Gary
GaryVasco
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Post by Gary on Oct 3, 2017 16:13:18 GMT
Vasco,
Exercise 2 is based on Liouville's Theorem, which is explained on p. 360. At the end of the continuation paragraph it says But if f compresses the whole plane to a region lying inside the disc of radius M, then the above result will continue to hold true no matter how large we make N. Therefore f(p) = 0 for all p, and we are done. I believe you used this conclusion in your answer to ex. 2. How does it follow that crushing the whole plane with increasing N implies f(p) = 0? Is it because f must be 0 for points inside the disc, for otherwise they would be sent by f to the exterior, and because the same function must send both interior and exterior points to the inside of the disc of radius M?
Gary
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Post by Admin on Oct 4, 2017 14:24:22 GMT
Vasco, Exercise 2 is based on Liouville's Theorem, which is explained on p. 360. At the end of the continuation paragraph it says But if f compresses the whole plane to a region lying inside the disc of radius M, then the above result will continue to hold true no matter how large we make N. Therefore f(p) = 0 for all p, and we are done. I believe you used this conclusion in your answer to ex. 2. How does it follow that crushing the whole plane with increasing N implies f(p) = 0? Is it because f must be 0 for points inside the disc, for otherwise they would be sent by f to the exterior, and because the same function must send both interior and exterior points to the inside of the disc of radius M? Gary Gary On page 360 about halfway down we have $\displaystyle |f(p)|\leq\frac{M|p|}{N}$ Needham says that this inequality still holds if $f$ is such that it compresses the whole plane to to a region inside the disc of radius $M$, and so it will continue to hold as we let the disc of radius $N$ get bigger and bigger to enclose more and more of the plane. I follows from the inequality above that as $N$ goes to $\infty$ the value of $|f(p)|$ goes to 0, and since $p$ can be any point in the plane then we can say $f(p)=0$ for all $p$ and so, saying this another way, we can say $f(z)=0$ for all $z$. Vasco |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Oct 4, 2017 17:13:05 GMT
Vasco, Exercise 2 is based on Liouville's Theorem, which is explained on p. 360. At the end of the continuation paragraph it says But if f compresses the whole plane to a region lying inside the disc of radius M, then the above result will continue to hold true no matter how large we make N. Therefore f(p) = 0 for all p, and we are done. I believe you used this conclusion in your answer to ex. 2. How does it follow that crushing the whole plane with increasing N implies f(p) = 0? Is it because f must be 0 for points inside the disc, for otherwise they would be sent by f to the exterior, and because the same function must send both interior and exterior points to the inside of the disc of radius M? Gary Gary On page 360 about halfway down we have $\displaystyle |f(p)|\leq\frac{M|p|}{N}$ Needham says that this inequality still holds if $f$ is such that it compresses the whole plane to to a region inside the disc of radius $M$, and so it will continue to hold as we let the disc of radius $N$ get bigger and bigger to enclose more and more of the plane. I follows from the inequality above that as $N$ goes to $\infty$ the value of $|f(p)|$ goes to 0, and since $p$ can be any point in the plane then we can say $f(p)=0$ for all $p$ and so, saying this another way, we can say $f(z)=0$ for all $z$. Vasco Vasco, I still have a hard time grasping this. I think it is the clause as $N$ goes to $\infty$ the value of $|f(p)|$ goes to 0 that is obscuring my vision. My persistent thought is, as the value of $|f(p)|$ goes to 0, then there must be $|f(p)|$'s (image points) that are not zero. But I think I see it now: M is fixed at any real number; p is fixed at any complex number; N can become very large, so |f(p)| must be very small to account for the whole disc. The value of $f(p)$ doesn't actually go to zero---it is zero. I wonder if I am the only reader who stumbles on this. Gary |
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Oct 4, 2017 18:06:26 GMT
Subtitle: unresolved doubts regarding Liouville's Theorem, p. 360
Vasco,
Since $f(z)$ depends on $z$, $M = max|f(z)|$ on $K$ must depend on $N = max|z|$. Since $f$ is analytic, $M = max|f(z)| = |f(N e^{i \theta})|$ on $K$. Then
$\hspace{2em}|f(p)| \leq \frac{|f(N e^{i \theta})|}{N}|p|$
This just says that $f(p)$ compresses points inside $K$. As $N \rightarrow \infty$, it is possible that $M = |f(N e^{i \theta})|$ also goes to infinity slightly compressed. Since $M$ is not fixed, how does one conclude that $f(p) = 0$ for all $p$? Or do we begin with a finite $N$, obtain an $M$ which we leave fixed, and only then allow $N$ to go to infinity?
Gary
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Post by Admin on Oct 5, 2017 13:57:40 GMT
Gary
Consider a function $f$ that compresses the whole plane to a region lying inside the disc of radius $M$. Then any disc will be compressed to a region inside the disc of radius $M$.
Since $\frac{|p|}{N}<1$ it follows that $|f(p)|\leq\frac{M|p|}{N}<M$ and this is true no matter how large we make $N$.
Letting $N$ go to $\infty$ we can see that $f(p)=0$ for any value of $p$.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Oct 5, 2017 16:53:27 GMT
Gary Consider a function $f$ that compresses the whole plane to a region lying inside the disc of radius $M$. Then any disc will be compressed to a region inside the disc of radius $M$. Since $\frac{|p|}{N}<1$ it follows that $|f(p)|\leq\frac{M|p|}{N}<M$ and this is true no matter how large we make $N$. Letting $N$ go to $\infty$ we can see that $f(p)=0$ for any value of $p$. Vasco Vasco, The only way that I can see this is to add that we may make $p$ as small as we like, $N$ as large as we like, $M \leq N$ always, but otherwise unrestricted. Since this means that $|f(z)| \leq |p|$ always, $f(z)$ must be 0. Does that work? Gary |
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Post by Admin on Oct 5, 2017 17:00:02 GMT
Gary
I don't think so, $p$ can be any point inside $N$ and so as $N$ gets bigger $p$ can also get bigger.
Vasco
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Gary
GaryVasco
Posts: 3,352
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Post by Gary on Oct 5, 2017 19:13:05 GMT
Gary I don't think so, $p$ can be any point inside $N$ and so as $N$ gets bigger $p$ can also get bigger. Vasco Vasco, I think I see it now. N can go to infinity, but M is fixed by f or a prior choice of M. p can any point where $|p| \leq N$, but f(p) is still inside M, and by the inequality, |f(p)| goes to 0 as N expands the pre-image circle to include the whole plane, so f(p) = 0 for all p. But it seems we must leave p fixed while letting N go to infinity, for otherwise |f(p)| would not be squashed to 0. Gary |
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