Inverse of the adjoint of the shift operator

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

Thanks in advance

 

[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 $H^2$.

 

[Likemath2014]

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

 

[micromass]

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

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

 

[blue_raver22]

for your help!

Hello,

First, let's define the adjoint of the shift operator. The adjoint, denoted by S^*, is the operator on the Hardy space H^2 such that for any two functions f and g in H^2, we have \langle Sf,g\rangle = \langle f,S^*g\rangle, where \langle \cdot,\cdot \rangle denotes the inner product on H^2.

Now, to prove the given identity, we will use the fact that the inverse of an operator A is given by (1-A)^{-1}=\sum_{n=0}^{\infty}A^n. So, we have

(1-\lambda S^*)^{-1}S^*f (z) = \sum_{n=0}^{\infty}(\lambda S^*)^nS^*f(z)

= \sum_{n=0}^{\infty}\lambda^n S^{*n}S^*f(z)

= \sum_{n=0}^{\infty}\lambda^n S^{*n+1}f(z)

Now, using the definition of the adjoint and the fact that (Sf)(z)=zf(z), we have

= \sum_{n=0}^{\infty}\lambda^n S^{*n+1}f(z)

= \sum_{n=0}^{\infty}\lambda^n zS^{*n}f(z)

= \sum_{n=0}^{\infty}\lambda^n zf(z)

= \frac{f(z)-f(\lambda)}{z-\lambda}

which proves the given identity.

I hope this helps. Let me know if you have any further questions.