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How can we represent a continuously differentiable function as a difference of two continuous strictly increasing functions?

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2 Answers 2

Hint: Let our function be $F(x)$. Let $p(x)=F'(x)$ wherever $F'(x)\ge 0$, and let $p(x)=0$ elsewhere. Let $q(x)=-F'(x)$ wherever $F'(x) \le 0$, and let $q(x)=0$ elsewhere. Note that $F'(x)=p(x)-q(x)$. Integrate.

Note that this does not necessarily produce strictly increasing functions. However, once we have obtained non-decreasing functions with the right property, strictly increasing is easy.

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Since $f$ is continuously differentiable, $f$ is locally Lipschitz, i.e. for every $x\in\mathbb R$ there is an open neighbourhood $U_x$ of $x$ and a number $K_x\in\mathbb R$ such that $y,z\in U$ implies $|f(y)-f(z)|\le K_x|y-x|$. That is all we need.

Consider first the case of compact domain $[n,n+1]$, $n\in\mathbb Z$. Then $[n,n+1]\subset \bigcup_{x\in\mathbb R} U_x$ and by compactness, a finite subcover suffices, $[n,n+1]\subseteq U_{x_1}\cup\ldots \cup U_{x_m}$, and if we choose $M_n>\max\{K_{x_1},\ldots,K_{x_m}\}\ge0$, then we see that (with $c_n\in\mathbb R$ arbitrary) the functions $$\begin{align}p_n(x)&=f(x)+M_nx+c_n,\\q_n(x)&=M_n+c_n\end{align}$$ are continuous and strictly increasing on $[n,n+1]$ and their difference equals $f$. To find a decomposition if the domain is $\mathbb R$, adjust the $c_n$ accordingly (by recursion in both directions) so that no problems occur at the interval boundaries, i.e. such that $p_n(n+1)= p_{n+1}(n+1)$ and $q_n(n+1)= q_{n+1}(n+1)$. This boils down to $$c_{n+1}=(M_n-M_{n+1})(n+1)+c_n.$$ With such a choice of $c_n$, let $$\begin{align}p(x)&=p_{\lfloor x\rfloor}(x)=f(x)+M_{\lfloor x\rfloor}+c_{\lfloor x\rfloor},\\ q(x)&=q_{\lfloor x\rfloor}(x)=M_{\lfloor x\rfloor}+c_{\lfloor x\rfloor}.\end{align} $$

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Hi Hagen, how was the locally lipschitz property used in your construction? I can't see why it's important here. –  anegligibleperson Jun 26 '13 at 5:33

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