Let $\mathcal{C}$ be a [pointed, salient](https://en.wikipedia.org/wiki/Convex_cone#Blunt.2C_pointed.2C_flat.2C_salient.2C_and_proper_cones) cone in $\mathbb{R}^d$. We may also assume that $\mathcal{C}$ is full-dimensional. Consider the set of binary classifiers $$\mathcal{H} = \{\boldsymbol{x}\mapsto\boldsymbol{1}\hspace{-0.9ex}\mathrm{1}(\boldsymbol{h}^\mathsf{T}\boldsymbol{x}\ge 0)\mid \boldsymbol{h} \in \mathcal{C} \}.$$
Is there any bound known for the VC-dimension of $\mathcal{H}$ that depends on $\mathcal{C}$ and can be significantly better than the VC-dimension of the set of all (linear) classifiers in $\mathbb{R}^d$ (i.e., $d$)? 


**Edit:** Above, $\boldsymbol{1}\hspace{-0.9ex}\mathrm{1}(\cdot)$ denotes the binary indicator function. So a more explicit definition of $\mathcal{H}$ is the set of functions $$f\_\boldsymbol{h}(\boldsymbol{x})=\begin{cases} 0 & \text{if }\boldsymbol{h}^\mathsf{T}\boldsymbol{x}<0 \\ 1 & \text{if }\boldsymbol{h}^\mathsf{T}\boldsymbol{x}\ge0\end{cases},$$ parametrized by $\boldsymbol{h}\in \mathcal{C}$.



VC dimension of cone-restricted linear classifiers
No, I claim that the set of $d$ unit vectors, $e\_1=(1,0,\ldots,0)$, $e\_2=(0,1,0,\ldots,0)$, $\ldots,$ $e\_d=(0,\ldots,0,1)$, along the axes can be shattered, except that for every $\mathcal H$ either the empty set, or all $d$ of the won't be in any $h\in \mathcal H$, so the VC-dimension is $d-1$. (For similar reasons, it cannot be $d$ for any point set.)


Indeed, if $\mathcal H$ is the *opposite* of the cone spanned by the above $d$ vectors, i.e., the all-negative hyperoctant, and $h\_I=-\sum\_{i\in I} e\_i$, then $f\_h(e\_i)=0$ if and only if $i\in I$.


