Prove that f is constant on omega

[docnet]

Homework Statement

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Relevant Equations

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$f(z)$ is holomorphic on $\Omega$ so f(z) satisfies the C.R. equations, i.e.,

for $f(z)=u+iv$

$u_x=v_y$
$u_y=-v_x$

and for $-f(z)=-u-iv$

$u_x=v_y \Rightarrow -u_x=-v_y$
$u_y=-v_x\Rightarrow -u_y=v_x$

so -f(z) satisfies the C.R. equations and hence $-f(z)$ is holomorphic on $\Omega$.

$|f(z_0)|\leq |f(z)|\Rightarrow |-f(z_0)|\geq |-f(z)|$

$|-f(z_0)|=sup_{z\in\Omega}|f(z)| \Rightarrow$ -f(z) is onstant in $\Omega$ by the maximum principle. f(z) is constant in $\Omega$.

 

[Office_Shredder]

Putting a minus sign*inside* the absolute values doesn't flip the inequality.

Hint: you are told the function doesn't vanish anywhere. You should make use of this fact. Which mathematical operation isn't valid when you plug in zero?

 

[docnet]

Putting a minus sign*inside* the absolute values doesn't flip the inequality.

Hint: you are told the function doesn't vanish anywhere. You should make use of this fact. Which mathematical operation isn't valid when you plug in zero?

Well, the division operation isn't valid when you plug in zero.

so $$\frac{f_0}{f}\leq 1$$ but I'm not sure where to go from here.another way I tried to prove this without needing $f>0$, but most likely some steps are incorrect.

$f$ is holomorphic on $\Omega$. The mean value theorem says that the average value of $f$ on any disc $B_R$ which is a subset of $\Omega$ is equal to $f_0$, i.e.,

$$|f_0|=|\frac{1}{2\pi}\int_{B_R(z_0)}f(z)ds|$$

If one assumes that there exists $z_0$ such that $|f_0|\leq |f|$ for any z, the mean value theorem requires that $f=f_0$ everywhere on the ball. It is obvious that this holds for the case $|f_0|=|f|$, since the average of a constant function $f_0$ over its domain is just $f_0$. We can show that this does not hold for the strict inequality case. Suppose $|f|>|f_0|$ for at some $z^*\neq z_0$ on the disc. The mean value theorem would yet say that the average of $f$ on the disc is $f_0$. Yet the average value would be greater than $f_0$.

$$| \frac{1}{2\pi}\int_{B_R(z_0)}(f)ds- \frac{1}{2\pi}\int_{B_R(z_0)} (f_0) ds|=|\frac{1}{2\pi}\int_{B_R(z_0)}(f-f_0) ds|$$

where $0<|f|-|f_0| \leq |f-f_0|$ on $z^*$ assures that $\frac{1}{2\pi}\int_{B_R(z_0)}(f-f0) ds>0$, leading to a contradiction. So it must be true that $f=f_0$ everywhere on the disc. In fact, open discs form a basiss of $\Omega$ so $f=f_0$ on $\Omega$.

 

[Office_Shredder]

You can also just apply the maximum principle to 1/f.

I don't think what you've done here is valid, since you've written things like $0 < f-f_0$, but the right hand side is a complex number, not a real number.

 

[docnet]

You can also just apply the maximum principle to 1/f.

I don't think what you've done here is valid, since you've written things like $0 < f-f_0$, but the right hand side is a complex number, not a real number.

TY so much

$f$ is defined over $\Omega$.

since $f\neq 0$, $\frac{1}{f}$ is defined over $\Omega$

Let $\frac{1}{f}=h$ and $h \cdot f = 1$. holomorphic functions form a Borel algebra so $h$ is holomorphic on $\Omega$.

$|f_0|$ is the minimum value of $f$ at $z_0\in \Omega$. So $\frac{1}{f}=h$ attains its maximum value $\frac{1}{|f_0|}$ at $z_0\in\Omega$.

$h$ is holomorphic and attains its maximum on the interior of $\Omega$, so the maximum principle leads to $h$ being constant on $\Omega$, hence $f$ is constant, which leads to $f=f_0$.

 

[docnet]

finally edited the mistakes and the proof is finally right... ?

 

[Office_Shredder]

It looks good to me.