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I have not yet seen something similar before. Let $f:D\times [0,1]\to \mathbb{R}$ be a function, and for every $x\in D\subset\mathbb{R}^n$ the function $f(x,s)$ is increasing in $s$. In other words: For every $x$ the function $f(x,\cdot):[0,1]\to \mathbb{R}; s\mapsto f(x,s),$ is increasing and thus invertible. Now the inverse of that function should be e.g. written as $(f(x,\cdot))^{-1}$ for every $x$.

However, I'm also interested in the behaviour of this "inverse" in $x$. Hence, probably the safest way is to define a function $h$ that maps from $G:\{(x,u):x\in D, f(x,0)\leq u\leq f(x,1)\}$ to $[0,1]$. Is this correct so far?

But since mathematicians are lazy and I want to emphasize that $h$ depends on $f$, I would rather write (with a lot of abuse of notation) $f^{-1}(x,u):G\to [0,1]$. At first glance this looks really horrible but somehow it is clear that not the standard inverse is meant.

Of course I know the implicit function theorem but somehow the notation from there seems to be inappropriate here. Any comments or suggestions?

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My suggestion is as follows.

Consider a function $f\colon D\times [0,1]\to \mathbb{R}$ and, given $x\in D$ define a new function $f_x\colon [0,1]\to \mathbb R$. This $f_x$ coincides with what you call $f(x,\cdot)$ (a notation which I abominate, because if this is a function, then the image at a point $t$ should be denoted by $f(x,\cdot)(t)$ and not by $f(x,t)$).

If you're worried about $f_x$ being mistaken with the partial derivative, then change the notation a bit, maybe something like $f_{(x)}$.

If $f_{(x)}$ happens to be an invertible function, that is, if $f_{(x)}^{-1}$ exists, and if I understand you correctly, you want to study the function $(x,u)\mapsto f_{(x)}^{-1}(u)$ which is only defined on $G$.

Finally and to name the function $(x,u)\mapsto f_{(x)}^{-1}(u)$ (which is $f$-depedent), you can introduce a new symbol, for instance, as you suggested on the comments, $I_f$, yielding $I_f\colon G\to\mathbb R, (x,u)\mapsto f_{(x)}^{-1}(u)$, with the hypothesis that $f_{(x)}$ is invertible for all $x\in D$.

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  • $\begingroup$ Thanks, I'll take your idea into account. My boss, somehow likes the $\dot$ notation and uses it very often :) $\endgroup$ Jan 3 '14 at 9:41
  • $\begingroup$ @Quickbeam2k1 'Everyone' uses it, that doesn't mean it's right, though. $\endgroup$
    – Git Gud
    Jan 3 '14 at 9:42
  • $\begingroup$ Sorry, I want to study the function you mention. So you could write $h:G\to [0,1], (x,u)\mapsto f^{-1}_{(x)}(u)$. However, I would prefer to write $f^{-1}$ instead of $h$ which is clearly awful. I'm looking for a better notation there, maybe $I_f: G\to [0,1]$, $(x,u)\mapsto f^{-1}_{(x)} (u)$. Other suggestions? Or would you simply identif $I_f$ with $f^{-1}_{(x)}$ $\endgroup$ Jan 3 '14 at 9:57
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    $\begingroup$ @Quickbeam2k1 I completed the answer with another (yours actually) suggestion. $\endgroup$
    – Git Gud
    Jan 3 '14 at 10:08

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