It's a really one of the simplest properties you could imagine for a function. But I haven't been able to find a name for it. What do you call a function $f$ with the following property:
$$f(x) \ge x$$
Note, this is not monotonicity. Monotonicity is the following property:
$$x \ge y \implies f(x) \ge f(y)$$
On
The following is probably not what you're after, but perhaps it's mildly educational.
A somewhat convoluted way to describe your function would be to say:
Every element is a post-fixed point.
An element $x$ is a fixed point of $f$ if $f(x)=x$. It is a pre-fixed point if $f(x)\leq x$, and it is a post-fixed point if $f(x)\geq x$.
On
I call it the opposite of a contraction; ie the concave variant; it would follow from star concavity at $x_0=0$ where $F(x_0=0)\geq 0$with now as $$F(1)=1 \land \text{dom(F)}=[0,1]$$.
star concavity at 0 $$\forall (t\in [0,1]):\forall (x \in [0,1]):F((1-t)\times x_0+t\times y )\geq t\times F(y)+(1-t)\times F(x_0)$$.
ie $$\to F(t\times y)=F( (1-t)\times 0+ t\times y)\geq (1-t)\times F(0)+t\times f(y)$$
As $(t,1-t)\in [0,1]\land F(x_0=0)\geq 0$
$$\to F(t\times y)\geq (1-t)\times F(0)+t\times f(y)\geq t \times F(y)$$
star concavity at $0$ with $F(0)\geq 0(B)$
$$\to (B)F(t\times y)\geq t\times F(y)$$
(2) now as $$F(1)=1 \land \text{dom(F)}=[0,1]$$.
$$t=x \to F(x)=F(t)$$ $$F(x)=F(t)= F(1 \times t)\to F(x)=F(1\times t)$$ concavity
$$\to F(x)=F(1\times t)\geq t\times F(1)$$
substitution of $F(1)=1$ $$\to F(x)\geq t\times F(1)=1\times t=t$$ $$\to F(x)\geq t $$=
where t=x so: $$\to F(x)\geq x$$
As mentioned, this property is called expansion, as the element $x$ can be expanded, or stretched-out and thus contained in, the element $f(x)$.
However, this property is also termed reflexivity.
Why? Well, recall that a relation is reflexive if it relates identical elements. More precicely,
$\;\;\; R \text{ reflexive} \\\equiv \forall x:: \; x \, R \, x \\\equiv \forall x,y:: \; x = y \implies x \, R \, y \\\equiv \forall x,y:: \; x \, I \, y \implies x \, R \, y \text{, where $I$ is the identity relation: $I=\{(x,y)\,|\,x=y\}$} \\\equiv I \subseteq R $
Even more so, recall that an ordering $\leq$ on elements can be lifted to functions by $$f \overset{.}{\leq} g \,:\equiv\, (\forall x :: f \ x \leq g \ x)$$ With this, and letting $1$ be the identity function $x \mapsto x$, we have $$f \text{ reflexive } \ \equiv \ 1 \overset{.}{\leq} f$$
Hope that helps!