Proving |f(z)|≤ M in a Closed Contour C

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In summary, the problem from Mathematical methods for physicists by Arfken, problem 6.4.7 discusses the boundedness of a function f(z) within a closed contour C. If f(z) is analytic within C and continuous on C, and if f(z) ≠ 0 within C and |f(z)|≤ M on C, it can be shown that |f(z)|≤ M for all points within C. The hint is to consider w(z) = 1/f(z). However, the author also mentions a subsequent problem where it is shown that this result does not hold if f(z) = 0 within C. This could potentially be due to the minimum modulus principle, which states that if a hol
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Anowar
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This problem is from Mathematical methods for physicists by Arfken, problem 6.4.7.

A function f(z) is analytic within a closed contour C (and continuous on C). If f(z) ≠ 0 within C and |f(z)|≤ M on C, show that |f(z)|≤ M for all points within C.

The hint is to consider w(z) = 1/f(z).

I have tried to solve this in this way using Gauss' mean value theorem. Suppose, there is a point within C where |f(z)|≥ M. As |f(z)|≤ M on C, there has to be a local maximum of |f(z)| within C. We call that point of local maximum z°. Then we can take a circle with a however small radius r around z°. From Cauchy's integral theorem,

f(z°) = (1/2πi)∫c[f(z)/(z-z°)] dz (here c is the small circle of radius r)

Now taking, z = z° + r*exp(iθ)
f(z°) = (1/2π)∫ f(z° + r*exp(iθ)) dθ

Taking the modulus, |f(z°)| = 1/2π |∫ f(z° + r*exp(iθ)) dθ| ≤ 1/2π ∫ |f(z° + r*exp(iθ))| dθ
(in all cases above, the integral is taken from 0 to 2π).

Now, m(f) = 1/2π ∫ |f(z° + r*exp(iθ))| dθ is the mean value of |f(z)| on the circle of radius r. From above, we rewrite,
|f(z°)| ≤ m(f).

Again, there must be at least one point z1 on the circle where |f(z1)| is larger than m(f) or the |f(z)| is equal to m(f) for all the points on the circle. So, |f(z1)| ≥ m(f) ≥ |f(z°)|. We can take r as small as we want to make z1 as nearer to z°. So, there is point z1 in the neighborhood of z° for which, |f(z1)| ≥ |f(z°)|, so |f(z°)| can't be a local maximum, and thus there is no point inside C such that |f(z)| ≥ M.

Now, my problem is that, I didn't use the condition f(z) ≠ 0 within C for this proof. So, there should be a problem with my solution, but I couldn't find any. Now, thinking about the hint, taking w(z) = 1/f(z), we don't know whether w(z) is continuous on C or not. For, f(z) = 0 for some point on the contour C, w(z) is not continuous and we can't take Cauchy integral of w(z) on C to be zero. But, for f(z), as it is continuous on C and analytic inside, it's contour integral on C would be zero from Cauchy-Goursat theorem. I think I could use this to solve the problem, but I failed to find any path to do so. I would appreciate it very much if someone can point out where I am doing wrong or give some more hints to solve this. I am sorry for the bad format of equations as I am not yet used to using latex.
 
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Its late so maybe I am missing something but you don't need f≠0. This follows from the maximum modulus principle that a holomorphic function on a compact subset of ℂ attains its maximum value on the boundary.
 
  • #3


deluks917 said:
Its late so maybe I am missing something but you don't need f≠0. This follows from the maximum modulus principle that a holomorphic function on a compact subset of ℂ attains its maximum value on the boundary.

But, in the text, the right next problem is,

If f(z) = 0 within the contour C, show that the foregoing result (the statement of the previous problem) does not hold and that it is possible to have |f(z)| = 0 at one or more points in the interior with |f(z)| > 0 over the entire bounding contour.

As it says, it does not hold if f(z) = 0 then there have to be some problem with the text then. Any clue?
 
  • #4


Anowar said:
But, in the text, the right next problem is,

If f(z) = 0 within the contour C, show that the foregoing result (the statement of the previous problem) does not hold and that it is possible to have |f(z)| = 0 at one or more points in the interior with |f(z)| > 0 over the entire bounding contour.

I don't see how that contradicts your first statement:

A function f(z) is analytic within a closed contour C (and continuous on C). If f(z) ≠ 0 within C and |f(z)|≤ M on C, show that |f(z)|≤ M for all points within C.

I think what the question is trying to get at is the "minimum modulus principle." That is if f is never zero then |f| attains its minimum on the boundary as well.
 
  • #5


deluks917 said:
I don't see how that contradicts your first statement:

The first problem asks to show that |f(z)|≤ M if f(z)≠0 within C and the second one asks to show that, |f(z)|≤ M does not hold if f(z)= 0 within C.

If the f(z) has its minimum on the boundary, how can it help me to show that |f(z)| is bounded within the region? Thanks for you help, I really appreciate your kind replies.
 
  • #6


I think the question is written wrong. If |f| > 0 on the boundary but f = 0 at some point inside this doesn't contradict the fact f has its "maximum modulus" on the boundary.
 
  • #7


I think the question doesn't want to arrive at maximum modulus principle, it is asking to proof that a function is bounded within the region if its bounded on the boundary. Am I missing something here?
 

1. How do you define |f(z)|?

|f(z)| refers to the absolute value of the complex function f(z). In other words, it is the magnitude or modulus of f(z), which is a measure of the distance from the origin to the point on the complex plane that represents f(z).

2. What is a closed contour C?

A closed contour C is a continuous curve that starts and ends at the same point, forming a closed loop. It can be any shape or size, as long as it does not intersect itself.

3. What does it mean for |f(z)| to be ≤ M?

This means that the absolute value of f(z) is always less than or equal to a constant value M. In other words, the distance between f(z) and the origin is always within a certain range defined by M.

4. How can you prove that |f(z)|≤ M in a closed contour C?

To prove this, you need to show that for any point z on the contour C, the absolute value of f(z) is always less than or equal to M. This can be done by using various techniques such as the maximum modulus principle or the Cauchy integral formula.

5. Why is proving |f(z)|≤ M in a closed contour C important?

This is important because it allows us to make conclusions about the behavior of a complex function f(z) within a given region. It also helps us to evaluate contour integrals and solve various problems in complex analysis.

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