Given a continuous and deriviable function of many variables, how do I know when this function is equals to zero on all the corners (or vertices) of a unit hypercube, i.e. all points, where each coordinate is equals either to 1 or 0.


I think that if I explore the function inside or/and around the unit hypercube, I will be able to tell if:


∀x1...xn: x1∈{0,1} ∧ ... ∧ xn∈{0,1} then
f(x1, ... , xn)=0 where $f: \Bbb{R}^n\rightarrow\Bbb{R}$


By exploring the function, I can compute the partial derivatives of each variable and/or multiple integrals and/or gradients and/or normal vectors, etc but I don't know what exploration should I do exactly to find out the answer.


If I know how the function behaves on the half point, i.e. (0.5, ... , 0.5), which is the center of the unit hypercube, can I find the answer to my question?


Any idea?


**Note:** you can assume that the given function is analytic.



Given a continuous function of many variables, how do I know when this function is equals to zero on all the corners of a unit hypercube?
Since $x^m = x$ for $x \in \{0,1\}$ and $m \ge 1$, we may assume wlog all exponents in your polynomial are $0$ or $1$. Thus your polynomial can be written as $$P(x) = \sum\_{\alpha \in A} c\_\alpha x^\alpha$$
where $A$ is a collection of not too many subsets of $\{1,\ldots, n\}$ and 
 $x^\alpha = \prod\_{j \in \alpha} x^j$.
Thus $P$ is $0$ at every point of the hypercube iff $P(\chi\_\beta) = \sum\_{\alpha \in A: \alpha \subseteq \beta} c\_\alpha = 0$ for every subset $\beta$ of $\{1,\ldots,n\}$. Of course we remove all terms where $c\_\alpha = 0$. Now unless $A$ is empty, there must be some $\alpha \in A$ that is minimal (i.e. has no proper subset in $A$). If so, $P(\chi\_\alpha) = c\_\alpha \ne 0$. So the only case where $P(x)$ is $0$ at all corners of the hypercube is where $A$ is empty, i.e. the polynomial is identically $0$.


