bound of an analytic function, and a bit on uniform convergence

**hatsoff** · June 6th 2011, 08:28 AM

Another practice exam exercise:

Suppose $f$ is an analytic function in $D = \{z:|z|<1\}$ for which $\lVert f\rVert=\sup_{z\in F}\{(1-|z|^2)|f'(z)|\}$ is finite.

(a) Prove for $z\in D$ that $|f(z)-f(0)|\leq\frac{1}{2}\lVert f\rVert \log\frac{1+|z|}{1-|z|}$ .

(b) If $\{f_n\}$ is a sequence of analytic functions in $D$ for which

$\lim_{|z|\to 1^-}[(1-|z|^2)|f'_n(z)|]=0$ for $n=1,2,3,\cdots$ , and

$\lVert f_n-f\rVert \to 0$ as $n\to\infty$ , prove $\{f_n\}$ converges to $f$ uniformly on compact subsets of $D$ .

I can't do part (a) at all, and I'm not much further along on part (b). If I use the result of (a), all I need is to show that $f_n(0)\to f(0)$ , and then part (b) follows from Weierstrass. But I cannot show that $f_n(0)\to f(0)$ , nor can I show part (a) at all.

I also notice that $F$ is not explicitly defined. I presume it's just some arbitrary (but fixed) subset of $D$ . But perhaps it's a misprint...? I don't know...

As usual, any help would be appreciated!

**Opalg** · June 6th 2011, 11:29 AM

Originally Posted by hatsoff

Another practice exam exercise:

I can't do part (a) at all, and I'm not much further along on part (b). If I use the result of (a), all I need is to show that $f_n(0)\to f(0)$ , and then part (b) follows from Weierstrass. But I cannot show that $f_n(0)\to f(0)$ , nor can I show part (a) at all.

I also notice that $F$ is not explicitly defined. I presume it's just some arbitrary (but fixed) subset of $D$ . But perhaps it's a misprint...? I don't know...

As usual, any help would be appreciated!

On the assumption that F is a misprint for D, it follows from the definition of $\|f\|$ that $|f'(z)|\leqslant\frac{\|f\|}{1-|z|^2}$ . Therefore

$|f(z)-f(0)| = \Bigl|\int_{[0,z]}f'(\zeta)\,d\zeta\Bigr| \leqslant\int_{[0,z]}|f'(\zeta)|\,d\zeta \leqslant \|f\|\int_{[0,z]}\frac1{1-|\zeta|^2}\,d\zeta.$

Now do the integral (partial fractions).

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