A condition for 'similarity' of subgradients of convex proper functions

Question

Motivated by the answer in Under which conditions does small uniform norm imply 'similarity' of subgradients. I have the following question.

Let $\mathcal{S}$ be the set of proper convex functions functions from $X$ to $\mathbb{R}$, where $X$ is a open and convex subset of $\mathbb{R}^{n}$. I was wondering under which conditions on $\mathcal{S}$ we have \begin{gather} \forall \epsilon>0\ \exists \delta>0 \text{ such that for } f,h \in \mathcal{S} ,\ ||f-h||_{\infty} < \delta \\ \Rightarrow \underset{ x \in X}{\sup} \underset{ v \in \partial f(x), w \in \partial h(x)}{\inf}||v-w||_2 <\epsilon \end{gather}

In particular, I'm more interested in an answer to my 'additional question' below, but an answer to any of my two questions would be fine.

Motivation: Intuitively, the fact that $||f-h||_{\infty}$ is small, means that the shape of the graphs of the functions are similar and hence also their suporting hyperplanes might be similar. Of course this is just a picture that I have in mind for the 1-dimensional case.

From where the problemm comes: I have a function $F$ that maps convex functions to elements of their subgradient at any given point. I would like to show that if the mapped functions are similar, i.e. $||f-h||_{\infty}$ is sufficiently small, then we can say that $||F(h)-F(f)||_{2}$ is small.

Some references: A similar question was asked here [Hausdorff Distance between Subdifferential sets. The problem is that the theory defines more general types of convergences and analyzes the convergence of the subdifferentials in that framework. I have almost no background in functional analysis so I would really just like to know under which conditions my type of 'convergence'; holds, as defined above.

Regarding my qustion: regarding the conditions on the set $\mathcal{S} \ $: The set I'm working with contains positive concave functions (nothing more is assumed), and $X$ is the open unit simplex. Hence, It would be great if we would have the implication for negative convex functions on the open unit simplex. If the implication does not hold, can we make the conditions on the set $\mathcal{S}$ stronger to make sure that the implication hodls?

additional question: what if we consider two Sets, i.e the set $\mathcal{S}$ of uniformly bounded and continously differentiale functions such that the gradient at any point is uiformly bounded and $\mathcal{M}$ the set of piecewise affine functions? It is well known that for all $f$ in the first set there exists a function in the Second set such that the supremum norm is small (as small as we want). Would we then have the Implication for the subgradients of the two functions?

Answer 1

If $S$ contains uniformly lipschitz functions (As it holds for the sets $S$ and $M$ you mentioned in additional part of your question) Then:

If $h$ monotonically approaches to $h$ then $\partial h$ approaches to $\partial f$ (in terms of the following distance ) $$d(\partial h , \partial f) := \underset{ x \in X}{\sup} \underset{ v \in \partial f(x), w \in \partial h(x)}{\inf}||v-w||_2 $$

In another words, $$h_n \uparrow f \quad \quad \Rightarrow \quad \quad \partial h_n \to \partial f $$

Proof Assume $h_n \uparrow f $. Since $h_n$ are uniformly Lipschitz then there exist a $M > 0$ such that $$\| h_n(x) - h_m(y)\| \leq M \| x -y\|$$ this implies for all $x \in X$, $n \in N$, and $ v \in \partial h_n (x)$ we have $\| v \| \leq M.$

Now pick $x , y \in X $ and fix $x \in X$, assume $\epsilon > 0$ is arbitrary. Since $h_n \uparrow f $, large enough $n$ we have $$ h_n(y) \leq f(y), \quad f(x) -\epsilon \leq h_n (x) $$

Pick $v_n \in \partial h_n(x)$ therefore: $$ \langle v_n , y- x\rangle \leq h_n(y) - h_n(x) \leq f(y) - f(x) + \epsilon . $$

Since $\{ v_n \}$ is bounded sequence, without loss of generality we may assume $v_n \to v .$ So by letting $ n \to \infty$ in above we get

$$ \langle v , y- x\rangle \leq f(y) - f(x) + \epsilon $$

Now because $\epsilon > 0$ was arbitrary, let $\epsilon \to 0$ in above so we arrive $$ \langle v , y- x\rangle \leq f(y) - f(x) $$ which means $ v \in \partial f(x)$

Answer 2

Here is a small example with $X = (-1,1)$, showing that you need some assumptions on $\mathcal S$.

We take $f(x) = |x|$ and $h_\delta(x) = |x - \delta/2|$. Then, $\|f - h_\delta\| = \delta/2 < \delta$, but $$\partial f(\delta/4) = \{1\},\qquad \partial h_\delta(\delta/4) = \{-1\}.$$ Thus, for $\varepsilon < 2$ there is no $\delta > 0$ such that your conclusion holds for the set $$\mathcal S = \{f, h_\delta \mid \delta > 0\}.$$