Consider parameterized linear programming $V(\theta) = \max_x \langle c(\theta),x\rangle$ s.t. $A(\theta)x\leq b(\theta)$, $x\geq 0$. Let's also assume $c,A,b$ are infinitely differentiable with respect to $\theta$.
It is known (the envelope theorem) that under suitable conditions (one sufficient condition is that the set of binding constraints does not change for all $\theta\in B_\delta(\theta_0)=\{\theta: \|\theta-\theta_0\|\leq\delta\}$ for some $\delta>0$) the objective $V$ is differentiable in $\theta$ for $\theta\in \mathrm{Int}B_\delta(\theta_0)$ and furthermore $$ \nabla_\theta V(\theta) = \langle\nabla c(\theta), x^*(\theta)\rangle-\langle\lambda^* (\theta) ,\nabla b(\theta) -\partial_\theta A(\theta)x^*(\theta)\rangle, $$ where $x^*(\theta)$ is the optimal solution at $\theta$ and $\lambda^*(\theta)$ is the optimal Lagriangian multipler for all constraints $A(\theta)x\leq b(\theta)$.
My question is: under the condition that the set of binding constraints does not change for all $\theta\in B_\delta(\theta_0)=\{\theta: \|\theta-\theta_0\|\leq\delta\}$ for some $\delta>0$, is $V$ also twice differentiable in $\theta$? If so, can we compute $$ \nabla_{\theta\theta^\top}^2V(\theta) $$ for all $\theta\in\mathrm{Int}B_\delta(\theta_0)$? If the hessian doesn't necessarily exist, what additional conditions should be imposed?
While a general solution to the problem is certainly welcome, for my particular application it is sufficient to handle the case when $c$ is a constant function and both $A(\theta),b(\theta)$ are linear functions of $\theta$.
