Is this sum of nonexpansive maps itself nonexpansive? For Hilbert spaces $\mathcal{H}_X$, $\mathcal{H}_Y$ and $\mathcal{K}$, consider a linear map $f\colon \mathcal{K} \oplus \mathcal{H}_X \to \mathcal{K} \oplus \mathcal{H}_Y$ that is given as a matrix
$$f = \begin{pmatrix} A \colon \mathcal{K} \to \mathcal{K} & B\colon \mathcal{H}_X \to \mathcal{K} \\ C\colon \mathcal{K} \to \mathcal{H}_Y & D\colon \mathcal{H}_X \to \mathcal{H}_Y \end{pmatrix}.$$

Conjecture. Assume $f$ is nonexpansive (i.e. $\lvert\!\lvert f \rvert\!\rvert_{op} \leq 1$) and $A$ is strictly contractive (i.e. $\lvert\!\lvert A \rvert\!\rvert_{op} < 1$). Then, the map $$ \Uparrow\!(f) = D + C \sum^\infty_{n=0} A^n B $$ is nonexpansive.

Notice that $\Uparrow\!(f)$ is well-defined because $A$ is assumed strictly contractive and therefore the series converges.
I know the conjecture above is true when $f$ is an isometry; in that case $\Uparrow\!(f)$ turns out to be an isometry too. This is proven in https://arxiv.org/pdf/1404.0074.pdf, in particular, in the first half of page 8 (numbered page 24 in the document). It seems natural that if it's true for isometries it should hold for any nonexpansive map, however I have not been able to work out the technical details to prove it.
I would expect this to be a well-known result; I'd be grateful if you could reference some text where this is proven or give me a hint on how to proceed.
 A: The answer to your conjecture is yes, and you are completely right that the result for isometries implies that result for nonexpansive mappings (which I will simply all contractions here).
This follows from Sz.-Nagy's dilation theorem for contractions on Hilbert spaces, which says the following:
Theorem. Let $T$ be a linear contraction on a Hilbert space $H$. Then there exists a Hilbert space $V$ that contains $H$ and a unitary $U$ on $V$ such that
\begin{align*}
  T^n = P U^nP
\end{align*}
for all integers $n \ge 0$, where $P$ denotes the orthogonal projection from $V$ onto $H$.
Now you can argue as follows:
Step 1. First assume that $\mathcal{H}_X = \mathcal{H}_Y$. Then $f$ is a contraction from a Hilbert space into itself, so we can apply the above dilation theorem. This theorem yields another Hilbert space $\mathcal{L}$ (which is the range of $\operatorname{id}-P$ in the theorem) and a unitary $U$ on $\mathcal{K} \oplus \mathcal{H}_X \oplus \mathcal{L}$ which is given by
\begin{align*}
 U = 
\begin{pmatrix}
A & B & F_1\\
C & D & F_2 \\
F_3 & F_4 & F_5
\end{pmatrix}.
\end{align*}
Now you apply the known result for isometries to see that
\begin{align*}
\begin{pmatrix}
D & F_2\\
F_4 & F_5
\end{pmatrix}
+
\begin{pmatrix}
C \\
F_3
\end{pmatrix}
\sum_{n=0}^\infty A^n
\begin{pmatrix}
B & F_1
\end{pmatrix}
\end{align*}
is an isometry. By projecting orthogonally onto $\mathcal{H}_X$ one concludes that $D + C \sum_{n=0}^\infty A^n B$ is contractive.
Step 2. Now let us consider spaces $\mathcal{H}_X$ and $\mathcal{H}_Y$ which might be different. Then define $\mathcal{H} = \mathcal{H}_X \oplus \mathcal{H}_Y$, as well as
\begin{align*}
\tilde B & = 
\begin{pmatrix}
B & 0
\end{pmatrix}
: \mathcal{H} \to \mathcal{K}, \\
\tilde C & =
\begin{pmatrix}
0 \\ C
\end{pmatrix}
: \mathcal{K} \to \mathcal{H}, \\
\tilde D & =
\begin{pmatrix}
0 & 0 \\
D & 0
\end{pmatrix}
: \mathcal{H} \to \mathcal{H}. 
\end{align*}
By applying what we have learned in Step 1 to the operator
\begin{align*}
\begin{pmatrix}
A & \tilde B \\
\tilde C & \tilde D
\end{pmatrix}
\end{align*}
we deduce that $\tilde D + \tilde C \sum_{n=0}^\infty A^n \tilde B$ is contractive. But this operator is equal to 
\begin{align*}
\begin{pmatrix}
0 & 0 \\
D + C \sum_{n=0}^\infty A^n B & 0
\end{pmatrix},
\end{align*}
so $D + C \sum_{n=0}^\infty A^n B$ is contractive, too.
Remark.  In the above argument we only used the Sz.-Nagy dilation theorem for the case $n=1$. For this special case, the Hilbert space $V$ and the unitary $U$ can be constructed in a particularly simple way. See for instance the article arxiv.org/abs/1803.09329 for details.
