# generalised inverse function

Let $f:\mathbb{R} \rightarrow [0, 1]$ be increasing (edit: i.e., non-decreasing).

Define $f^-(y) = \inf \{x \in \mathbb{R} : f(x) \geq y \}$, $y \in [0, 1]$.

Is the following line true?

$$x \leq f^-(y) \quad\leftrightarrow\quad f(x) \leq y$$

-
Do you allow constant (or locally constant) functions? –  Siminore Jun 20 '12 at 12:08
@Siminore: I understand increasing in the sense of non-decreasingness, if this is what you mean... thx –  user7064 Jun 20 '12 at 12:17
Thank you for the link; i think it will help a lot! –  user7064 Jun 20 '12 at 13:29

The definition doesn't make sense for $y$ less than or equal to the infimum of the image of $f$ (for example, $0$). Unless you allow $-\infty$.
Even where defined, what you said is not true (in fact, it is equivalent to saying that $f$ is strictly increasing).
Choose $f(x)=1$ for $x\geq 0$, and $f(x)=e^{x}$ for $x<0$. Then $f^-(1)=0$, and $1>0$, but $f(1)=1\leq 1$, so $\leftarrow$ fails. (But $\rightarrow$ is of course true.)
The equivalence is true if you reverse both inequalities: $$x\geq f^-(y)\iff f(x)\geq y$$