# Inverse of the adjoint of the shift operator

1. Mar 14, 2014

### Likemath2014

Hi there,

Let $S$ denote the shift operator on the Hardy space on the unit disc $H^2$, that is $(Sf)(z)=zf(z)$.

My question is to show the following identity

$(1-\lambda S^*)^{-1}S^*f (z)=\frac{f(z)-f(\lambda)}{z-\lambda},$

where $\lambda,z\in\mathbb{D}$

2. Mar 14, 2014

### micromass

First of all, can you figure out what $S^*$ does exactly? This can be made easy if you can figure out an orthonormal basis of $H^2$.

3. Mar 14, 2014

### Likemath2014

Yes, $S^*=\frac{f(z)-f(0)}{z}$. But my problem is with the term $(1-\lambda S^*)^{-1}$.

4. Mar 14, 2014

### micromass

Just put it on the other side. So you need to prove

$$S^*f(z) = (1-\lambda S^*)\frac{f(z)-f(\lambda)}{z-\lambda}$$