For $\lambda\in \mathbb{D}$ fixed, let $C_{\phi}$ denote the composition operator on Hardy space $H^2(\mathbb{D})$ (I.e. $C_{\phi}f:=f(\phi)$) with symbol $\phi=\frac{\lambda-z}{1-\bar{\lambda}z}$ (a Mobius transform). And $k_{\lambda}=\frac{1}{1-\bar{\lambda}z}$ is the reproducing kernel of $H^2$ at ${\lambda}$ (I.e. $\forall f \in H^2$, the inner product $(f,k_{\lambda})=f(\lambda)$). The question is: Does $||C_{\phi}(f)\cdot k_{\lambda}||=||f||\cdot ||k_{\lambda}||$ always hold, for $f\in H^2$?
This is a claim (without proof) in the paper I’m reading, but I don’t know why. I made the following calculation:
By the proposition of “reproducing kernel”, we have $$||C_{\phi}f\cdot k_{\lambda}||^2=(C_{\phi}f\cdot k_{\lambda}, C_{\phi}f\cdot k_{\lambda})=(\overline{C_{\phi}f}C_{\phi}f \cdot k_{\lambda},k_{\lambda})=|C_{\phi}f(\lambda)|^2\cdot k_{\lambda}(\lambda)=|f(\phi(\lambda))|^2(k_{\lambda}, k_{\lambda})=|f(0)|^2||k_{\lambda}||^2.$$ But this is not equal to $||f||\cdot||k_{\lambda}||$.
Is there something wrong with me? Or something wrong with the claim? Any help will be appreciated!
The third “=” is wrong. For the proof, just check $f\mapsto f\circ \phi \cdot k_{\lambda}$ is a self-adjoint unitary operator.