Let $f:\mathbb{R}_+\rightarrow \mathbb{R}_+$ be a given monotonically increasing, continuous function. For example, consider $f(x) = x$.

How can I determine (if one exists) a monotonically increasing function $g:\mathbb{R}_+\rightarrow \mathbb{R}_+$ such that, for every $t\ge 0$,

  • $g(t) \le f(t)$ and
  • the area of $\{(x,y) : x \ge t, g(x) \le y \le f(t)\}$ equals $f(t)$?

The diagram below illustrates the area in question, labeled $A$, for a single $t$. It is the intersection of an axis-parallel rectangle (with upper-left corner $(t, f(t))$, and upper and left boundaries shown with dashed lines in the figure) and the area above the curve $\{(x, g(x)) : x \ge 0\}$.

enter image description here

I would accept an answer for just the special case when $f(x) = x$. In this case, surely $g(x) = f(x) - \Theta(\sqrt x)$ (for large $x$), but I'm looking for a more precise answer.

For the case when $f(x) = e^x$, the answer appears to be $g(x) = e^{x-c}$ for $c\approx 1.84$.


The question is interesting. Here are some developments. At the end of the answer, I found a closed-form expression for $g$ in the case $f(t)=t$.

Let us define the function $$ h(t) = g^{-1}(f(t)), $$ which defines the argument for which $g$ intersects with $f(t)$, see the figure below. (Just in case, $g^{-1}$ is the inverse of $g$. If $g$ is strictly monotone, then $g^{-1}$ truly exists. But the situation when $g$ has some "flat" parts is also ok, in fact.)

enter image description here

Let us denote the considered set as $A(t)$, i.e., $$ A(t) := \{(x,y): x \geq t,~ g(x) \leq y \leq f(t)\}, $$ and denote its area as $|A(t)|$. By the assumption, we require $|A(t)| = f(t)$ for each $t \geq 0$.

Let us explicitly calculate $|A(t)|$. We have $$ f(t) = |A(t)| = f(t)(h(t)-t) - \int_t^{h(t)} g(s) \, ds, $$ i.e. $|A(t)|$ is the area of the rectangle with sides $f(t)$ and $(h(t)-t)$, minus the area of the set which lies below $g$. Therefore, $$ \tag{1} \boxed{f(t)(h(t)-t-1) = \int_t^{h(t)} g(s) \, ds \quad \text{for all } t \geq 0.} $$

This equation already gives some information about $g$. Namely, we necessarily have $$ h(t) = g^{-1}(f(t)) \geq t+1, $$ which implies (by the monotonicity of $g$) that $$ f(t) \geq g(t+1). $$ In particular, $f(0) \geq g(1)$. For example, if $f(t)=t$, then $g(t)=0$ for all $t \in [0,1]$, and only for some $t >1$ it starts to grow.

A bit more developed bound can be obtained by estimating the integral on the right-hand side of (1) from below (by the monotonicity of $g$): $$ f(t)(h(t)-t-1) \geq g(t) (h(t)-t). $$ In particular, we see the following property:

If $f(0)>0$, then $f(0) > g(0)$. Or, equivalently, if $f(0)=g(0)$, then $f(0)=g(0)=0$.

Suppose now that $f$ and $h$ are differentiable. Since the equality (1) holds for all $t \geq 0$, we can differentiate both sides and obtain $$ f'(t)(h(t)-t-1) + f(t)(h'(t)-1) = g(h(t)) h'(t) - g(t). $$ Note that, by definition of $h$, we have $g(h(t))=f(t)$. That is, $$ \tag{2} \boxed{f'(t)(h(t)-t-1) - f(t) + g(t) = 0 \quad \text{for all } t \geq 0.} $$ More or less, this is the equation for $g$, under the assumption that $f$ and $h$ are differentiable. But it is not explicit, since it contains the inverse function $g^{-1}$. Note that $h$ can be nondifferentiable. Indeed, if we assume that (2) holds all the time, then if $f'(0)=0$, we have $f(0)=g(0)$. However, if, additionally, $f(0)>0$, then we get a contradiction from the property stated above. So, the nondifferentiability of $h$ occurs when $f'=0$.

However, I conjecture that if $f'(t)>0$ for all $t$, then (2) is valid. Suppose, this is true. Then, in the particular case $f(t)=t$, we get the following elegant equation for $g$: $$ g(t) + g^{-1}(t) = 2t+1, $$ which gives the recurrence formula $$ t = g(2t+1-g(t)). $$ It is possible to resolve it step by step. For instance, here are first few steps (see the figure below): $$ g(t) = \left\{ \begin{aligned} &0, &&t \in [0,1];\\ &\frac{t-1}{2}, &&t \in [1,3];\\ &\frac{2t-3}{3}, &&t \in [3,6];\\ &... \end{aligned} \right. $$ It can be seen that such $g$ is indeed what we need. That is, the application of (2) is ok.

enter image description here

  • $\begingroup$ Very nice. To continue your calculation, the closed form seems to be that $g(t) = t\, i\, / (i+1) - i/2$ for $i$ such that $$t\in \Big[{i+1 \choose 2}, {i+2 \choose 2}\Big].$$ $\endgroup$ – Neal Young May 10 '18 at 0:22
  • $\begingroup$ Yeah, it could to be something like that. Moreover, if we consider $f(t)=e^t$ and use the ansatz $g(t)=e^{t-c}$, then (2) shows that $c$ satisfies the following equations: $e^{-c}=2-c$. That is, $c=1.8414...$, as you wrote. $\endgroup$ – Voliar May 10 '18 at 8:50

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