# Lipschitz continous of matrix functions

If a matrix function $g(\bf X)$ can be shown to be Lipschitz continuous in the whole space $\mathbb{R}^{N\times N}$. Then can we say that this function is also Lipschitz continuous in the real symmetric space $\mathbb{S}^N$?

Intuitively, since $\mathbb{S}^N$ is a subspace of $\mathbb{R}^{N\times N}$, this conclusion seems to be natural. However, the derivatives of symmetric matrices are different from the general matrices, which makes me a bit confused.