Wikipedia says:
In functional analysis and operator theory, a bounded linear operator is a linear transformation ${\displaystyle L:X\to Y}$ between topological vector spaces (TVSs) ${\displaystyle X}$ and ${\displaystyle Y}$ that maps bounded subsets of ${\displaystyle X}$ to bounded subsets of ${\displaystyle Y.}$.
If ${\displaystyle X}$ and ${\displaystyle Y}$ are normed vector spaces (a special type of TVS), then ${\displaystyle L}$ is bounded if and only if there exists some ${\displaystyle M>0}$ such that for all ${\displaystyle x\in X,}$
$${\displaystyle \|Lx\|_{Y}\leq M\|x\|_{X}.}$$
Since bounded subset concept exists in TVS and in metric space, can bounded mapping concept exist in metric space? For example,
Given metrics spaces $M_1 := (S_1, d_1)$ and $M_2 := (S_2, d_2)$, a mapping $f:M_1\to M_2$ is said to be bounded, if $\exists M \geq 0$, so that $\forall x, y \in M_1$, $d_2(f(x), f(y)) \leq M \times d_1(x, y)$.
Is a hash function with the diffusion property (https://www.cs.cornell.edu/courses/cs312/2008sp/lectures/lec21.html) an example of a non-bounded operator on a metric space:
Diffusion (stronger than injection): if k1 ≠ k2, knowing h(k1) gives no information about h(k2). For example, if k2 is exactly the same as k1, except for one bit, then every bit in h(k2) should change with 1/2 probability compared to h(k1). Knowing the bits of h(k1) does not give any information about the bits of h(k2).
Thanks.