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Recall the following definitions:

  • We say a set $E\subseteq\mathbb{R}^n$ is semi-algebraic if there exist real polynomials $g_{ij},h_{ij}:\mathbb{R}^n\rightarrow\mathbb{R}$ such that

$$E=\bigcup_{j=1}^p\bigcap_{i=1}^q\{x\in\mathbb{ R}^n:g_{ij}(x)=0\text{ and }h_{ij}(x)<0\}.$$

  • A function $f:\mathbb{R}^n\rightarrow(-\infty,\infty]$ is called semi-algebraic, if its graph \begin{equation*} \{(x,y)\in\mathbb{R}^{n+1}:f(x)=y\} \end{equation*} is semi-algebraic.

Literature says real polynomials are semi-algebraic, which to me is a natural result. To further understand this concept, I am wondering the following:

  • Is the exponential function $x\mapsto e^x$ semi-algebraic?

Unfortunately I have no idea of how to prove or disprove it, so any hint or comment will be appreciated. Thanks a billion!

Update: I am an optimizer and optimization people care about this concept because semi-algebraic functions enjoy Kurdyka-\L{}ojasiewicz property, a key assumption in many convex/non-convex optimization problems.

RamenHunter
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  • If you think about it, you are allowed to intersect finitely many zero-sets and pre-images of the open half plane of real polynomials and then take a finite union of such things. Now the zero-sets are basically points and line-segments, whereas the half-planes are open. Hence your graph has to look around some points like the zero-set of a polynomial. Now you can find out in some way or another that the zero-set of a polynomial can be parametrised by a polynomial map. This is impossible for a segment of the graph of the exponential map as you can see by taking derivatives. – s.harp Mar 30 '20 at 18:11

1 Answers1

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If $e^x$ were semi-algebraic, its graph would be a union of finitely many semialgebraic sets defined by $\{(x,y)\in\Bbb R^2\mid p_i(x,y)=0\}$. By taking the products of all the $p_i$, this means that there is a polynomial $p(x,y)$ which vanishes on all points of the form $(x,e^x)$. Expanding and collecting terms, we get $\sum_{j=1}^{n} h_j(x)e^{jx}=g(x)$ for some polynomials $g,h_j$. Taking the derivative $\deg_x(g)+1$ times, we see that we would have an equation of the form $\sum_{j=1}^n e^{jx}q_j(x)=0$ holds on $\Bbb R$ with $q_j(x)$ nonzero polynomials. But this is impossible: $e^x\neq 0$, and a nonzero polynomial only has finitely many roots.

There are also plenty of other contradictions we could derive depending on how much semialgebraic geometry you know. For instance, Lojasiewicz's inequality states that if $f,g:K\to \Bbb R$ are continuous semialgebraic functions on a compact semialgebraic set $K$ so that $f^{-1}(0)\subset g^{-1}(0)$, then we have that there exist $C,N>0$ so that $$|f(x)|\geq C|g(x)|^N$$ for all $x\in K$. Taking $K=[-1,1]$, $f=e^{-1/x^2}$ and $g=|x|$, we see that this inequality is not satisfied, but $f$ would be semialgebraic if $e^x$ was.

KReiser
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  • Is $h(x)$ in your answer necessarily a polynomial? Let's say $p(x,y)=x^2y+2x^3+y^2$. Then $p(x,e^x)=e^x(x^2+e^x)+2x^3=0\Rightarrow(x^2+e^x)e^x=-2x^3$ which means $h(x)=x^2+e^x$. What did I miss here? – RamenHunter Mar 30 '20 at 19:33
  • I guess what's always true is we get $h(x,e^x)e^x=g(x)$ for some polynomial $h(x,y)$ and $g(x)$, then taking derivative gives the contradiction? – RamenHunter Mar 30 '20 at 19:33
  • That was a minor mistake - I have fixed it. – KReiser Mar 30 '20 at 19:39
  • Splendid. Thank a billion! – RamenHunter Mar 30 '20 at 19:44
  • It looks like this proof fails when the function has a root, say $e^x-1$? – RamenHunter Mar 30 '20 at 19:50
  • What? There's no failure there. If that's the first step, then you write $e^x=1$, take derivatives of both sides, and then you have $e^x=0$, which doesn't work (just like the proof says). – KReiser Mar 30 '20 at 19:56
  • You are right. I really need to have lunch now. Thanks! – RamenHunter Mar 30 '20 at 20:06
  • I don't understand why the $q_j$ have to be nonzero. Do we know that the $h_j$ are of higher degree than $g$? Edit: I think I understand it now, it's because we retain an original $h_j(x)$ when we differentiate $h_j(x)e^{jx}$ right? – Grinsekotze Dec 18 '20 at 17:49
  • @Grinsekotze Suppose $h$ is of degree $d$. Look at the term of the form $x^de^{jx}$ in $\frac{d}{dx} h(x)e^{jx} = jh(x)e^{jx} + h'(x)e^{jx}$. – KReiser Dec 18 '20 at 20:06