A sufficient condition for a relaxation of Lipschitz continuity

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I am interested in a property that, I believe, is closely related to Lipschitz continuity. Lipschitz continuity is defined as:

Given two metric spaces $(X, d_X)$ and $(Y, d_Y)$, where $d_X$ denotes the metric on the set $X$ and $d_Y$ is the metric on set $Y$, a function $f: X\rightarrow Y$ is called Lipschitz continuous if there exists a real constant $K \geq 0$ such that, for all $x_1$ and $x_2$ in $X$, $d_{Y}(f(x_{1}),f(x_{2}))\leq Kd_{X}(x_{1},x_{2}).$

I focus on the case $K=1$ in which case $f$ is called a metric map and I take this relaxation:

$d_{Y}(f(x_{1}),f(x_{2}))\leq d_{X}(x_{1},x_{2}) + \varepsilon$

for a $\varepsilon \geq 0$.

I am interested in the condition:

$D_{X,Y}(x, f(x)) \leq \varepsilon$, for $D$ a metric between spaces $X$ and $Y$ and $\varepsilon\geq 0$. I'm assuming that the metric $D$ exists, for example, when $X=Y$.

That is, we are limiting where $f$ can send $x$.

I believe that this property would imply my relaxation of Lipschitz. For example, if we assume $d = D = d_Y = d_Y$, by the triangle inequality:

$d(f(x_{1}),f(x_{2}))\leq d(f(x_{1}), x_1) + d(x_2,f(x_{2})) + d(x_2,x_1)\leq 2\,\varepsilon + d(x_2,x_1).$

I am trying to find more information related to this.