Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose we have some function $f(x)$ with local extrema at $x_1, x_2, \dots$, and a second function $g(x)$ which is continuous, strictly increasing and non-zero everywhere over the range of the $x_i$. Will $g(f(x))$ have its local extrema at the same $x_i$ and no others?

If so, are there any obvious loosenings of the constraints on $g$ for which this will remain true?

(I'm really thinking of this in the context of signal processing, looking at transformations that preserve the visual structure of an image, but it seems like a general question that must have been trivially proved by someone 250 years ago...)

share|cite|improve this question
up vote 4 down vote accepted

You don't need to assume that $g$ is non-zero, and it could be strictly decreasing as well. Furthermore, the conditions on $g$ only need to hold on the image of $f$ (which doesn't need to the be whole of $\mathbb{R}$, for example).

On the other hand, if $g$ has a local extremum at $y=f(z)$ and $f$ is strictly increasing around $z$, then you're obviously in trouble, because $g\circ f$ will have a local extremum at $z$. But having no local extrema is equivalent to being strictly monotonic.

The only that might relax the conition on $g$ is that it has local extrema exactly where $f$ does and they "cancel each other out" or "amplify each other".

share|cite|improve this answer

Since g is continuous and strictly increasing, its inverse $g^{-1}$ is a function and strictly increasing. Since both are strictly increasing, $a<b\Leftrightarrow g(a)<g(b)\Leftrightarrow g^{-1}(a)<g^{-1}(b)$. From this, it follows that $x_i$ is a local max (min) of g(f(x)) iff it is a local max (min) of f(x).

If g were continuous and strictly decreasing, it would exchange local maximums and minimums (because $a<b\Leftrightarrow g(a)>g(b)$ and $a<b\Leftrightarrow g^{-1}(a)>g^{-1}(b)$).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.