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Let $f: X \to [ -\infty, \infty]$ be some function,

Can someone provide a non-trivial example where the subdifferential evaluated at a point $x$, $$\partial f(x)$$ is "unbounded"? (trivial examples included the improper functions)

A rough definition of an unbounded subgradient is that there exists some sequence $v_n \in \partial f(x)$ such that $\|v_n\| \to \infty$.

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  • $\begingroup$ Observation: If a function has unbounded subdifferential, suppose without loss it occurs at $0$ and $f(0) = 0$ (you can always translate so that this is the case); then $f(v_n) \geq \|v_n\|_2^2$, for all $n$. Hence $\liminf f(v_n) \geq +\infty$. $\endgroup$
    – Drew Brady
    Commented Oct 30, 2019 at 4:31
  • $\begingroup$ This should happen at zero for $\sqrt{|x|}$. $\endgroup$
    – pseudocydonia
    Commented Oct 30, 2019 at 7:39

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A simple example is the convex function $$ f(x) = \begin{cases} \infty, & \text{if}\ x<0\\ x & \text{if}\ x\geq 0. \end{cases} $$ It holds that $\partial f(0) = ]-\infty,1]$.

There are no examples without the value $\infty$: If $f$ is convex and bounded in a neighborhood at some point, then $f$ is locally Lipschitz at that point and hence, the subgradient at that point is bounded.

If you do not assume convexity, you can get unbounded subgradients even for bounded functions as a comment to the question shows.

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  • $\begingroup$ Thank you. If I visualize the graph it really makes a lot of sense. $\endgroup$
    – Sin Nombre
    Commented Oct 30, 2019 at 15:00

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