Suppose that $f: \mathbb{R}^n \to \mathbb{R}$ is convex and globally Lipschitz continuous.
Let $(x^k) \subset \mathbb{R}^n$ be a sequence such that $ x^k \to \bar{x} $ for some $\bar{x} \in \mathbb{R}^n$.
Assume $(\varepsilon_k) \subset \mathbb{R}_{+}$ a sequence of positive scalars such that $\varepsilon_{k} \to 0 \, .$
Does the following hold:
$$ \operatorname{dist} \left( \, \partial_{\varepsilon_k} f(x^k) \, , \, \partial f (\bar{x}) \, \right):= \sup_{g \in \partial_{\varepsilon_k} f(x^k)} \, \, \inf_{h \in \partial f (\bar{x})} \, \| g – h \|_2 \to 0 $$
where
- $\partial_{\varepsilon_k} f(\cdot)$ denotes the $\varepsilon_k$-subdifferential of $f$
- $\partial f(\cdot)$ denotes the regular convex subdifferential of $f$
- $\| \cdot \|$ denotes the Euclidean distance.
The following is all from Convex Analysis and Minimization Algorithms II and might be of use:
Definition 1.1.1 Given $x \in \mathbb{R}^n$ the vector $s \in \mathbb{R}^{n}$ is called an $\varepsilon$ -subgradient of $f$ at $x$ when the following property holds:
$$
f(y) \geqslant f(x)+\langle s, y-x\rangle-\varepsilon \quad \text { for all } y \in \mathbb{R}^{n}
$$
It follows immediately from the definition that
- $\partial_{\varepsilon} f(x) \subset \partial_{\varepsilon^{\prime}} f(x)$ whenever $\varepsilon \leqslant \varepsilon^{\prime}$
- $\partial f(x)=\partial_{0} f(x)=\cap\left\{\partial_{\varepsilon} f(x): \varepsilon>0\right\}\left[=\lim _{\varepsilon \downarrow 0} \partial_{\varepsilon} f(x)\right] $
Theorem 1.1.4 For $\varepsilon \geqslant 0, \partial_{\varepsilon} f(x)$ is a closed convex set, which is nonempty and bounded.
Proposition 4.1.1 Let $\left\{\left(\varepsilon_{k}, x_{k}, s_{k}\right)\right\}$ be a sequence converging to $(\varepsilon, x, s),$ with $s_{k} \in$ $\partial_{\varepsilon_{k}} f\left(x_{k}\right)$ for all $k .$ Then $s \in \partial_{\varepsilon} f(x)$.
Proposition 4.1.2 Let $\delta>0$ and $L$ be such that $f$ is Lipschitzian with constant $L$ on some ball $B(x, \delta),$ where $x \in \mathbb{R}^n$. Then, for all $\delta^{\prime}<\delta$
$$
\|s\| \leqslant L+\frac{\varepsilon}{\delta-\delta^{\prime}}
$$
whenever $s \in \partial_{\varepsilon} f(y),$ with $y \in B\left(x, \delta^{\prime}\right)$.
As a result, the multifunction $\partial_{\varepsilon} f$ is outer semi-continuous, just as is the exact subdifferential.
Theorem 4.1.3 Let $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ be a convex Lipschitzian function on $\mathbb{R}^{n} .$ Then there exists $K>0$ such that, for all $x, x^{\prime}$ in $\mathbb{R}^{n}$ and $\varepsilon, \varepsilon^{\prime}$ positive:
$$
\Delta_{H}\left(\partial_{\varepsilon} f(x), \partial_{\varepsilon^{\prime}} f\left(x^{\prime}\right)\right) \leqslant \frac{K}{\min \left\{\varepsilon, \varepsilon^{\prime}\right\}}\left(\left\|x-x^{\prime}\right\|+\left|\varepsilon-\varepsilon^{\prime}\right|\right)
$$
This result implies the inner semi-continuity of $(x, \varepsilon) \longmapsto \partial_{\varepsilon} f(x)$ for a Lipschitz-continuous $f$.
In particular, for fixed $\varepsilon>0$
$$\partial_{\varepsilon} f(y) \subset \partial_{\varepsilon} f(x)+\|y-x\| B(0, \frac{K}{\varepsilon}) \quad \text{for all } x \text{ and } y \, .$$
Corollary 4.1.5 Let $f: \mathbb{R}^{n} \rightarrow \mathbb{R}$ be convex. For any $\delta \geqslant 0,$ there is $K_{\delta}>0$ such that
$$\Delta_{H}\left(\partial_{\varepsilon} f(x), \partial_{\varepsilon} f\left(x^{\prime}\right)\right) \leqslant \frac{K_{\delta}}{\varepsilon}\left\|x-x^{\prime}\right\| \quad \text{ for all } x \text{ and } x^{\prime} \in B(0, \delta) \, .$$
Theorem 4.2.1 Let be given $f: \mathbb{R}^n \to \mathbb{R}$, $x \in \mathbb{R}^n$ and $\varepsilon \geqslant 0 .$ For any $\eta>0$ and $s \in \partial_{\varepsilon} f(x),$ there exist $x_{\eta} \in B(x, \eta)$ and
$s_{\eta} \in \partial f\left(x_{\eta}\right)$ such that $\left\|s_{\eta}-s\right\| \leqslant \varepsilon / \eta$.
This result can be written in a set formulation:
$$
\partial_{\varepsilon} f(x) \subset \bigcap_{\eta>0} \bigcup_{\|y-x\| \leqslant \eta}\left\{\partial f(y)+B(0, \frac{\varepsilon}{\eta})\right\} .
$$
The following one is not from the above mentioned source but easy to prove:
Theorem Assume $f: \mathbb{R}^n \to \mathbb{R}$ is convex and let $x \in \mathbb{R}^n$. Then for every $\varepsilon > 0$ there is a $\delta > 0$ such that
$$
\bigcup_{y \in B_{\delta}(x)} \partial f(y) \subset \partial_{\varepsilon} f(x) \,.
$$
Best Answer
The Clarke subdifferential is upper semi-continuous. By definition, for each $\epsilon>0$ and $k$ big enough we must have that $\partial f(x^k) \subset \partial f(x) + B(0, \epsilon).$ Hence, for such $k$ big enough,
$$\sup_{g_k \in\partial f(x^k)} \inf_{g \in\partial f(x)} \| g_k - g \| \leq \epsilon. $$ Consequently,
$$\lim \sup_{g_k \in\partial f(x^k)} \inf_{g \in\partial f(x)} \| g_k - g \| = 0$$