Let $X \subset \mathbb R^d$ be closed and convex and $f: X \to \mathbb R$ be convex and Lipschitz. That means there is $L > 0$ such that $|f(x)-f(y)| \le L \|x-y\|$ for all $x,y \in X$. Here $\|\cdot\|$ is the Euclidean norm. We define the subgradient sets:
$$\partial_f(x) = \{v \in \mathbb R^d: f(y) \ge f(x)+ v \cdot(y-x) \ \forall y \in X\}.$$
It seems common practice (For example Lemma 2.6 of this survey ) to claim each subgradient of $f$ has norm at most $L$. Unfortunately this is not true $-$ the proof in the survey only works under the assumption that $x$ is an interior point of $X$.
If we allow boundary points here is a counterexample: Let $X=[0,1]$ and $f(x) = x$ and $L=1$. Then at each interior point $\partial_f(x) = \{1\}$ which is clearly bounded by $L$. However at the boundary we have $\partial_f(0) = (-\infty,1]$ since for any $y \in [0,1]$ and $v \le 1$ we have $y \ge vy$ and so $$f(y) = y \ge 0 + vy = 0+v(y-x) =f(x)+v(y-x).$$ Likewise $\partial_f(0) = [1,\infty)$.
However I notice there still exists a small subgradient at each endpoint, even if there are many more large ones. Namely $1$ is still a subgradient at the endpoints. I wonder is this fact true in general?
Conjecture: Let $X \subset \mathbb R^d$ be closed and convex and $f: X \to \mathbb R$ be convex and Lipschitz with constant $L$. For each $x \in X$ there is $v \in \partial_f(x)$ with $\|v\| \le L$.
Yes. We may assume that $X$ has nonempty interior (otherwise it's contained in an affine subspace, and we can extend the function by making it constant on the orthogonal complement...). Now any $x \in X$ is the limit of a sequence $x_n$ in the interior of $X$, and at each $x_n$ there is a subgradient $v_n$ with norm $\le L$. The $v_n$ have a limit point, which is a subgradient at $x$ and has norm $\le L$.