Let $K$ be a bounded convex set of $\mathbb{R}^n$ and $x_1,\ldots,x_k\in K$. Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ be a $C^\infty$ function that cancels on points $x_1,\ldots,x_k$ . When $n=1$, we have the following classical estimate for $f$. For $x\in K$, one has $$f(x) = \frac{1}{k!}\left(\prod_{i=1}^k(x-x_k)\right)f^{(k)}(\xi),\qquad\qquad(1)$$ for some $\xi\in K$, from which we deduce the subsequent bound $$\|f\|_{\infty,K} \leq \frac{1}{k!}\mathrm{Diam}(K)^k\|f^{(k)}\|_{\infty,K}.\qquad\qquad(2)$$ The proof can be deduced from Lagrange interpolation at point $x_1,\ldots,x_k$. Since $f$ cancels on these points, Lagrange interpolation polynomial is zero and $(1)$ is the usual remainder formula for Lagrange interpolation. Heuristically, if $f$ is zero at many points and sufficiently regular then the infinite norm of $f$ is small.
In higher dimensional setting the question is more difficult since there are no analog of Lagrange interpolation. Kergin interpolation (see here) asserts the existence of a polynomial $Pf$ that cancels on points $x_1,\ldots,x_k$ and such that $$\|f-Pf\|_{\infty,K} \leq \frac{1}{k!}\mathrm{Diam}(K)^k\sum_{|\alpha|=k}\binom{k}{\alpha}\|\partial^\alpha f\|_{\infty,K}.\qquad\qquad(3)$$ But contrary to Lagrange interpolation, the polynomial $Pf$ is not constructed only from the values of $f$ at points $x_1,\ldots, x_k$ and thus $Pf$ is not necessarly zero on $K$. We cannot then deduce that a multidimensional version of $(2)$ holds.
I was wondering whether a similar inequality as $(1)$ or $(2)$ holds true in all dimension, maybe with some constraints of the points $x_1,\ldots,x_k$ (I know that some interpolation schemes require that the points $x_1,\ldots,x_k$ are in general position in the sense that they do not lie on zero set of a polynomial of degree $k-1$).
For instance, if all the points $x_1,\ldots,x_k$ lies on a line $D$ then there is no hope for an estimate such as $(2)$ (take $f(x,y) = y$ and $K$ a small ball centered at zero).
