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I saw in my lecture not about this theorem of elementary function continuity:” All elementary functions are C1 in the interior of their maximal domain except the norm when the argument is zero and the powers with exponent $\alpha \in (0,1)$ when the argument is 0 and 0 is not at the boundary of the maximal domain” I really don’t know what does it means saying “ except the norm when the argument is zero and the powers with exponent $\alpha \in (0,1)$ when the argument is 0 and 0 is not at the boundary of the maximal domain” what is it argument of the norm? And can anyone give me specific examples on these exceptions? Thank you so much!

Eileen
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    It means except $f(x)=|x|$ at $x=0$ (because that absolute value function is continuous but not differentiable at $x=0$). – Gerry Myerson Dec 31 '21 at 22:15
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    Similarly for $x^a$ with $0\le a\le1$ at $x=0$ – think about $y=\sqrt x$ at $x=0$. – Gerry Myerson Dec 31 '21 at 22:16
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    are C1 in the interior of their maximal domain except --- Wouldn't this be $C^{\infty},$ or even real-analytic? It seems strange to use such a relatively weak smoothness property in the presence of so many technical and advanced terms. But maybe I'm overlooking some examples where $C^1$ is about the best we can claim. – Dave L. Renfro Dec 31 '21 at 22:24
  • @GerryMyerson Thank you so much this is very clear! – Eileen Jan 01 '22 at 07:42
  • @DaveL.Renfro I’m not sure if we should consider$C^\inf$, this is from my master mathematical method courses and we use the book Mathematics for Economists, so maybe the level is not hard since it is for master and economic student. – Eileen Jan 01 '22 at 07:44
  • Actually, I think $C^{\infty}$ is easier to understand than $C^1$ -- continuity of the derivative function vs. being differentiable is a bit more subtle, I think, than simply being differentiable as many times as one desires. However, your observation seems reasonable for real-analytic. On the other hand, maybe the property of being $C^1$ is a common "smoothness hypothesis" in your subject, in which case stating the result as $C^1$ becomes a more natural observation. – Dave L. Renfro Jan 01 '22 at 18:02

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For example $ |x|$ is not differentiable at 0, nor are $x^p, p\in (0,1)$

Annika
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