Let $f\in\mathbb{Z}[x,y]$ and fix $k\in\mathbb{N}$. Suppose that for all $v=(v_1,v_2)\in S^1$, $$ \left|\frac{d^k}{dt^k}f(tv)\right|_{t=0} $$ is bounded by $C$. (The value of $C$ that I use comes from the geometry of the zero set of the variety for $f$, but it is unlikely to be useful here).
Is it possible to derive a (reasonable) bound on $$ \left|\frac{\partial^k}{\partial x^i\partial y^{k-i}}f(x,y)\right|_{(x,y)=(0,0)} $$ in terms of $C$? (I am not quite sure how to define reasonable as this is a step in a paper that I'm working on.)
My thoughts:
One can note that $$ \frac{d^k}{dt^k}f(tv)=\sum_{i=0}^k\begin{pmatrix}k\\i\end{pmatrix}\frac{\partial^k f(0,0)}{\partial x^i\partial y^{k-i}}v_1^iv_2^{k-i}. $$ Therefore, for certain choices of $v_1$ and $v_2$, the bound is easy. The problem is that for any fixed $v$, there may be cancellation in the sum.
I also considered looking at this with Fourier, by writing $v=(\cos(\theta),\sin(\theta))$. In this case, the question becomes recovering $\cos(\theta)^i\sin(\theta)^{k-i}$ from the Fourier coefficients, which are known to be bounded. This then becomes a linear system because, for example, when $k=2$, $$\cos(x)^2=\frac{1}{2}+\frac{1}{2}\cos(2x)$$ $$\cos(x)\sin(x)=\frac{1}{2}\sin(2x)$$ and $$\sin(x)^2=\frac{1}{2}-\frac{1}{2}\cos(2x).$$ The corresponding matrix is $$ \begin{pmatrix}\frac{1}{2}&0&\frac{1}{2}\\0&\frac{1}{2}&0\\\frac{1}{2}&0&-\frac{1}{2} \end{pmatrix} $$ (Here, the columns represent $1$, $\sin(x)$, and $\cos(2x)$, respectively). I noticed that, at least for the first few cases, the smallest Eigenvalue of these matrices was bounded below by a "reasonable" quantity that did not grow too quickly (like $1/2^k$). I, however, couldn't prove a bound.