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$f$ is analytic in $D$: $|z| < 1$ such that $|f(z)| ≤ 1$. $g$: $D → C$ is given by $g(z) = f(z)/z$ when $z \neq 0$ & $f'(0)$ when $z = 0$. Then which of the followings are true:

1) $g$ is analytic on $D$.

2) $|g(z)| ≤ 1$ on $D$.

3) $|f'(z)|≤ 1$ on $D$.

4) $|f'(0)| ≤ 1$.

I used the Taylor series representation of $f$ on $D$ but couldn't come to any conclusion.

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Did you mean to also say that $f(0)=0$? Otherwise $g$ is obviously unbounded and discontinuous. –  Jonas Meyer Nov 24 '12 at 3:16
No. Such condition is not given. –  Sugata Adhya Nov 24 '12 at 3:23
Sugata: Consider an example where $f(0)\neq 0$ and see what happens then. –  Jonas Meyer Nov 24 '12 at 3:25
I've shown (3) is incorrect by using the holomorphic function f(z) = z^4 on D & using the point z = cube root of 1/2. –  Sugata Adhya Nov 24 '12 at 3:26
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2 Answers

up vote 4 down vote accepted

For (i) and (ii), consider the function $f(z) = z/2+1/2$ on the unit disc, certainly $|f(z)|\le |z|/2+1/2 \le 1$, but $g(z)=1/2+1/(2z)$ has a pole at the origin and so cannot be analytic and $|g(z)|$ is not bounded.

For (iii) Sugata Adhya provides a counter example.

So (iv) has to be correct, and it follows form lee's suggestion:

recall the Cauchy integral formula for the derivatives: $$ f^{'}(0) = \frac{n!}{2\pi i }\int_{\partial B(0,r)} \frac{f(z)}{z^2}dz. $$ where $r<1$ so the ball $B(0,r)$ lies in the unit circle.

Using the standard estimation, we have $$ |f'(0)|\le \frac{1}{2\pi}(2\pi r) \max_{|z|=r}\{\frac{|f(z)|}{|z|^2}\} \le \frac{1}{r} $$ since $|f(z)|\le 1$.

Because this is true for every $r < 1$, passing to the limit shows $|f'(0)|\le 1$.

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Thanks. That (1) is incorrect can also be shown using the function f(z) = cos z on D. But no conclusion on (2) can be made out of it. You approached in a better way. –  Sugata Adhya Nov 24 '12 at 9:10
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Use Schwartz(I'm not sure about this name)'s theorem...

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Would you please state the theorem ? –  Sugata Adhya Nov 24 '12 at 2:57
@SugataAdhya:Schwartz Lemma –  y zhao Nov 24 '12 at 3:10
Yeah.. It is Schwartz lemma.. You can google it. –  lee Nov 24 '12 at 3:20
lee, there's no reason $|F(z)|$ would be bounded by $1$, and if it were, how would that relate back to the original problem? –  Jonas Meyer Nov 24 '12 at 3:30
And the proposition4.. Use the fact that integration of f(z)/z^2 over a circle around 0 is exactly f'(0)(I remember this but my memory isn't reliable TAT) –  lee Nov 24 '12 at 3:43
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