Is a one-one non-decreasing function always increasing?

Question

Suppose $f:[a,b]\rightarrow[c,d]$ is one-one and non-decreasing, then $f$ is always increasing.

Is this statement correct? If not, give an example.

The sketch of the proof is as follows: If any flaw in this, please point out.

Now,

Let $x,y \in [a,b]$, We need to prove that $x<y \implies f(x)<f(y)$.

Since $f(x), f(y) \in [c,d]$ either $f(x)=f(y)$ or $f(x)>f(y)$ or $f(x)<f(y)$. Since $f$ is non-decreasing, $f(x)>f(y)$ is not possible. If $f(x)=f(y)$, then $x=y$ as $f$ is one-one, but we assumed that $x<y$, therefore, only possbility is that $f(x)<f(y)$.

For example $f(x)=x$ when $0\le x<1$ and $f(x)=-x+3$ when $1\le x\le 1.5$ — Jo Jomax, Feb 08 '20 at 17:46
It is not satisfying the hypothesis. First of all, the function is not defined at 1. — panch, Feb 08 '20 at 17:53
@JoJomax It is not decreasing as in $\neg(\forall x,y\colon x<y\to f(x)>f(y))$, but that doesn't make it non-decreasing as in $\forall x,y\colon (x\le y\to f(x)\le f(y))$ — Hagen von Eitzen, Feb 08 '20 at 17:54

mathematics2x2life · Accepted Answer · 2020-02-08T17:52:27.533

1

The proof is fine. Although it can be cleaner, i.e. shorter. For instance, ()<(), ()=(), or ()>() is always true for real valued functions, regardless of the fact that the image is in [,]. So you can simply say:

Let ,∈[,] with <.

By assumption, ()≤(), otherwise is not non-decreasing. But if ()=(), because is 1-to-1, then =, impossible. Therefore, ()<() whenever <, i.e. is increasing.

edited Feb 08 '20 at 17:52

answered Feb 08 '20 at 17:42

mathematics2x2life

13,322

So what is my example? – Jo Jomax Feb 08 '20 at 17:49
@JoJomax Your function is not non-decreasing, take $f(1.2)$ and $f(1.3)$ as an example. $f(1.2)>f(1.3)$, which should not happen if $f$ is non-decreasing. – mathematics2x2life Feb 08 '20 at 17:56
@JoJomax The point is that your function does decrease. In your head you've split it into an interval where it increases (so non-dec) and then where it does and compared the two. But you need it non-dec on the entire interval, and on the second half of your piecewise function, it is most certainly decreasing, again: $f(1.3)<f(1.2)$ which cannot happen if $f$ is nondecreasing. – mathematics2x2life Feb 08 '20 at 17:59
@JoJomax Or alternatively, find a flaw in the proof the OP or I give.... – mathematics2x2life Feb 08 '20 at 18:01
So "$f$ is Non-decreasing" is not equivalent to "$f$ in not decreasing" – Jo Jomax Feb 08 '20 at 18:01
@JoJomax https://en.wikipedia.org/wiki/Monotonic_function – mathematics2x2life Feb 08 '20 at 18:02
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The second half of your function is already decreasing, as its slope is negative. – panch Feb 08 '20 at 18:02

Is a one-one non-decreasing function always increasing?

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