0
$\begingroup$

Let $f$ be a holomorphic function in the Hardy space $H^2(U)$ with $\parallel f\parallel_{H^2}\leq 1$. For a given $\alpha\in U$(the open unit disk) Find the upper bound of $|f'(\alpha)|$.

We know that $\parallel f\parallel_{H^2}=(\int_0^{2\pi}|f(re^{i\theta})d\theta|)^{1/2}$. Consider $\varphi_{-\alpha}(z)=\frac{z+\alpha}{1+\bar{z}\alpha}$ and let $g=f\circ\varphi_{-\alpha}$.

If $\parallel g\parallel_{H^2}\leq 1$, then $$|f'(\alpha)|=\frac{|g'(0)|}{|\varphi'_{-\alpha}(0)|}\leq \frac{|g(0)|}{r(1-|\alpha|^2)}$$ by Cauchy's estimate for some $0<r<1$. Since $$|g(0)|=\frac{1}{2\pi}|\int_0^{2\pi}g(re^{i\theta})d\theta|\leq \frac{1}{2\pi}\int_0^{2\pi}|g(re^{i\theta})|^2rd\theta \cdot \int_0^{2\pi}rd\theta$$ by mean value principle and Cauchy-Schwartz inequality.

Then $$|f'(\alpha)|\leq \frac{1}{1-|\alpha|^2}$$.

I want to ask whether $\parallel g\parallel_{H^2}\leq 1$ is true. If not, how can I get the upper bound of $|f'(\alpha)|$?

$\endgroup$
0
$\begingroup$

Your estimation $$ |f'(\alpha)|\leq \frac{1}{1-|\alpha|^2} $$ is wrong.

Let $f(z)\in H^2(U)$ with $\parallel f\parallel_{H^2}\leq 1$ and it's Taylor expansion be $\sum_{n=0}^\infty a_nz^n \,(|z|<1)$.
Then \begin{align} &\int_0^{2\pi} |f(re^{i\theta }|^2\,d\theta=\int_0^{2\pi} \left(\sum_{n=0}^\infty a_nz^n\right)\overline{\left(\sum_{n=0}^\infty a_nz^n\right)}d\theta =2\pi\sum_{n=0}^\infty |a_n|^2r^{2n},\\ &\parallel f\parallel_{H^2}=\sup_{r<1}\left(\frac{1}{2\pi}\int_0^{2\pi} |f(re^{i\theta }|^2\,d\theta \right)^\frac{1}{2}=\left(\sum_{n=0}^\infty |a_n|^2\right)^\frac{1}{2}\le 1. \end{align}

Using Cauchy-Schwarz inequality we have \begin{align} |f^\prime(\alpha )|&=\left| \sum_{n=1}^\infty na_n\alpha ^{n-1}\right|\le \left( \sum_{n=1}^\infty |a_n|^2\cdot\sum_{n=1}^\infty n^2|\alpha |^{2(n-1)}\right)^\frac{1}{2}\\ &\le \left(\sum_{n=1}^\infty n^2|\alpha |^{2(n-1)}\right)^\frac{1}{2}=\frac{\sqrt{1+|\alpha |^2}}{(1-|\alpha |^2)^\frac{3}{2}},\tag{1} \end{align} since $$ \sum_{n=1}^\infty n^2x^{n-1}=\frac{1+x}{(1-x)^3}\quad (|x|<1).$$

The estimation $(1)$ is sharp. Fix $\alpha $ with $0<\alpha <1$ and let $\beta =\frac{\sqrt{1+\alpha ^2}}{(1-\alpha ^2)^\frac{3}{2}}$. Define $$ f(z)=\frac{1}{\beta }\sum_{n=1}^\infty n\alpha ^{n-1}z^n.$$ Then $f\in H^2(U), \,\parallel f\parallel_{H^2}\leq 1$ and $$ |f^\prime(\alpha )|=\frac{\sqrt{1+\alpha ^2}}{(1-\alpha ^2)^\frac{3}{2}}.$$ Note that $$\frac{\sqrt{1+\alpha ^2}}{(1-\alpha ^2)^\frac{3}{2}}>\frac{1}{1-|\alpha|^2}.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.