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Let $f$ be a continuous strictly-increasing function that maps $\mathbb{R}_+$ to $\mathbb{R}_+$. Define a function $g$ on $\mathbb{R}_+$ as follows:

$$g(x) := f^{-1}(f(x)+1).$$

For example, if $f(x) = x^2$, then $g(x) = \sqrt{x^2+1}$. In words, $g(x)$ describes to what value you should increase $x$, so that $f(x)$ increases by $1$.

The function $g$ clearly satisfies two properties:

  1. $g(x)>x$ for all $x$;
  2. $g(x)$ is strictly increasing.

QUESTION: Given a function $g$ that satisfies these two properties, does there always exist a corresponding function $f$? If not, what other conditions are required?

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  • $\begingroup$ "The quantity $\epsilon=g(x)-x$ describes how much you should increase $x$, so that $f(x+\epsilon) = f(x)+1$" seems a bit more accurate to me if I'm interpreting your definition $g$ correctly. $\endgroup$
    – Snared
    Feb 25 at 17:44
  • $\begingroup$ @Snared I changed the explanation to "to what value you should increase $x$". $\endgroup$ Feb 25 at 17:52

1 Answer 1

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If $g$ is continuous, strictly increasing with $g(x)>x$, then there exist infinitely many strictly increasing $f$ with a well-defined inverse $f^{-1}$ that satisfy the definition of $g$. To see this, we may substitute $x=f^{-1}(y-1)$ into the definition of $g$: $$g\circ f^{-1}(y-1) =f^{-1}(y),\quad y\geq f(0)+1\geq 1. $$ Therefore, for any given $g$, if we can find some strictly increasing function $h:\mathbb{R}_+\mapsto \mathbb{R}_+$ such that $g\circ h(x-1) =h(x)$ for all $x\geq 1$, we can simply set $f=h^{-1}$ which is the required function.

Let us show that such $h$ is not unique. Pick an arbitrary strictly increasing function $h_0:[c_0,c_0+1]\mapsto [0,g(0)]$ which is always possible for any $c_0\geq0$ and $g$ since $g(0) > 0$, so there are infinitely many possible choices. The required function $h:[c_0,\infty)\to\mathbb{R}_+$ can be constructed by the following piecewise design: $$h(x) =g^n\circ h_0(x-n),\quad x\in[c_n,c_{n+1}], \quad c_n=c_0+n,\quad n\in\mathbb{N},$$ where $g^0(x) = x$ and $g^n = \underbrace{g\circ \ldots \circ g}_{n\text{ times}}$. By the continuity and strict monotonicity of $h_0$ and $g$, it is clear that $h$ is continuous and strictly increasing on every interval of the form $[c_n,c_{n+1}]$. One can take limits from left and right to conclude that $h$ is continuous at each $h=c_n$ with $h(c_n) = g^n(0)>g^{n-1}(0)>\ldots >g(0)>0$. Therefore, $h$ is further continuous and strictly increasing on $\mathbb{R}_+$, thus it must have a well-defined inverse function $h^{-1}$. Since $h_0$ and $c_0$ are arbitrary, we must have infinitely many $f\equiv h^{-1}$ with $f(0) = c_0$, and we are done. One can write out the expression of $f$ explicitly in terms of $h_0^{-1}$ and $(g^{n})^{-1}$, but those inverse functions may not have an explicit form thus $f$ may need to be determined numerically from $h$.


To verify the above construction, you can try $g(x) = \sqrt{x^2+1}$, so $g(0)=1$. If we pick $h_0=\sqrt{x}$ defined on $[0,1]$ with $c_0=0$, then on $[1,2]$, $h(x) = \sqrt{\sqrt{x-1}^2+1 }=\sqrt{x}$, on $[2,3]$, $h(x) = \sqrt{\sqrt{\sqrt{x-2}^2+1}^2+1}=\sqrt{x}$ and so on, which leads to the result $h(x) = \sqrt{x}$ and hence $f=x^2$.

But you don't have to pick $h_0(x) =\sqrt{x}$. Picking $h_0(x)=x$ instead yields the solution: $$f(x) = \sqrt{x^2-n}+n, \quad x\in[\sqrt{n}, \sqrt{n+1}], n=0,1,2,\ldots$$ Its inverse function is: $$f^{-1}(x)\equiv h(x) = \sqrt{(x-n)^2+n},\quad x\in[n,n+1],n=0,1,2\ldots$$ To verify the desired property of $g(x)$, notice that, for any $x\in[\sqrt{n},\sqrt{n+1}]$, $f(x)+1\in[n+1,n+2]$, and thus $$f^{-1}(f(x)+1) = \sqrt{ (\underbrace{\sqrt{x^2-n}+n}_{f(x)}+1-n-1)^2 +n+1}=\sqrt{x^2+1}=g(x),$$ as desired. You can visualize the two different functions and their inverses below. By design, the values of the two inverse functions agree on $n=0,1,2,\ldots$ and equal $g^n(0)=\sqrt{n}$. enter image description here

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  • $\begingroup$ How do you know you can always pick an $h$ satisfying that? And also, yes, if $f$ works, then $f+\lambda$ works for $\lambda \in \mathbb R$. Also, you used $f^{-1}$, but how do you know that any $f$ would be invertible simply based on the two conditions on $g$? Can you give an example like $g(x)=1+x+x^2+x^3+x^4+\cdots+x^k$? What does $f$ look like in that case? I'm not sure how your answer solves the question except build a bit of semantic modeling around the logical framework of it. $\endgroup$
    – Snared
    Feb 26 at 23:01
  • $\begingroup$ See edited for a different example. A function $f$ corresponding to your choice of $g(x)$ can easily be constructed following the same construction (say, picking $h_0(x)=x$ and $c_0=0$, but it won't look nice because $g^n(x)$ takes a rather complicated form when $n$ becomes large. $\endgroup$
    – Fred Li
    Feb 26 at 23:29
  • $\begingroup$ You assume that $g$ is continuous - but what if it is not continuous? My guess is that a similar construction would work, but $f$ might be only weakly-increasing. Is it correct? $\endgroup$ 2 days ago
  • $\begingroup$ @ErelSegal-Halevi if $g$ is discontinuous but strictly monotone, then there can only be countable jump discontinuities in $g$. The same construction should still work, and $f$ is still strictly increasing on its domain by the same principle. But the domain of $f$ becomes a countable union of disconnected intervals due to the discontinuities in $g$, which creates `holes' in the range of $g$ and hence the domain of $f$. $\endgroup$
    – Fred Li
    2 days ago

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