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Let $\mathbb{N}\equiv \{1,2,3,...\}$. Could you suggest two functions $f_1: \mathbb{N}\rightarrow \mathbb{R}$, $f_2: \mathbb{N}\rightarrow \mathbb{R}$ such that

$f_1$ is monotone decreasing,

$f_2$ is monotone increasing, and

$f\equiv f_1+f_2$ is monotone decreasing.

I managed to find $f_1, f_2$ such that $f\equiv f_1+f_2$ is monotone increasing (e.g., $f_1(n)=n^2$, $f_2(n)=1/n$) but not the other way around.


Also, is there any theorem saying that any monotone (increasing or decreasing) function $f$ can be written as as the sum of a monotone increasing function $f_1$ and a monotone decreasing function $f_2$?

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  • $\begingroup$ Your $f_2$ is not having a codomain in $N$ $\endgroup$
    – nonuser
    Dec 8, 2017 at 13:27
  • $\begingroup$ I assume you want these functions to be strictly increasing and decreasing, as otherwise any two constant functions would work, but then I'm a bit confused about how to have a strictly decreasing function on the naturals since you can only decrease so far... Also, your $f_2$ is not a function from $\mathbb{N}$ to $\mathbb{N}$. $\endgroup$
    – cjohnson
    Dec 8, 2017 at 13:27
  • $\begingroup$ @JohnWatson, yes, my mistake, question edited $\endgroup$
    – Star
    Dec 8, 2017 at 13:29
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    $\begingroup$ So, what about $f_1(n)=-2n$, $f_2(n)=n$? $\endgroup$
    – lulu
    Dec 8, 2017 at 13:31
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    $\begingroup$ Sure. $f_1(n)=-n^2$, $f_2(n)=n$ (shift by $1$ if you are worried about $n=0$). $\endgroup$
    – lulu
    Dec 8, 2017 at 13:40

2 Answers 2

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Let $f$ denote any monotone decreasing function from $\mathbb N\to \mathbb R$. Then let $f_1=2f$, $f_2=-f$. And similarly if $f$ is monotone increasing.

Taking, say, $f(n)=-n$ gives a particular example, but the general construction establishes the theorem you requested.

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  • $\begingroup$ Thanks. Does the theorem has well known name in math? Can it be restricted by requiring $f_1, f_2$ of different degree? $\endgroup$
    – Star
    Dec 8, 2017 at 13:51
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    $\begingroup$ I don't suppose this has a name. In general, the degree of a function is not defined. If you specify that $f$ is a polynomial, then of course there is the notion of a degree and I think you should be able to play with examples to show that you can always vary the degree (hint: you can always add and subtract $n^k$ from your function). $\endgroup$
    – lulu
    Dec 8, 2017 at 13:55
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Let $f_1(x)$ have slope $m$. The general idea is to have $f_2(x)$ increasing but have a slope less than $|m|$, so that overall $f_1(x) + f_2(x)$ is decreasing.

You can do this with straight lines: Let $f_1(x) = -x$, and $f_2(x)$ be $\frac{1}{2}x$. Then $f_1(x)+f_2(x) = -\frac{1}{2}x$. Here's an illustration of this on Desmos.

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  • $\begingroup$ Thank you. What about my last question? $\endgroup$
    – Star
    Dec 8, 2017 at 13:32
  • $\begingroup$ We can let $f_1(x)$ be $-x$ when $x≥0$, and $0$ when $x<0$. We can also let $f_2(x)$ be $\frac{1}{2}x$ when $x≥0$, and $0$ when $x<0$. $\endgroup$
    – Toby Mak
    Dec 8, 2017 at 13:36
  • $\begingroup$ I meant from the main text "Also, is there any theorem saying that any monotone (increasing or decreasing) function $f$ can be written as as the sum of a monotone increasing function $f_1$ and a monotone decreasing function $f_2$?" $\endgroup$
    – Star
    Dec 8, 2017 at 13:37
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    $\begingroup$ @STF For your second question, when $\frac{dy}{dx} = kx$, it has the solution $\frac{kx^2}{2} + c$. However, this function is not monotone. However, when $\frac{dy}{dx} = kx^2$, the solution is $\frac{kx^3}{3} + c$ which is monotone. (Look, I'm really bumping into calculus here..., if you don't know calculus I'm sorry) $\endgroup$
    – Toby Mak
    Dec 8, 2017 at 13:42
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    $\begingroup$ @STF I really can't prove your first question - you'll really have to leave it to someone else (unless one of the functions can be constant). $\endgroup$
    – Toby Mak
    Dec 8, 2017 at 13:49

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