This question has probably been asked before, but I don't know how to phrase the property in English so as to use the right search terms to find it on Google. If this question is a duplicate, please feel free to close this question and link to the original.
For any $f: \mathbb{R}\to \mathbb{R}$, and any $\vec{x} \in \mathbb{R}^n$, let $f(\vec{x}):=(f(x_1), \dots, f(x_n))$. The question is:
What is the name of the following property for a function $f: \mathbb{R} \to \mathbb{R}$? For any $\vec{x} \in \mathbb{R}^n$, for any $\vec{a},\vec{b} \in \mathbb{R}^n$ such that $\vec{a}\cdot\vec{x} \le \vec{b}\cdot\vec{x}$, one has that $\vec{a}\cdot f(\vec{x}) \le \vec{b} \cdot f(\vec{x})$.
Attempt: I think it might be related to convexity, since a function $f$ is convex if and only if for any $\vec{c} \in \mathbb{R}^n$ whose entries are non-negative and sum to $1$, for any $\vec{x} \le \mathbb{R}^n$, $f(\vec{c}\cdot \vec{x}) \le \vec{c} \cdot f(\vec{x})$. If in addition $f$ is non-decreasing, that implies $\vec{c}\cdot \vec{x} \le \vec{c} \cdot f(\vec{x})$. Then for any $\vec{v} \in \mathbb{R}^n$ whose entries are non-negative, since $(\vec{v}/||\vec{v}||_1) \cdot \vec{x} \le (\vec{v}/||\vec{v}||_1)\le f(\vec{x})$, one also has $\vec{v} \cdot \vec{x} \le \vec{v} \cdot f(\vec{x})$.
However that wouldn't extend to vectors with any negative entries, whereas $\vec{a}$ and $\vec{b}$ can be arbitrary. And in any case the goal is to compare $\vec{a} \cdot f(\vec{x})$ and $\vec{b} \cdot f(\vec{x})$, not $\vec{a} \cdot f(\vec{x})$ and $\vec{a} \cdot \vec{x}$.
Obviously $f:x \mapsto x$ satisfies the property.
Are there any other functions satisfying the property?
Any function $f$ provided $f(\vec{x})$ is always pointing in the same direction as $\vec{x}$.
Specifically, if $g : \mathbb{R}^n \to \mathbb{R}^{\geq 0}$, then $f(\vec{x}) = g(\vec{x}) \vec{x}$ works. Specifically, $g$ associates a nonnegative scalar to every point in $\mathbb{R}^n$.
In particular,
$$\vec{a} \cdot \vec{x} \leq \vec{b} \cdot \vec{x}$$
Multiply both sides by $g(\vec{x})$:
$$g(\vec{x}) \vec{a} \cdot \vec{x} \leq g(\vec{x})\vec{b} \cdot \vec{x}$$
Scalar multiple property of dot products:
$$\vec{a} \cdot (g(\vec{x})\vec{x}) \leq \vec{b} \cdot (g(\vec{x})\vec{x})$$
By the definition of $f$:
$$\vec{a} \cdot f(\vec{x}) \leq \vec{b} \cdot f(\vec{x})$$
In fact, this is all such functions. Suppose that we have such an $f$ and $\vec{x}$. By the $\cos$ definition of the dot product:
$$f(\vec{x}) \cdot \vec{x} \leq \left(\frac{\vec{x} |f(\vec{x})|}{|\vec{x}|}\right) \cdot \vec{x}$$
(The left side is just the right side times $\cos(\theta)$ where $\theta$ is the angle between $\vec{x}$ and $f(\vec{x})$.)
Applying the property of $f$ with $\vec{a} = f(\vec{x})$ and $\vec{b} = \left(\frac{\vec{x} |f(\vec{x})|}{|\vec{x}|}\right)$:
$$f(\vec{x}) \cdot f(\vec{x}) \leq \frac{|f(\vec{x})|}{|\vec{x}|} \vec{x} \cdot f(\vec{x})$$
The left side is $|f(\vec{x})|^2$ and the right side is $|f(\vec{x})|^2 \cos(\theta)$, so $\theta$ must be $0$.