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
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 $X$.