Cauchy's Integral Formula problem

[Gavins]

Homework Statement

Suppose f is a holomorphic function on U = C - {0} and that
| f(z) | \leq | z |^{\frac{1}{2}}
for all z in U. Show that f is everywhere zero.

Homework Equations

That question is supposed to require Cauchy's Integral Formula.

The Attempt at a Solution

Choose R>0 and take points a in the disc D(0,R) such that by Cauchy's Integral Formula,
|f(a)| = | \frac{1}{2\pi i} \int_{\partial D} \frac{f(z)}{z-a} dz| \leq \max |\frac{f(z)}{z-a}| 2\pi R
I'm stuck on finding the max. I'm guessng I use the assumption but some help would be good. Thanks.

 

[snipez90]

You don't want to pick a point a in D(0,R). Rather, you want consider a in U and the disc D(a,R) where R > 0 (sufficiently small so D(a,R) does not intersect 0). Parametrize the boundary circle of the disc D(a, R) and you'll see why this takes care of the z-a in the denominator of the integrand. Then apply the given condition.

Also you need to determine f(0) separately, but why is f(0) = 0?

 

[Gavins]

Thanks. So doing what you've said,
f(a) = \frac{1}{2\pi i} \int_{\partial D} \frac{f(z)}{z -a}dz = \frac{1}{2\pi i} \int_0^{2\pi} \frac{f(a + Re^{i\theta})}{Re^{i\theta}} iRe^{i\theta} d\theta = \frac{1}{2\pi} \int_0^{2\pi} f(a + Re^{i\theta}) d\theta
where I have parametrised the contour as c = a + Re^iO.

The estimation lemma gives,
|\frac{1}{2\pi} \int_0^{2\pi} f(a + Re^{i\theta}) d\theta| \leq \max |f(a+Re^{i\theta})| 2\pi R = |a + Re^{i\theta}|^{\frac{1}{2}} 2 \pi R

Now what?

 

[snipez90]

If that were true, that would be great since we could let R tend to 0, so that f(a) = 0. But unfortunately, you can't apply the estimation lemma to a real integral, so that extra 2\pi R term should not be there. The final bound should have been just |a+ Re^{i\theta}|^{1/2} but actually now it's apparent this bound is useless (we can't make |f(a)| arbitrarily small). But at least now you get the idea of what we're trying to do.

Actually showing f(0) = 0 first is probably best, because then from http://en.wikipedia.org/wiki/Removable_singularity#Riemann.27s_theorem" we see that f is actually holomorphic on C, not just U = C - {0}. Then if you know the general Cauchy formula, you can show f' is identically equal to zero, which implies f is constant and hence identically equal to 0 (since f(0) = 0).

Your original estimate seems promising since you can bound the 1/|z-a| term using the reverse triangle inequality |z-a| \geq |z| - |a| and then let R go to zero, but the problem is that you fixed a point a already so it doesn't make sense to make the circles smaller and smaller by letting R to go zero.

 

[xeno_gear]

I've been following the thread.. seems interesting. Will the f(0) = 0 argument work? We only know that |f(z)| \le |z|^{1/2} on \mathbb{C} \backslash \{0\}.. or does the fact that f is holomorphic make it work?

 

[Dickfore]

Maybe you should follow an approach similar to the proof of Liouville's Theorem:

http://en.wikipedia.org/wiki/Liouville's_theorem_(complex_analysis)#Proof"

 

[snipez90]

@xeno_gear, I think it works though this is the first time I've actually used Riemann's theorem. You can check that criterion 3 and 4 hold, so we have a holomorphic extension to C. I don't know I was anticipating a flaw in this argument.

@Dickfore, yes this is the approach outlined in my last post, but obviously to use any of this you need to know how f behaves at 0, which is the only tricky part about this problem. That estimate in the proof of Liouville is Cauchy's estimate, and basically takes care of problems like these in under a minute provided you know f is well-behaved on your domain.

 

[Dickfore]

Hint: It's holomorphic in any neighborhood of 0.