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

• Jun 6th 2011, 08:28 AM
hatsoff
bound of an analytic function, and a bit on uniform convergence
Another practice exam exercise:

Quote:

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

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

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

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

$\displaystyle \lVert f_n-f\rVert \to 0$ as $\displaystyle n\to\infty$, prove $\displaystyle \{f_n\}$ converges to $\displaystyle f$ uniformly on compact subsets of $\displaystyle D$.
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 $\displaystyle f_n(0)\to f(0)$, and then part (b) follows from Weierstrass. But I cannot show that $\displaystyle f_n(0)\to f(0)$, nor can I show part (a) at all.

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

As usual, any help would be appreciated!
• Jun 6th 2011, 11:29 AM
Opalg
Quote:

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 $\displaystyle f_n(0)\to f(0)$, and then part (b) follows from Weierstrass. But I cannot show that $\displaystyle f_n(0)\to f(0)$, nor can I show part (a) at all.

I also notice that $\displaystyle F$ is not explicitly defined. I presume it's just some arbitrary (but fixed) subset of $\displaystyle D$. 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 $\displaystyle \|f\|$ that $\displaystyle |f'(z)|\leqslant\frac{\|f\|}{1-|z|^2}$. Therefore

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

Now do the integral (partial fractions).