Entire function with positive real part is constant (no Picard)

Question

A problem asks to show that an entire function on $\mathbb{C}$ with positive real part must be a constant. I spoke to a professor, and asked why not just use the Picard theorem. He said that we should try to aim the solution at the level of the problem, and that Picard was a little too high-powered for this problem.

How would I solve it in the absence of Picard's theorem? A related question: suppose we have an isolated singularity, near which $\Re (f)$ (alternately, $\Im (f)$) is bounded. How can we show the singularity is removable?

Yet another related problem: why is a positive harmonic function on Rn a constant? Mean value property seems not to be the way...

score 13 · Answer 1 · edited Oct 29 '17 at 20:21

13

Write $f = u + iv$, where by assumption $u \geq 0$. Consider the analytic function

$$\exp(-f), \,\,\,\,\,\,\, |\exp(-f)| = |\exp(-u)|$$

What can we say about $\exp(-f)$? (Hint: Louiville's Theorem)

edited Oct 29 '17 at 20:21

amWhy

209,954

answered Nov 10 '13 at 18:37

Elchanan Solomon

30,807

2

I am just curious. If hypothesis was negative real part then one could take the function $exp (f)$ to show that $f$ is constant? And if hypothesis is positive imaginary part then $exp(if(z))$ and negative imaginary part then $exp(-if(x))$? Am I correct? – Babai Jul 26 '17 at 19:59
@Babai yes, you are correct. – john doe Oct 29 '17 at 19:50

score 6 · Answer 2 · answered Nov 10 '13 at 18:38

6

Compose your map with $e^{-z}$, and you get a bounded entire function.

answered Nov 10 '13 at 18:38

Per Erik Manne

11,106

score 4 · Accepted Answer · answered Nov 10 '13 at 18:36

4

Hint: a fractional linear transformation (aka Möbius transformation) takes a half-plane to a disk.

answered Nov 10 '13 at 18:36

Robert Israel

448,999

So it seems that for g the FLT, g circ f is entire and bounded, so constant, but then f is constant since g is bijective? Would you say that is sound logic? – user88203 Nov 10 '13 at 18:47
Yes, that is sound. – Robert Israel Nov 10 '13 at 19:16
1

@ Robert Israel: Unfortunately, the composition of a fractional linear transformation and an entire function is not an entire function in the general case. – user64494 Nov 10 '13 at 20:16
1

@user64494: In general, the composition is a meromorphic function. But in this case it is a bounded meromorphic function, i.e. an entire function. – Robert Israel Nov 10 '13 at 21:25
@ Robert Isriel: Could you explain why a bounded meromorphic function is an entire function? – user64494 Nov 11 '13 at 10:20
Its singularities are all removable. – Robert Israel Nov 11 '13 at 17:50

Styles · Answer 4 · 2017-10-21T05:56:03.973

0

Let $f(z) $ be an entire and bounded function with real part bounded,then

$f(z)$ is entire$\implies \phi(z)=e^{f(z)}$ entire

$\implies \vert\phi(z)\vert=\vert e^{f(z)}\vert= e^{Re(f(z))} \le e^M$ ,Where $Re(z) \le M$ for some fixed $M \in \mathbb R$

$\implies \phi(z)$ is constant,by Liouville's theorem

$\implies\phi'(z)=e^{f(z)}f'(z)=0\forall z\in \mathbb C$

$\implies f'(z)=0 $ in $\mathbb C$,since $e^{f(z)}\neq 0$ in $\mathbb C$

$\implies f(z)$ is constant

edited Oct 21 '17 at 05:56

answered Oct 20 '17 at 20:53

Styles

3,539

Why this answer is downvoted? – Styles Oct 21 '17 at 05:39
1

It says real part is positive, real part may or may not be bounded. But you assume it to be bounded – john doe Oct 29 '17 at 19:45

Entire function with positive real part is constant (no Picard)

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