I have some confusion in difference between monotone increasing function and Increasing function. For example

$$f(x)=x^3$$ is Monotone increasing i.e, if $$x_2 \gt x_1$$ then $$f(x_2) \gt f(x_1)$$ and some books give such functions as Strictly Increasing functions.

But if

$$f(x)= \begin{cases} x & x\leq 1 \\ 1 & 1\leq x\leq 2\\ x-1 & 2\leq x \end{cases} $$

Is this function Monotone increasing?


I'm used to the following: $f$ is

increasing iff $x\le y \Rightarrow f(x)\le f(y)$

strictly increasing iff $x< y \Rightarrow f(x)< f(y)$

decreasing and strictly decreasing: similar, with the inequalities for $f$ reversed.

Monotonic: either increasing or decreasing

strictly monotonic: either strictly increasing or strictly decreasing.

In particular, monotonically increasing is the same as increasing, strictly monotonically increasing the same as strictly increasing.

(In general there is always some freedom when it comes to definitions, this is why I wrote 'I'm used to'. In case you are reading a textbook on analyisis the author should define these terms and then stick to them).

  • 5
    $\begingroup$ a downvote after 18 month? without a comment or hint what the problem is? I'd appreciate an explanation. Downvoting without giving a hint what could be improved is as useless as it can get. $\endgroup$ – Thomas Nov 7 '17 at 18:36
  • $\begingroup$ Never heard of monotonic and strictly monotonic 🙉 $\endgroup$ – Fawad Feb 26 '18 at 11:38
  • $\begingroup$ $f$ is increasing – $x≤y \Rightarrow f(x)≤f(y)$ does the converse hold? That is if $f(x)≤f(y) \implies x≤y$ then $f$ is increasing? Similarly does the converse hold for strictly increasing function too? $\endgroup$ – William Nov 9 '18 at 6:31
  • $\begingroup$ @William no. If $f$ is constant, then $f(x)\le f(y)$ for every $x,y$, but there is no way to conclude any relation between $x$ and $y$. In case of strict monotonicity you can argue by contradiction: if $f(x) < f(y)$ but $x>y$ then the condition on strict monotonicity immediately implies that you arrive at a contradiction. $\endgroup$ – Thomas Nov 9 '18 at 16:42
  • $\begingroup$ why do you say $x\leq y$ this doesn't add anything to the definition. $x< y$ -> $f(x)\leq f(y)$ is sufficient $\endgroup$ – Allan Henriques Dec 12 '20 at 17:18

As I have always understood it (and various online references seem to go with this tradition) is that when one says a function is increasing or strictly increasing, they mean it is doing so over some proper subset of the domain of the function. To say a function is monotonic, means it is exhibiting one behavior over the whole domain. That is, a monotonically increasing function is nondecreasing over its domain and is also an increasing function since it is non-decreasing over any subset of the domain. Similarly, a strictly monotonically increasing function is a function that is strictly increasing over its whole domain, rather than simply increasing over a subset of the domain (as determined from the increasing/decreasing test in Calculus). One can say similar things about a monotonically decreasing function vs. a decreasing function. This largely echoes what was said by Thomas above, but, taking monotone as a term referring to the behavior of a function over the whole domain, one does not need to say "I'm used to." That said, one should always be clear on what definitions are being used as consistency is not a human's strong point.


Let $y=f(x)$ be a differentiable function on an interval $(a,b)$. If for any two points $x_1,x_2 \in (a,b)$ such that $x1 \lt x2$, there holds the inequality $f(x_1) \leq f(x_2)$, the function is called increasing on ths interval.

If there holds the inequality $f(x_1) \lt f(x_2)$, the function is called strictly increasing on the interval.

  • $\begingroup$ i like this clear & concise answer. one thing which has always bugged me about this definition tho is that "x = k" is technically both increasing and decreasing. intuitively i feel the definition should include something like $\exists x1, x2$ where $x1 < x2$ such that $f(x1) < f(x2)$. $\endgroup$ – orion elenzil May 21 '19 at 21:04

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