I have a multidimensional function $A(h, u)$, that is linear in $u$ for any $h$, and convex in $h$. How can I prove that $\mathcal{A}(u) = \min_h A(h, u)$ is Lipschitz? If this is some standard theorem, then I would be grateful for textbook reference.
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This is not true in general. For example, $A(h,u)=hu$ (both $h,u$ in $\mathbb R$) satisfies the conditions, but taking the infimum over $h$ results in discontinuous function $$\mathcal {A}(u) = \begin{cases} 0 ,\quad & u=0\\ -\infty, \quad & u\ne 0\end{cases}$$
Here is what can be said:
- $\mathcal{A}$ is a concave function of $u$ (standard fact: concavity is preserved by taking infimum of a family of functions).
- A concave function on $\mathbb R^n$ is locally Lipschitz on the set where it is finite (i.e., not equal to $-\infty$). (See, e.g., A convex function that is bounded on a neighborhood is Lipschitz. A standard reference is Convex Analysis by Rockafellar.)
So, if in your case $\mathcal{A}$ never turns into $-\infty$, then $\mathcal{A}$ is locally Lipschitz.
You don't get global Lipschitz continuity without extra assumptions. Example: if $A(h,u) = hu+h^2$ (again, both $h,u$ are in $\mathbb R$), then the minimum is attained at $h=-u/2$, and therefore $\mathcal A(u) = -u^2/4$.
