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Let $f:\mathbb R\to\mathbb R$ be a lower semicontinuous function, we define the Fréchet subdifferential of $f$ at $x\in\mathbb R$ by $$\partial^F f(x):=\left\{L\in\mathbb R: \liminf_{v\to0}\frac{f(x+v)-f(x)-Lv}{|v|}\geq0\right\}.$$ According to this paper, in the third page, authors claimed that this subdifferential satisfies a additive property, namely $$\partial^F(f+g)(x)\subset \partial^Ff(x)+\partial^Fg(x)\tag{$*$}$$ for any convex function $g:\mathbb R\to\mathbb R$ (here I choose the Banach space $\mathbb R$, then $g$ is automatically continuous and $\partial^F g(x)\neq\emptyset$ for all $x\in\mathbb R$.)

As usual, the summation in $(*)$ should be the Minkowski sum, hence in particular we have $\emptyset+S=\emptyset$ for all $S\subset\mathbb R$.

I'm trying to prove $(*)$, but here is a counterexample: Let $f(x)=-|x|, g(x)=|x|$ for all $x\in\mathbb R$, then $g$ is a convex function and $\partial^Fg(0)=[-1,1]$ and $\partial^F(f+g)(0)=\{0\}$; however, $\partial^Ff(0)=\emptyset$: if $L\in\partial^Ff(0)$, then $\liminf_{v\to0}\frac{-|v|-Lv}{|v|}\geq0$, i.e., $-1-|L|\geq0$, which is a contradiction. Now it seems that I've got a counterexample of $(*)$, but $(*)$ is used everywhere in literature, so what am I missing now?

In the same paper, at the end of the proof of Theorem 3.1, authors used an argument like this: if $f+h$ takes minimum at $\bar x$ with $h$ convex and $\partial^F h(\bar x)\neq\emptyset$, then $$0\in \partial^F(f+h)(\bar x)\subset \partial^Ff(\bar x)+\partial^Fh(\bar x);$$ combining this with $\partial^Fh(\bar x)\subset (-\delta,\delta)$ gives that $\partial^Ff(\bar x)\cap(-\delta,\delta)\neq\emptyset$. But my example above still indicates the failure of this argument.

Theorem 3.1 impiles the fact that $\operatorname{dom}(\partial^Ff)$ is dense in $\mathbb R$ if $f:\mathbb R\to\mathbb R$ is a lower semicontinuous function, which is a well-known result. So I must miss something here.

Thanks for any help!

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In part (P3) of Definition 2.1 in the paper you linked, it is also required that $g$ be $\partial$-differentiable at $x$ (meaning that both $\partial g(x)$ and $\partial(-g)(x)$ are nonempty), which is not the case in your example (with $\partial=\partial^F$ and $g(\cdot)=|\cdot|$).

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  • $\begingroup$ Thanks for your answer! I understand now. I didn't realize the stronger condition of $\partial$-differentiable, which is also the reason why the authors require $p=2>1$ in the proof of Theorem 3.1, since $|x|^2$ is smoother than $|x|$. However, I've seen in another paper, in the proof of Theorem 7, Rockafellar invoked the ordinary Ekeland's variational principle, in which case $h(x)=A|x|(A>0)$ and he still used the property $(*)$. Is there any other particular reason in Rockafellar's proof? $\endgroup$ Nov 26, 2023 at 2:33
  • $\begingroup$ @Fergns : I suggest you post this additional question separately. $\endgroup$ Nov 26, 2023 at 3:17

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