Let $f(x)$be a polynomial with variables in $\mathbb{R}^n$
We note $$g(x)=\sum_{3\le |\alpha|\le r} |\partial_x^{\alpha}f(x)|^{\frac{1}{|\alpha|}}$$ $$E=\left\{x\in \mathbb{R}^n, \;|\nabla f(x)|^{\frac{4}{3}}\ge 2\Big(|Hess f(x)|+(g(x))^4\Big)\right\}$$ and $$F=\left\{x\in \mathbb{R}^n, dist(x,\mathbb{R}^n \backslash E)\le \frac{2}{g(x)}\right\}$$
I want to prove that $F$ is a semialgebraic set.
Please help me to do so. Thanks
For any rational number $r \in\Bbb Q$, the function $x\mapsto |x|^r$ is semialgebraic. As $\partial^{\alpha} f(x)$ is again a polynomial, this shows that $g$ is semialgebraic.
Since magnitude of a vector with semialgebraic entries is again semialgebraic, $|\nabla f(x)|^{\frac43}$ is a semialgebraic function, and since $|Hess f(x)|$ is the absolute value of a polynomial, this shows that both sides of the equation in $E$ are semialgebraic, and so the set of things satisfying this inequality is again semialgebraic.
Since the distance function to a semialgebraic set is semialgebraic (and semialgebraic sets are closed under complements), $F$ is semialgebraic.
All of the statements above are very basic - I would recommend reviewing an overview article (like http://www.journalofsing.org/volume13/article4.html) or beginning book if they're not sticking.