# Difference between strictly increasing and increasing functions

What is the precise difference between strictly increasing and increasing functions?? I see these terms being thrown around a lot My guess is that strictly increasing mean that derivative is only greater than 0 and in case of just increasing derivative can be greater than or equal or 0?

• Something not noted in the answers is that, even in the differentiable case, it is not the case that a function is strictly increasing iff its derivative is positive everywhere. Indeed, the function $f(x)=x^3$ is strictly increasing, but $f'(0)=0$. Aug 25, 2019 at 16:59
• The definitions also imply that $f(x) = 0$ is both a (non-strictly) increasing and decreasing flat-line function. Sep 11, 2022 at 22:43

You're basically correct (update: as stated in this answer and question comment by Wojowu, I forgot to add that strictly increasing differentiable functions do not require the derivative to always be positive), except the concept of "strictly increasing" and "increasing" functions also applies to non-differentiable functions. In particular, for all $$x \gt y$$, where $$x$$ and $$y$$ are in the function domain, strictly increasing means $$f(x) \gt f(y)$$ while just increasing means $$f(x) \ge f(y)$$.

• To give a specific example, a constant function is increasing but not strictly increasing.
– user694818
Aug 25, 2019 at 5:08

Strictly increasing means that $$f(x)> f(y)$$ for $$x>y$$. While increasing means that $$f(x)\geq f(y)$$ for $$x>y$$.

Example:

$$f:\mathbb{R}\to \mathbb{R}, x\mapsto x$$ is strictly increasing.

And the function $$f:\mathbb{R}\to\mathbb{R}, x\mapsto \lfloor x\rfloor$$ is increasing.

$$\lfloor x\rfloor$$ notes the flooring-function: https://en.wikipedia.org/wiki/Floor_and_ceiling_functions#Definition_and_properties

And looks like this:

Note that this function stays constant on certain intervals, but increases over time.

Note that your inituition about derivatives is only $$3$$ quarters correct: If the function $$f$$ under consideration is differentiable in the considered interval open $$I$$, then

$$f \text{ is increasing in } I \Longleftrightarrow f'(x) \ge 0\; \forall x \in I.$$

OTOH, under the same conditions only one of the implications holds for strictly increasing functions:

$$f \text{ is strictly increasing in } I \Longleftarrow f'(x) > 0\; \forall x \in I.$$

The other implication is wrong, for example $$f(x)=x^3$$ is strictly increasing in $$I=(-1,1)$$, while the derivative is zero at $$x=0$$.