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Are all multivariate elementary functions $C^\infty$ on any open ball for which they are defined?

I believe they are, and can find no counterexample, but do not know how to prove this. (All the candidate counterexamples I found were on boundaries, not in open balls).

Update

Please use this definition of elementary functions.

If not, what restrictions on the set of elementary functions make them all $C^\infty$?

SRobertJames
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    What is your definition of an elementary function? Typically the absolute value is considered an elementary function and it is not even differentiable. – Severin Schraven Jan 18 '24 at 21:38
  • If you don't like the absolute value, look at the cubic root, which is also not differentiable (and which is defined on all of $\mathbb{R}$). – Severin Schraven Jan 18 '24 at 21:40
  • Can you include your definition of elementary functions? Also, it's a bit rude to completely change the question after somebody posted an answer to your original question. – Severin Schraven Jan 19 '24 at 01:46

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Let $n\in \mathbb{N}$ be odd. Then the $n$th root $\mathbb{R}\rightarrow \mathbb{R}, x \mapsto x^{1/n}$ is an elementary function (it is the inverse of the polynomial map $x\mapsto x^n$) and it is not differentiable at the origin (and in particular it is not $C^\infty$). Here I used the definition of wikipedia which includes inverses of polynomial functions.

The absolute value is yet another elementary function ($\vert x \vert=\sqrt{ x^2}$) which is not differentiable and hence also not smooth.

Added: If we want a class of functions which are all smooth on their maximal domain, we can exclude the logarithm in the definition What makes elementary functions elementary? The problematic functions arise from the fractional powers, which disappear if we exclude the logarithm.

Severin Schraven
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