For an <absolutely one-homogeneous functional> $J$, the subdifferential has the characterization
$$
p\in\partial J(u)
\quad\Longleftrightarrow\quad
\langle p,u\rangle=J(u)
\ 
\text{and}
\ 
\langle p,v\rangle\leq J(v)\quad\text{for every }v.
$$
If $c>0$, then $J(cu)=cJ(u)$, so
$$
\langle p,cu\rangle=c\langle p,u\rangle=cJ(u)=J(cu).
$$
The second condition does not involve the base point. Therefore
$$
\boxed{\partial J(cu)=\partial J(u)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 326 / 2 / 2 / 3 / a / Solution

A functional $J$ is absolutely one-homogeneous when $J(cu)=|c|J(u)$ for every scalar $c$. For a convex such functional,
$$
p\in\partial J(u)
\quad\Longleftrightarrow\quad
\langle p,u\rangle=J(u)
\text{ and }
\langle p,v\rangle\leq J(v)\text{ for every }v.
$$


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