Hypercontractions and automorphisms of the unit disc Recall that an bounded operator $T$ on a Hilbert space $\mathcal H$ is said to be $n$-hypercontraction for $n\in\mathbb N$ if 
$$ I- {n \choose 1} T^*T + {n \choose 2} {T^*}^2T^2-\cdots+ (-1)^{n}{n \choose n}{T^*}^nT^n \geq 0.$$
Let $T$ be an bounded operator on a Hilbert space $\mathcal H$ such that its spectrum $\sigma(T) \subseteq \mathbb D$ and let $\varphi$ be an automorphism of unit disc given by $\varphi(z)= \frac{z-a}{1-\bar{a}z},\,\,a\in \mathbb D.$ It is easy to see that 
\begin{align}
I-\varphi(T)^* \varphi(T) &= (1-|a|^2)(1- a T^*)^{-1}(I-T^*T)(1-\bar{a}T)^{-1}\\
I- 2\varphi(T)^*\varphi(T)+ {\varphi(T)^*}^2 \varphi(T)^2&=(1-|a|^2)^2(1- a T^*)^{-2}( I- 2T^*T +{T^*}^2T^2) (1-\bar{a}T)^{-2}.
\end{align}
From there it follows that if $T$ is a $n$-hypercontraction then $\varphi(T)$ is also a $n$-hypercontraction, for $n=1,2.$
My question is if $T$ is $n$-hypercontraction then does it follow that $\varphi(T)$ is also a $n$-hypercontraction for every $n\in \mathbb N?$ My guess is that the answer is yes and similar kind of expression as above for $n$-hypercontraction is also true. But I am unable to find a simple proof. Any help or comment is welcome. 
 A: Over a year has passed since the original question was posted. Hopefully the OP is still interested in the answer.
By the recursive formula for binomial coefficients one has
\begin{align*}
\left(\sum_{k=0}^n (-1)^k{n\choose k}T^{*k}T^k\right) + T^*\left(\sum_{k=0}^n  (-1)^k{n\choose k}T^{*k}T^k\right)T = \sum_{k=0}^{n+1} (-1)^k{n+1\choose k}T^{*k}T^k.
\end{align*}
Consider an automorphism of the disc $\varphi$ as defined in the question. Since we assume that the spectrum is in the open disc then we can use holomorphic functional calculus to get $\varphi(T) = (T - aI)(I-\bar aT)^{-1}$. As the OP notes 
\begin{align*} I - \varphi(T)^*\varphi(T) 
& = (I-aT^*)^{-1}\left( (I-aT^*)(I-\bar a T) - (T^* - \bar aI)(T-a I) \right)(I-\bar a T)^{-1}
\\ & = (1-|a|^2)(I-aT^*)^{-1}(I - T^*T)(I-\bar a T)^{-1}.
\end{align*}
For $n\geq 1$, assume that 
$$ \sum_{k=0}^n (-1)^k{n\choose k}\varphi(T)^{*k}\varphi(T)^k = (1-|a|^2)^n(I - aT^*)^{-n}\left(\sum_{k=0}^n (-1)^k{n\choose k}T^{*k}T^k\right)(I-\bar aT)^{-n}.$$
Thus, by the recursive formula
\begin{align*}
\sum_{k=0}^{n+1} (-1)^k{n+1\choose k}\varphi(T)^{*k}\varphi(T)^k 
\\
 = \left(\sum_{k=0}^n (-1)^k{n\choose k}\varphi(T)^{*k} \varphi(T)^k\right) &+ \varphi(T)^*\left(\sum_{k=0}^n (-1)^k{n\choose k}\varphi(T)^{*k}\varphi(T)^k\right)\varphi(T)
\\ = (1 - |a|^2)^n(I- aT^*)^{-(n+1)}&\left[(I-aT^*)\left(\sum_{k=0}^n (-1)^k{n\choose k}T^{*k}T^k\right)(I-\bar aT)\right.\\ + (T^* - \bar aI)&\left.\left(\sum_{k=0}^n (-1)^k{n\choose k}T^{*k}T^k\right)(T-aI) \right](I-\bar aT)^{-(n+1)}
\\ = (1-|a|^2)^{n+1}(I - aT^*)^{-(n+1)}&\left(\sum_{k=0}^{n+1} (-1)^k{n+1\choose k}T^{*k}T^k\right)(I-\bar aT)^{-(n+1)}.
\end{align*}
Hence, by induction we have for all $n\in\mathbb N$
$$
\sum_{k=0}^n (-1)^k{n\choose k}\varphi(T)^{*k}\varphi(T)^k = (1-|a|^2)^n(I - aT^*)^{-n}\left(\sum_{k=0}^n (-1)^k{n\choose k}T^{*k}T^k\right)(I-\bar aT)^{-n}.
$$
Therefore, if $\sum_{k=0}^n (-1)^k{n\choose k}T^{*k}T^k \geq 0$ then $\sum_{k=0}^n (-1)^k{n\choose k}\varphi(T)^{*k}\varphi(T)^k \geq 0$, that is, $T$ is an n-hypercontraction if and only if $\varphi(T)$ is an n-hypercontraction.
