# Constructing a function from a function of its inverse

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?

• "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. Feb 25 at 17:44
• @Snared I changed the explanation to "to what value you should increase $x$". Feb 25 at 17:52

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 $$g^n(0)=\sqrt{n}$$.

• 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. Feb 26 at 23:01
• 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. Feb 26 at 23:29
• 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? 2 days ago
• @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 $f$. 2 days ago