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My question is in the title. I'm a little lost in how to solve this problem. There is a hint associated with the problem that states the following:

Use a line integral to show that $ |f(z)| \leq A + |z| $ where A = $\max\{1, |f(0)|\}$

Thanks for the help.

Elias Costa
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Max
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1 Answers1

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We may as well suppose $|f(z)| > 1$

Take the line $[0,z]$ and let $d = \sup \{ w : w\in (0,z), |f(w)| \leq 1\}$ and $d = 0$ if this set is empty. Then $|f'(z)| \leq 1$ on $[d,z]$.

Then $$ |z| \geq |z - d| \geq \int_d^z |f'(w)| dw \geq \left|\int_d^z f'(w) dw \right| = |f(z) - f(d)| $$

Now $|f(d)| = \max \{ |f(0)|, 1 \}$. Because of the inequality $|a - b| \geq ||a| - |b|| \geq |a| - |b|$, we therefore have $|z| + |f(d)| \geq |f(z)|$

Excuse the notation, once I restrict to the line $[0,z]$, I borrow a lot of the language from real numbers over.

muzzlator
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  • I was just going to update my post with something similar. The problem I am having is getting the signs to line up correct. How did you get $ |z| + |f(d)| \geq |f(z)| $? – Max Mar 26 '13 at 14:39
  • $|a - b| \geq ||a| - |b|| \geq |a| - |b|$ – muzzlator Mar 26 '13 at 14:46
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    Nice argument. I think there is a little problem. You should probably define $d$ as the sup (=max) of $w$ such that $|f(w)|\leq 1$. Then for all $w$ in $[d,z]$, except for $d$, we have $|f(w)|>1$ hence $|f'(w)|\leq 1$. You need $>1$ to deduce $\leq 1$ given the assumption. – Julien Apr 19 '13 at 12:57
  • Thanks, I'll edit it with the correction – muzzlator Apr 20 '13 at 02:37
  • @muzzlator Your answer is very nice. But I am tempted think about generalizations. Suppose, for each $z$, $|f(z)| \leq 1$ or $|f'(z)| \leq 1$ or $|f''(z)| \leq 1$. Can we show that $f$ is a polynomial? – Kavi Rama Murthy Apr 10 '19 at 07:57