Applying SC. Lemma: Conditions & Practical Uses

In summary: It is clearly analytic, and injective: it is a mobius map. g(0) is obviously 0, so all it remains is to show that it transforms delta inside itself. so what is the |g(z)| if |z|<=1?In summary, the book states that for the Schwarz lemma to work, the absolute value of f(z) must be less than or equal to 1 and z must be less than 1. However, the book uses the lemma in some problems even if one of the conditions is not satisfied. The theorem can be rephrased as: a function f that maps the unit complex disc inside itself and fixes the origin satisfies the conditions |f(z)|<=|z| and
  • #1
sweetvirgogirl
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so the way book states it, for the sc. lemma to work, |f(z)| has to be less than or equal to 1 and and z has to be less than 1. However, the book seems to use the lemma in some problems even if one of the conditions is not satisfied ... any help with gretaly appreciated

also ...if you could rephrase the theorem so that it makes more practical sense, then that would be great too ... like so I know when I should apply the theorem to achieve the desired result.
 
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  • #2
A function f that maps the unit complex disc inside itself and fixes the origin satisfies |f(z)|<=|z| and |f'(0)|<=1. there is also the fact that if there is any point where |f(z)|=|z| or if |f'(0)|=1 then f is a rotation.

Can't this just be seen by imagining a taylor series?

unless you post the other problems then no one can properly help you (ie how do they fail to meet the hypotheses). this is a purely local statement, so you can shift g(z) by a constant so that g(z)-c satisfies the hypotheses and apply to g(z)-c and get info about g, or you can scale g so that it maps the disc into the disc by replacing g by a contsnat times g, or you can restrict g to some region where you can apply the lemma.

Eg if g mapped the disc of radius 2 about the origin into itself and g(0)=0 set f(z)=g(2z)/2 to get something you can apply schwarz's lemma too.
 
  • #3
matt grime said:
A function f that maps the unit complex disc inside itself and fixes the origin satisfies |f(z)|<=|z| and |f'(0)|<=1. there is also the fact that if there is any point where |f(z)|=|z| or if |f'(0)|=1 then f is a rotation.

Can't this just be seen by imagining a taylor series?

unless you post the other problems then no one can properly help you (ie how do they fail to meet the hypotheses). this is a purely local statement, so you can shift g(z) by a constant so that g(z)-c satisfies the hypotheses and apply to g(z)-c and get info about g, or you can scale g so that it maps the disc into the disc by replacing g by a contsnat times g, or you can restrict g to some region where you can apply the lemma.

Eg if g mapped the disc of radius 2 about the origin into itself and g(0)=0 set f(z)=g(2z)/2 to get something you can apply schwarz's lemma too.
f (z) = lamdba (a - z)/(1 - abar*z) |lambda| = 1; |a| <1
where f is any one-to-one analytic function mapping delta = {z:|z| < 1} onto itself
set g(z) = (a - f(z))/(1-abar*f(z))
now they say that g is one-to-one analytic function mapping delta onto delta (why?) and g(0) = 0
and then they applied schwarz lemma to the function g(z) (why??)

i might be missing something fundamental about properties of conjugate of a complex number (oh i know what conjugate is though ... so don't restate the obvious lol) that I don't see how |g(z)| <=1

and also why is g(z) is one-to-one analytic function mapping delta onto delta?

thanks
 
  • #4
sweetvirgogirl said:
f (z) = lamdba (a - z)/(1 - abar*z) |lambda| = 1; |a| <1
where f is any one-to-one analytic function mapping delta = {z:|z| < 1} onto itself
set g(z) = (a - f(z))/(1-abar*f(z))
now they say that g is one-to-one analytic function mapping delta onto delta (why?)

Work it out: f is a mobius transformation and g(z) is f(f(z).

and g(0) = 0
and then they applied schwarz lemma to the function g(z) (why??)

presumably because it shows them something useful. as we have no idea what they want to prove we cannot explain it.

I don't see how |g(z)| <=1. and also why is g(z) is one-to-one analytic function mapping delta onto delta?

Whic part don't you see? Which part have you tried to prove? It is clearly analytic, and injective: it is a mobius map. g(0) is obviously 0, so all it remains is to show that it transforms delta inside itself. so what is the |g(z)| if |z|<=1?
 
  • #5
sweetvirgogirl said:
f (z) = lamdba (a - z)/(1 - abar*z) |lambda| = 1; |a| <1
where f is any one-to-one analytic function mapping delta = {z:|z| < 1} onto itself
set g(z) = (a - f(z))/(1-abar*f(z))
now they say that g is one-to-one analytic function mapping delta onto delta (why?)

Work it out: f is a multiple of mobius transformation and g(z) is easy to work out.

Are you claiming that f is any analytic bijection? Or have you proved that all analytic bijections of the disc are of that form? What exactly is it you are trying to prove? It is very unclear from what you have written what is given and what is to be proven, if anything.

and g(0) = 0
and then they applied schwarz lemma to the function g(z) (why??)

presumably because it shows them something useful. as we have no idea what they want to prove we cannot explain it.

I don't see how |g(z)| <=1. and also why is g(z) is one-to-one analytic function mapping delta onto delta?

Whic part don't you see? Which part have you tried to prove?
 
Last edited:

1. What is the SC Lemma?

The SC Lemma, also known as the Syllogism Condition Lemma, is a mathematical theorem that states that if a set of premises entails a conclusion, then every subset of those premises also entails the same conclusion.

2. What are the conditions for applying the SC Lemma?

The conditions for applying the SC Lemma are that the premises must be true, the conclusion must be logically derived from the premises, and the premises must not contradict each other.

3. How is the SC Lemma used in practical applications?

The SC Lemma is used in practical applications to ensure that logical arguments and deductions are valid and sound. It is also used in computer programming and artificial intelligence to verify the correctness of logical reasoning.

4. Can the SC Lemma be applied to all types of logical arguments?

Yes, the SC Lemma can be applied to all types of logical arguments as long as they meet the conditions mentioned above. This includes deductive, inductive, and abductive arguments.

5. What are the limitations of the SC Lemma?

The SC Lemma is limited in its application to formal logic and may not be applicable to informal arguments or real-world situations. It also does not guarantee the truthfulness of the premises, only the validity of the argument.

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