Given a continuous monotonically-increasing function $f: [0,1]\to \mathbb{R}$ and a parameter $\epsilon>0$, does there exist a continuous monotonically-increasing function $h$ such that, for all $x\in[0,1]$:

$$h(x)+h(x+\epsilon) = f(x)?$$

If $\epsilon=0$ then $h(x)=f(x)/2$. But when $\epsilon>0$, the function $f$ should be split into two parts with a "phase difference" of $\epsilon$. It seems easy, but I could not find the formula for this $h$.

  • $\begingroup$ well, just a trivial thought, for linear functions, something like $h(x):=f(x-\frac{\epsilon}{2})$ does the job $\endgroup$ – alphaomega Jul 8 '20 at 20:23
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    $\begingroup$ When $f$ is a generic polynomial (even not increasing), there is unique polynomial solution $h$ of the same degree: drive.google.com/file/d/1aWaRJ-m9nRMccxbhGT9cx5pLhg2k_Nee/… $\endgroup$ – enzotib Jul 8 '20 at 20:24
  • $\begingroup$ Is $h(x)$ supposed to be independent of $\epsilon$? $\endgroup$ – herb steinberg Jul 8 '20 at 20:37
  • $\begingroup$ @herbsteinberg $h$ may depend on $\epsilon$. $\endgroup$ – Erel Segal-Halevi Jul 8 '20 at 20:41

No, this is not generally true. For any $\epsilon < 1/2$, we can construct a strictly increasing differentiable function $f$ such that no monotonically-increasing function $h$ satisfies your property.

Outline of construction: let $f$ be flat on the intervals $[0, \epsilon +\delta]$ and $[\epsilon +2\delta, 1]$ but steep in between.

Fix any $\epsilon<1/2$ and define $\delta>0$ such that $\delta < \min\{1/2 - \epsilon, \epsilon/2\}$. Construct $f$ to be linear for $x \leq \epsilon+\delta$ with slope $\gamma>0$:

  1. $f(x)=c + \gamma x$ for $x \leq \epsilon+\delta$.

Lemma 1: $(c - \gamma)/2 \leq h(x) \leq (c + \gamma)/2$ for $x \leq \epsilon+\delta$.

Proof: First observe that $h(x) \leq f(x)/2$ for all $x \in [0,1]$, otherwise $h(x) + h(x+\epsilon)>f(x)$ by monotonicity, which gives the upper bound for $x \leq \epsilon+\delta$. The lower bound follows by substituting this upper bound for $h(\epsilon)$ in the expression: $h(0) + h(\epsilon) = c$.

Lemma 2: $h(x) \leq c/2 + \gamma$ for $x \in [\epsilon+\delta, 2\epsilon+\delta]$.

Proof: This follows by substituting the lower bound from Lemma 1 for $h(x-\epsilon)$ in the expression: $h(x-\epsilon) + h(x) = c + \gamma(x-\epsilon)$.


Let $f$ be linear for $x \geq \epsilon+2\delta$ with slope $\gamma$:

  1. $f(x) = d + \gamma x$ for $x \geq \epsilon+2\delta$.

Lemma 3: $(d - \gamma)/2 \leq h(x) \leq (d + \gamma)/2$ for $x \in [\epsilon+2\delta, 1]$.

Proof: Same as in Lemma 1.


Notice that both Lemmas 2 and 3 apply to the point $x = \epsilon + 2\delta$.

  1. Choose $c$, $d$, and $\gamma$ such that:

$$c/2 + \gamma < (d - \gamma)/2$$

$$\Longleftrightarrow \gamma < (d-c)/3$$

This gives the contradiction that: $$h(\epsilon+2\delta) \leq c/2 + \gamma < (d - \gamma)/2 \leq h(\epsilon+2\delta)$$


Finally, it doesn't matter what $f$ is for $x \in (\epsilon+\delta, \epsilon + 2\delta)$; any valid (smooth strictly increasing) construction here would work.

Conjecture: There exists such an $h$ for all $f$ satisfying a bound on the ratio of derivatives: $f'(x)/f'(y) \leq M(\epsilon)$ for all $x,y \in [0,1]$. Basically, the slope cannot fluctuate too much.

(This trivially holds in the linear case where $M=1$, but a higher/the highest bound would be interesting.)

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    $\begingroup$ Interesting. But what if $f$ must be strictly monotonically-increasing? $\endgroup$ – Erel Segal-Halevi Jul 9 '20 at 0:48
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    $\begingroup$ @ErelSegal-Halevi I've updated the proof now to include that constraint. $\endgroup$ – Sherwin Lott Jul 9 '20 at 3:21
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    $\begingroup$ Great answer, and very interesting conjecture too! $\endgroup$ – Erel Segal-Halevi Jul 9 '20 at 7:25

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