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I am reading the paper "A. D. Ioffe, An invitation to tame optimization, SIAM J. Optim., 19 (2009), 1894–1917" and stuck at Proposition 3.1. The author claims (in the language of semi-algebraic geometry) that if $\varphi:\mathbb{R}^n\times\mathbb{R}^m\rightarrow\overline{\mathbb{R}}$ is a semi-algebraic function then the function $$\psi(x):=\sup_{\varepsilon>0}\inf_{0<\left\|h\right\|<\varepsilon}\varphi(x,h)$$ is also semi-algebraic. I tried to use the Tarski-Seidenberg Theorem to prove this claim but I could not. Although it is easy to show that $g(x)=\inf_{h\in S}\varphi(x,h)$ is semi-algebraic for any semi-algebraic set $S\subset \mathbb{R}^m$, it seems too hard for me to deal with $\inf_{0<\left\|h\right\|<\varepsilon}\varphi(x,h)$ where the infimum is taken respect to $\varepsilon$ which is not fixed.

I am very grateful to you if you could help me to overcome this obstacle. Thank you for your help!

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    If you accept $\inf_{0\leq |h|\leq \varepsilon} \varphi(x,h)$ for fixed $\varepsilon$ as semi-algebraic, then you should also accept the larger definition as semi-algebraic: the point is that you're adding an extra variable in to your expression and then letting that vary over some semi-algebraic domain and taking a semi-algebraic function of the result. You could also literally write out the whole logical formula for this, which would be a pain, but it would show the expression is definable and thus semi-algebraic. – KReiser Aug 21 '22 at 13:59
  • Could you please show it to me? – Dat Ba Tran Aug 22 '22 at 14:05

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