Dimension of a general partial derivative of a linear subspace of polynomials

Question

Let $U \subseteq \mathbb{C}[x_1,\dots, x_n]_d$ be a linear subspace of the vector space of homogeneous degree-$d$ polynomials (including zero). I would like a proof or counterexample of the claim that for a general linear partial derivative $\partial=\alpha_1 \frac{\partial}{\partial{x_1}}+\dots + \alpha_n \frac{\partial}{\partial{x_n}}$, it holds that the linear subspace $\partial U \subseteq \mathbb{C}[x_1,\dots, x_n]_{d-1}$ obtained by applying $\partial$ to every element of $U$ satisfies the following dimension bound: $$ \binom{n+d-1}{d}\dim(\partial U) \geq \binom{n+d-2}{d-1} \dim(U) $$

Answer 1

Change coordinates, or act by a linear transformation, so that $U$ is a general subspace and we are differentiating by $\partial = \frac{\partial}{\partial x_1}$. Since $U$ is general, it has a basis whose leading monomials, in lexicographic order, are an initial segment: $x_1^d, x_1^{d-1} x_2, \dotsc$.

If $\dim U \leq \binom{n+d-2}{d-1}$ then every basis element of $U$ has a leading term containing an $x_1$. Then $\partial$ is injective on $U$ (the derivatives of basis elements are nonzero and have pairwise distinct lex-leading terms, so they are linearly independent). In this case $\dim(\partial U) = \dim(U)$.

Otherwise, $\dim U > \binom{n+d-2}{d-1}$, but $\partial U$ consists of all the degree $d-1$ forms: as before, the first $\binom{n+d-2}{d-1}$ basis elements of $U$ map injectively, hence surjectively (the other basis elements have no $x_1$, and are annihilated). In this case $\dim(\partial U) = \binom{n+d-2}{d-1}$. Since $\dim(U) \leq \binom{n+d-1}{d}$, the inequality you seek is satisfied.