for all h ∈ Uδ(0). Let x 0 be an arbitrary vector representation that projects to X. Then there is a γ ∈ Γ with x 0 = γ(x). Since ˜f is invariant under the group actions of Γ, we have ˜f(x 0) = ˜f(x). Then for each h 0 ∈ Uδ(0), we find that

$${\tilde{f}}(\mathbf{x}^{\prime}+\mathbf{h}^{\prime})-{\tilde{f}}(\mathbf{x}^{\prime})={\tilde{f}}(\mathbf{x}+\mathbf{h})-{\tilde{f}}(\mathbf{x})=\left\langle\nabla{\tilde{f}}\left(\mathbf{x}\right),\mathbf{h}\right\rangle+o(\mathbf{h}),$$

where h ∈ X with γ(h) = h 0. Since the elements of Γ are isometries, we have khk = kh 0k giving h ∈ Uδ(0). In addition, from isometry of γ follows

$$\left\langle f_{\mathbf{x}},\mathbf{h}\right\rangle=\left\langle\gamma\left(\nabla{\tilde{f}}\left(\mathbf{x}\right)\right),\gamma(\mathbf{h})\right\rangle=\left\langle\gamma\left(\nabla{\tilde{f}}\left(\mathbf{x}\right)\right),\mathbf{h}^{\prime}\right\rangle.$$

We obtain
$$\tilde{f}(\pmb{x}'+\pmb{h}')-\tilde{f}(\pmb{x}')=\left\langle\gamma\left(\nabla\tilde{f}\left(\pmb{x}\right)\right),\pmb{h}'\right\rangle+o'(\pmb{h}'),$$) = e x $\gamma^{-1}(\pmb{h}')$ satisfies. 
where o 0(h 0) = o ◦ γ
−1(h 0) satisfies

$$\operatorname*{lim}_{{\boldsymbol{h^{\prime}\to0}}}{\frac{o^{\prime}({\boldsymbol{h^{\prime}}})}{\|{\boldsymbol{h^{\prime}}}\|}}=\operatorname*{lim}_{{\boldsymbol{h^{\prime}\to0}}}{\frac{o(\gamma^{-1}({\boldsymbol{h^{\prime}}}))}{\|{\boldsymbol{h^{\prime}}}\|}}=\operatorname*{lim}_{{\boldsymbol{h^{\prime}\to0}}}{\frac{o(\gamma^{-1}({\boldsymbol{h^{\prime}}}))}{\|\gamma^{-1}({\boldsymbol{h^{\prime}}})\|}}=0.$$

This proves that ˜f is differentiable at each vector representation that projects to X. In addition, from the proof follows that the gradient of ˜f at x 0 = γ(x) is of the form

$$\nabla{\bar{f}}\left(\mathbf{x}^{\prime}\right)=\gamma\left(\nabla{\bar{f}}\left(\mathbf{x}\right)\right).$$

ut

## Generalized Differentiable Orbifold Functions

Proposition 4. Let f : XΓ → R *be an orbifold function. Suppose that its lift*
˜f : X → R is generalized differentiable at a vector representation x that projects to X ∈ XΓ . Then ˜f is generalized differentiable at γ(x) for all γ ∈ Γ and

$$\partial\tilde{f}(\gamma(\mathbf{x}))=\gamma\left(\partial\tilde{f}(\mathbf{x})\right).$$

is a subdifferential of ˜f at γ(x) *for all* γ ∈ Γ. Proof. Since ˜f is generalized differentiable at x, there is a multi-valued mapping
∂ ˜f : Uδ(x) → 2 X defined on some neighborhood Uδ(x). Let γ ∈ Γ be an arbitrary permutation and x 0 = γ(x). Then

$$\partial\tilde{f}:{\mathcal U}_{\delta}(\mathbf{x}^{\prime})\to2^{\mathcal{X}},\quad\mathbf{y}^{\prime}=\gamma(\mathbf{y})\mapsto\gamma\left(\partial\tilde{f}(\mathbf{y})\right)$$

is a multi-valued mapping in a neighborhood of x 0.

Since γ is a homeomorphic linear map, we find that γ(∂ ˜f(x)) = ∂ ˜f(x 0) is a convex and compact set.