# Continuous functional on $H^2$

Let $\mathbb{D}=\{z: |z|<1\}$ and $a\in \mathbb{D}$ is fixed. Suppose $H^2=\{f(z)=\sum_{n=0}^{\infty}a_nz^n: \sum_{n=0}^{\infty}|a_n|^2<\infty\}$ and $\langle f,g\rangle_{H^2}=\sum_{n=0}^{\infty}a_n\overline{b_n}$. Define $$\begin{array}{l} \varphi_a:H^2\to \mathbb{C}\\ \varphi_a(f):=f'(a) \quad (f\in H^2)\end{array}$$ (a) Show that $\varphi_a$ is a continuous functional on $H^2$.

(b) Find the function $\psi_a\in H^2$ such that $$\varphi_a(f)=f'(a)=\langle f,\psi_a\rangle_{H^2}\quad (f\in H^2)$$

a) \begin{align*} |\varphi_{a}(f)|&\leq\sum_{n=1}^{\infty}n|a_{n}||a|^{n-1}\\ &\leq\left(\sum_{n=1}^{\infty}n^{2}|a|^{2(n-1)}\right)^{1/2}\left(\sum_{n=1}^{\infty}|a_{n}|^{2}\right)^{1/2}\\ &=\left(\sum_{n=1}^{\infty}n^{2}|a|^{2(n-1)}\right)^{1/2}\|f\|_{H}. \end{align*} b) Try $\psi_{a}(z)=\displaystyle\sum_{n=1}^{\infty}n\overline{a}^{n-1}z^{n}$.
• Thanks. Why $\left(\sum_{n=1}^{\infty}n^{2}|a|^{2(n-1)}\right)^{1/2}<\infty$ ? – Fair Mar 12 '18 at 18:48
• Try to use root test to just $\displaystyle\sum_{n=1}^{\infty}n^{2}|a|^{2(n-1)}$. – user284331 Mar 12 '18 at 18:48