I get stuck in checking something. Assume we have a linear subspace $V_j$ of $\mathbb{R}^m$. And we have an arbitrary function h: $V_j\to\mathbb{R}$. Denoting $\pi_j$ as the orthogonal projection from $\mathbb{R}^m$ to $V_j$. Then we can define a function $h \circ \pi_j$: $\mathbb{R}^m\to\mathbb{R}$. Let $w$ be any vector in $V_j^\bot$, the orthogonal complement of $V_j$. The assertion is that $h \circ \pi_j$ is annihilated by the constant coefficient differential operators $w\cdot\triangledown$. Here is what I have done. Let $\pi_j(x_1...x_m)=(y_1...y_m)$. Here I view each $y_j$ as a function: $\mathbb{R}^m\to\mathbb{R}$. So we want to show that $\triangledown (h\circ \pi_j)$ is in $V_j$.
I just write down one partial derivative as an example: $$\frac{\partial(h\circ \pi_j)}{\partial x_1}=\frac{\partial(h(y_1...y_m))}{\partial x_1}=\sum_i\frac{\partial h}{\partial y_i}\cdot\frac{\partial y_i}{\partial x_1}$$ $$w\cdot\triangledown (h\circ \pi_j)=\sum_k\sum_i\frac{\partial h}{\partial y_i}\cdot\frac{\partial y_i}{\partial x_k}\cdot w_k$$ Since h is arbitrary, we need $$\sum_k\frac{\partial y_i}{\partial x_k}\cdot w_k=0$$ for fixed i, which means $\triangledown y_i$ is orthogonal to $w$. I don't see why and I haven't used $\pi_j$ is the orthogonal projection. And I don't know how to use it. Thanks for any help!
Let $U$ be the matrix for $\pi_j$ (so $\pi_j(\vec{x})=U\vec{x}$). Here I'm using the standard basis of $\mathbb{R}^n$, but this is fine since the column space of $U$ is $V_j$. Let $\vec{h}$ be any vector in $\mathbb{R}^n$ such that $h(\vec{x})=\vec{h}^T\vec{x}$. Then $\triangledown(h\circ \pi_j)$ has the matrix $\vec{h}^TU$ since these are linear functions, and this matrix is a $1\times n$ row vector. The dot product between it and $w$ is $\vec{h}^TUw$, which is $0$ since $Uw=0$.
This carries over to what you were saying because each $\left(\frac{\partial y_i}{\partial x_k}\right)_k$ is a row of $U$.