Liouville’s Theorem

Spring 2020.3, Extras Fall 2009 #complex/qual/completed

problem (?):

    
  • Assume f(z)=n=0cnzn converges in |z|<R. Show that for r<R,

12π2π0|f(reiθ)|2dθ=n=0|cn|2r2n

  • Deduce Liouville’s theorem from (a).
solution:

Computing the LHS: [0,2π]|f(reiθ)|2dθ=[0,2π]f(reiθ)¯f(reiθ)dθ=[0,2π]k0ckrkeikθj0¯cjrjeijθdθ=[0,2π]k,j0ck¯cjrk+jei(kj)θdθ=k,j0ck¯cjrk+j[0,2π]ei(kj)θdθ=k,j0ck¯cjrk+jχi=j2π=k0ck¯ckr2k2π=2πk0|ck|2r2k, where we’ve used that the series converges uniformly in its radius of convergence to commute sums and integrals.

Now supposing |f(z)|M for all zC, if f is entire then k0ckzk converges for all r, so k0|ck|2r2k=12π[0,2π]|f(reiθ)|2dθ12π[0,2π]M2dθ=M2. Thus for all r, |c0|2+|c1|2r2+|c2|2r4+M2, and taking r forces |c1|2=|c2|2==0. So f(z)=c0 is constant.

FTA via Liouville #complex/exercise/completed

problem (?):

Prove the Fundamental Theorem of Algebra (using complex analysis).

solution:

    
  • Strategy: By contradiction with Liouville’s Theorem
  • Suppose p is non-constant and has no roots.
  • Claim: 1/p(z) is a bounded holomorphic function on C.
    • Holomorphic: clear? Since p has no roots.

    • Bounded: for z0, write P(z)zn=an+(an1z++a0zn).

    • The term in parentheses goes to 0 as |z|

    • Thus there exists an R>0 such that |z|>R|P(z)zn|c:=|an|2.

    • So p is bounded below when |z|>R

    • Since p is continuous and has no roots in |z|R, it is bounded below when |z|R.

    • Thus p is bounded below on C and thus 1/p is bounded above on C.

  • By Liouville’s theorem, 1/p is constant and thus p is constant, a contradiction.

Entire functions satisfying an inequality #complex/exercise/completed

problem (?):

Find all entire functions that satisfy |f(z)||z|zC. Prove this list is complete.

concept:

    
  • If f is bounded in a neighborhood of a singularity z0, then z0 is removable.
solution:

    
  • Suppose f is entire and define g(z):=zf(z).
  • By the inequality, |g(z)|1, so g is bounded.
  • g potentially has singularities at the zeros Zf:=f1(0), but since f is entire, g is holomorphic on CZf.
  • Claim: Zf={0}.
    • If f(z)=0, then |z||f(z)|=0 which forces z=0.
  • We can now apply Riemann’s removable singularity theorem:
    • Check g is bounded on some open subset D{0}, clear since it’s bounded everywhere
    • Check g is holomorphic on D{0}, clear since the only singularity of g is z=0.
  • By Riemann’s removable singularity theorem, the singularity z=0 is removable and g has an extension to an entire function ˜g.
  • By continuity, we have |˜g(z)|1 on all of C
    • If not, then |˜g(0)|=1+ε>1, but then there would be a domain ΩC{0} such that 1<|˜g(z)|1+ε on Ω, a contradiction.
  • By Liouville, ˜g is constant, so ˜g(z)=c0 with |c0|1
  • Thus f(z)=c10z:=cz where |c|1

Thus all such functions are of the form f(z)=cz for some cC with |c|1.

Entire functions with an asymptotic bound #complex/exercise/completed

problem (?):

Find all entire functions satisfying |f(z)||z|12 for |z|>10.

solution:

Since f is entire, take a Laurent expansion at z=0, so f(z)=k0ckzk where 2πik!ck=f(k)(0) by Cauchy’s integral formula. Take a Cauchy estimate on a disc of radius R>10: |ck|k!2π|z|=R|f(ξ)(ξ0)k+1|dξk!2π|z|=R|ξ|12|ξ|k+1dξ=k!2π1Rk+122πR=O(1/Rk12). So in particular, if k1 then k12>0 and ck=0. This forces f=c0 to be constant.

Tie’s Extra Questions: Fall 2009 #complex/exercise/completed

problem (?):

Let f(z) be entire and assume values of f(z) lie outside a bounded open set Ω. Show without using Picard’s theorems that f(z) is a constant.

solution:

We have |f(z)|M for some M, so |1/f(z)|M1 is bounded, and we claim it is entire as well. This follows from the fact that 1/f has singularities at the zeros of f, but these are removable since 1/f is bounded in every neighborhood of each such zero. So 1/f extends to a holomorphic function. But now 1/f=c is constant by Liouville, which forces f=1/c to be constant.

Tie’s Extra Questions: Fall 2015 #complex/exercise/completed

problem (?):

Let f(z) be bounded and analytic in C. Let ab be any fixed complex numbers. Show that the following limit exists: lim

Use this to show that f(z) must be a constant (Liouville’s theorem).

solution:

Apply PFD and use that f is holomorphic to apply Cauchy’s formula over a curve of radius R enclosing a and b: \begin{align*} \int_\gamma {f(z) \over (z-a)(z-b)}\,dz &= \int_\gamma f(z)\qty{{a-b \over z-a} + {b-a\over z-b} } \,dz\\ &= (a-b)^{-1}\int_\gamma {f(z) \over z-a} \,dz+ (b-a)^{-1}\int_\gamma {f(z) \over z-b}\,dz\\ &= (a-b)^{-1}\cdot 2\pi i f(a) + (b-a)\cdot 2\pi i f(b)\\ &= 2\pi i\qty{f(a) - f(b) \over a-b } .\end{align*} Since f is bounded, this number is finite and independent of R, so taking R\to\infty preserves this equality. On the other hand, if {\left\lvert {f(z)} \right\rvert}\leq M, then we can estimate this integral directly as \begin{align*} I \leq \int_{{\left\lvert {z} \right\rvert} = R} {M \over {\left\lvert {R-a} \right\rvert} \cdot {\left\lvert {R-b} \right\rvert} } = {M\cdot 2\pi R \over {\left\lvert {R-a} \right\rvert} \cdot {\left\lvert {R-b} \right\rvert}} \ll {1\over R} \to 0 ,\end{align*} which forces f(a) =f(b). Since a, b were arbitrary, f must be constant.

#complex/qual/completed #complex/exercise/completed