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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...)

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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".

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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 $a<b\Leftrightarrow g^{-1}(a)>g^{-1}(b)$).

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