# Splitting a continuous monotonically-increasing function $f(x)$ as $h(x)+h(x+\epsilon) = f(x)$

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 $$h$$.

• well, just a trivial thought, for linear functions, something like $h(x):=f(x-\frac{\epsilon}{2})$ does the job – alphaomega Jul 8 '20 at 20:23
• 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/… – enzotib Jul 8 '20 at 20:24
• Is $h(x)$ supposed to be independent of $\epsilon$? – herb steinberg Jul 8 '20 at 20:37
• @herbsteinberg $h$ may depend on $\epsilon$. – 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: $$h(x-\epsilon) + h(x) = c + \gamma(x-\epsilon)$$.

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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.

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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: $$h(\epsilon+2\delta) \leq c/2 + \gamma < (d - \gamma)/2 \leq h(\epsilon+2\delta)$$

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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.)

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