)$ is analytic on the closed unit disk $overline{mathbb{D}}$ and $1<|f(z)|<M$ for $|z|=1$, while $f(0)=1$. Show that $f(z)$ has a zero in $mathbb{D}$ and that such zero $z_{0}$ satisfies $|z_{0}|>frac{1}{M}$.
My attempt:
Suppose $f$ has no zero in the unit disk. Then, by minimum modulus principle, $f$ attains its minimum on the boundary. But maximum modulus principle suggests that $f$ attains its maximum on the boundary. This implies that $|f(z)|=1$ for all $z$, and this is a contradiction to our initial assumption that $1<|f(z)|<M$ for $|z|=1$. Hence,$f(z)$ has a zero in $mathbb{D}$.
Now consider the mapping $$phi_{M^{-1}}=frac{M^{-1}-z}{1-overline{M^{-1}}z}$$. Then $phi_{M^{-1}}left(frac{f(z)}{M}right):mathbb{D}rightarrowmathbb{D}$ and maps $0$ to $0$.
Then, by Schwarz Lemma, $$left|phi_{M^{-1}}left(frac{f(z_0)}{M}right)right|=left|frac{1}{M}right|leq|z_0|$$
Any clarification is appreciated.
complex-analysis
complex-analysis
asked Jan 13 at 18:19
Ya GYa G
541211
asked Jan 13 at 18:19
Ya GYa G
541211
asked Jan 13 at 18:19
Ya GYa G
541211
asked Jan 13 at 18:19
Ya GYa G
541211
asked Jan 13 at 18:19
Ya GYa G
541211
541211
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your invoking minimum modulus principle is fine, but the subsequent argument using maximum principle does not produce $|f(z)|=1$ on the boundary (if there's an argument that can prove it, then you should include it in your proof.) I suggest instead to note that $|f(0)|=1<|f(z)|$ for all $|z|=1$. It says that the minimum of $|f(z)|$ cannot occur on $|z|=1$. This leads to a contradiction.
About the second part, it looks fine except that the equality sign in
$$
left|frac{1}{M}right|leq|z_0|
$$ can be dropped since $|f(z)|=M$ cannot occur on the boundary.
Here, I will present another way in which the Schwarz lemma can be applied. Define
$$
g:overline{mathbb{D}}ni wmapsto fleft(frac{w+z_0}{1+overline{z_0}w}right).
$$ Then we can see that $g(0)=f(z_0)=0$ and $|g(w)|<M$ for all $|w|=1$. Schwarz's lemma implies $$
|g(w)|<M|w|
$$ for all $0<|w|le 1$. Finally the conclusion comes from noticing that
$$
|g(-z_0)|=|f(0)|=1<M|-z_0|=M|z_0|.
$$
answered Jan 13 at 18:36
SongSong
18.6k21651
$endgroup$
add a comment |
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3072349%2fshow-that-fz-has-a-zero-in-mathbbd-and-that-z-0-frac1m%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your invoking minimum modulus principle is fine, but the subsequent argument using maximum principle does not produce $|f(z)|=1$ on the boundary (if there's an argument that can prove it, then you should include it in your proof.) I suggest instead to note that $|f(0)|=1<|f(z)|$ for all $|z|=1$. It says that the minimum of $|f(z)|$ cannot occur on $|z|=1$. This leads to a contradiction.
About the second part, it looks fine except that the equality sign in
$$
left|frac{1}{M}right|leq|z_0|
$$ can be dropped since $|f(z)|=M$ cannot occur on the boundary.
Here, I will present another way in which the Schwarz lemma can be applied. Define
$$
g:overline{mathbb{D}}ni wmapsto fleft(frac{w+z_0}{1+overline{z_0}w}right).
$$ Then we can see that $g(0)=f(z_0)=0$ and $|g(w)|<M$ for all $|w|=1$. Schwarz's lemma implies $$
|g(w)|<M|w|
$$ for all $0<|w|le 1$. Finally the conclusion comes from noticing that
$$
|g(-z_0)|=|f(0)|=1<M|-z_0|=M|z_0|.
$$
answered Jan 13 at 18:36
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
Your invoking minimum modulus principle is fine, but the subsequent argument using maximum principle does not produce $|f(z)|=1$ on the boundary (if there's an argument that can prove it, then you should include it in your proof.) I suggest instead to note that $|f(0)|=1<|f(z)|$ for all $|z|=1$. It says that the minimum of $|f(z)|$ cannot occur on $|z|=1$. This leads to a contradiction.
About the second part, it looks fine except that the equality sign in
$$
left|frac{1}{M}right|leq|z_0|
$$ can be dropped since $|f(z)|=M$ cannot occur on the boundary.
Here, I will present another way in which the Schwarz lemma can be applied. Define
$$
g:overline{mathbb{D}}ni wmapsto fleft(frac{w+z_0}{1+overline{z_0}w}right).
$$ Then we can see that $g(0)=f(z_0)=0$ and $|g(w)|<M$ for all $|w|=1$. Schwarz's lemma implies $$
|g(w)|<M|w|
$$ for all $0<|w|le 1$. Finally the conclusion comes from noticing that
$$
|g(-z_0)|=|f(0)|=1<M|-z_0|=M|z_0|.
$$
answered Jan 13 at 18:36
SongSong
18.6k21651
$endgroup$
add a comment |
$begingroup$
Your invoking minimum modulus principle is fine, but the subsequent argument using maximum principle does not produce $|f(z)|=1$ on the boundary (if there's an argument that can prove it, then you should include it in your proof.) I suggest instead to note that $|f(0)|=1<|f(z)|$ for all $|z|=1$. It says that the minimum of $|f(z)|$ cannot occur on $|z|=1$. This leads to a contradiction.
About the second part, it looks fine except that the equality sign in
$$
left|frac{1}{M}right|leq|z_0|
$$ can be dropped since $|f(z)|=M$ cannot occur on the boundary.
Here, I will present another way in which the Schwarz lemma can be applied. Define
$$
g:overline{mathbb{D}}ni wmapsto fleft(frac{w+z_0}{1+overline{z_0}w}right).
$$ Then we can see that $g(0)=f(z_0)=0$ and $|g(w)|<M$ for all $|w|=1$. Schwarz's lemma implies $$
|g(w)|<M|w|
$$ for all $0<|w|le 1$. Finally the conclusion comes from noticing that
$$
|g(-z_0)|=|f(0)|=1<M|-z_0|=M|z_0|.
$$
answered Jan 13 at 18:36
SongSong
18.6k21651
$endgroup$
Your invoking minimum modulus principle is fine, but the subsequent argument using maximum principle does not produce $|f(z)|=1$ on the boundary (if there's an argument that can prove it, then you should include it in your proof.) I suggest instead to note that $|f(0)|=1<|f(z)|$ for all $|z|=1$. It says that the minimum of $|f(z)|$ cannot occur on $|z|=1$. This leads to a contradiction.
About the second part, it looks fine except that the equality sign in
$$
left|frac{1}{M}right|leq|z_0|
$$ can be dropped since $|f(z)|=M$ cannot occur on the boundary.
Here, I will present another way in which the Schwarz lemma can be applied. Define
$$
g:overline{mathbb{D}}ni wmapsto fleft(frac{w+z_0}{1+overline{z_0}w}right).
$$ Then we can see that $g(0)=f(z_0)=0$ and $|g(w)|<M$ for all $|w|=1$. Schwarz's lemma implies $$
|g(w)|<M|w|
$$ for all $0<|w|le 1$. Finally the conclusion comes from noticing that
$$
|g(-z_0)|=|f(0)|=1<M|-z_0|=M|z_0|.
$$
answered Jan 13 at 18:36
SongSong
18.6k21651
edited Jan 13 at 20:40
answered Jan 13 at 18:36
SongSong
18.6k21651
answered Jan 13 at 18:36
SongSong
18.6k21651
answered Jan 13 at 18:36
SongSong
18.6k21651
18.6k21651
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3072349%2fshow-that-fz-has-a-zero-in-mathbbd-and-that-z-0-frac1m%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
