Let the kernel be $f(\mathbf{x},\mathbf{y}) = \arccos(\mathbf{x}^T \mathbf{y})$, where $\mathbf{x}$ and $\mathbf{y}$ are $\ell_2$ normalized vectors of the same dimensionality, and $\arccos(\cdot): [-1,1] \to [0,\pi]$ is the inverse cosine function.

Question: Is $f$ conditionally negative definite? If yes, how can I prove it?

I know the definition of a conditionally negative definite kernel, but I find it difficult to apply.

A kernel $f: (\mathcal{X} \times \mathcal{X}) \to \mathbb{R}$ is called (conditionally) negative definite if it it symmetric and $\sum_{i,j=1}^m c_i c_j f(x_i,x_j) \leq 0$ for all $m \in \mathbb{N}$, $\{x_1,\cdots,x_m\} \subseteq \mathcal{X}$ and $\{c_1,\cdots,c_m\} \subseteq \mathbb{R}$ with $\sum_{i=1}^m c_i = 0$.