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

  1. #1
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    bound of an analytic function, and a bit on uniform convergence

    Another practice exam exercise:

    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!
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  2. #2
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    Quote Originally Posted by hatsoff View Post
    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).
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