Question on operator theory classes of operators on Hilbert spaces I was recently tackled by this in my class on operator theory dealing with operators on Hilbert spaces:

We are to find and prove the inclusion relations between the classes of operators: contractions, strict contractions, proper contractions

I have no idea how to show this, as I have no idea how to show the inclusion or even think about it, so I am asking this here for help Thanks
Sorry forgot definitions:

Contraction: An operator T satisfying $ ||T|| \leq 1 $
Strict contraction: An operator T satisfying $ ||T|| < 1 $
Proper contraction: An operator on a Hilbert space satisfying $||Tx|| < ||x|| $ for all vectors x in the Hilbert space

 A: If I understand correctly what you need to show that:


*

*every strict contraction is a proper contraction as well (hence the set of strict contractions is included in the set of proper contractions) and

*every proper contraction is a contraction as well (hence the set of proper contractions is included in the set of contractions).


Note that this automatically applies that the set of strict contractions is included in the set of contractions.
To prove this use the definition of the operator norm.
A: Note that for any operator $T \colon X \to X$ on the Hilbert space $X$, we have 
$$ \def\n#1{\left\|#1\right\|}\n{Tx} \le \n T \n x $$
and 
$$ \n T = \sup_{ x \ne 0} \n x^{-1}\n{Tx} $$
Hence, if $T$ is a strict contraction, we have for $x \ne 0$:
$$ \n{Tx} \le \n T \n x < \n x $$
so $T$ is a proper contraction.
If $T$ is a proper contraction, we have 
$$ \n T = \sup_{x \ne 0} \frac{\n{Tx}}{\n x} \le \sup_{x\ne 0} \frac{\n x}{\n x} = 1 $$
so $T$ is a contraction. Therefore 
$$ \{\text{strict contractions}\} \subseteq \{\text{proper contractions}\} \subseteq \{\text{contractions}\} $$
If $X$ is finite, dimensional, every proper contraction is strict. To see this, note that $S_X = \{x \in X : \n x = 1\}$ is compact, hence the continuous map $x \mapsto \n{Tx}$ attains is maximum $m = \n{Tx_0}$ on $S_X$, but then 
$$ \n T = \sup_{\n x = 1} \n{Tx} = m = \n{Tx_0} < \n{x_0} = 1. $$
The idenitity is always a non-proper contraction, hence the last inclusion is always strict. 
If $T$ is infinite-dimensional, the first inclusion is also, to see this, let $(e_n)$ a countable orthonormal set and $U := {\rm span}\{e_n : n \in \mathbf N\}$. Define $T$ on $X = U \oplus U^\bot$ by 
$$ T\left(\sum_n x_n e_n + y\right) = \sum_n \left(1 - \frac 1n\right)x_ne_n $$
Then $T$ is a proper contraction, as for $x \in X$ with $x \ne 0$ we have either $x_n \ne 0$ for some $n$, hence
\begin{align*}
  \n{Tx}^2 &= \sum_n \def\a#1{\left|#1\right|}\left(1 - \frac 1n\right)^2\a{x_n}^2 \\
   &< \sum_n \a{x_n}^2\\
   &\le \sum_n \a{x_n}^2 + \n y^2\\
   &= \n x^2
\end{align*}
or $y \ne 0$, in which case we have
\begin{align*}
  \n{Tx}^2 &= \sum_n \def\a#1{\left|#1\right|}\left(1 - \frac 1n\right)^2\a{x_n}^2 \\
   &\le \sum_n \a{x_n}^2\\
   &< \sum_n \a{x_n}^2 + \n y^2\\
   &= \n{x}^2 
\end{align*}
But we have
$$ \n T \ge \sup_n \n{T e_n} = \sup_n 1 - \frac 1n =  1 $$
so $T$ is not strict.
Hence, we have
$$ \{\text{strict contractions}\} \subsetneq \{\text{proper contractions}\} \subsetneq \{\text{contractions}\} $$
for infinite-dimensional $X$ and 
$$ \{\text{strict contractions}\} = \{\text{proper contractions}\} \subsetneq \{\text{contractions}\} $$
for finite-dimensional $X$.
