Strict inclusion in subdifferential sum rule $\partial{f(x)}+\partial{g(x)}\subseteq{\partial(f+g)(x)}$.

Question

I wish to find an example to show that the inclusion in the subdifferential sum rule

$$\partial{f(x)}+\partial{g(x)}\subseteq{\partial(f+g)(x)}$$

is strict. However, I have a problem understanding the way this inclusion is stated.

This inclusion in the subdifferential sum rule is taken from the book Convex Analysis by R. Tyrrell Rockafellar, Page 223, Theorem 23.8. This theorem says:

Let $f_1,f_2,\dots,f_m$ be proper convex functions on $\mathbb{R}^n$, and $f=f_1+f_2+\cdots+f_m$, then $\partial{f_1(x)}+\cdots+\partial{f_2(x)}\subset{\partial{f(x)}}~~~\forall{x}$.

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It also says if the domains of the $f_i$s have common point, then the inclusion turns into equality. From what I understand, if the domains of the $f_i$s do not have common point, then the strict inclusion should hold. One of the many examples that I tried was the pair $\sqrt{1-x}$ and $\sqrt{x-2}$. These two functions do not have any point in common in their domains. However, I do not understand how we can add two functions whose domains do not overlap.

I have two main issues:

1. Intuitively, the subset sign should be the other way around. In the last part of the answer to this question, there is an example where the inclusion is true for the case where the subset sign is the other way around. In addition, the example is in contradiction to the inclusion statement in the question.

2. This might be an easy question, but how can we add two functions whose domains do not intersect?

Any insight and example that can help me understand this are really appreciated!

Answer 1

Re 2: If domains don't overlap, then the sum function is $+\infty$ everywhere.

Re 1: Consider two disks in $\mathbb{R}^2$ that touch in exactly one point. To be concrete, $$A = \text{ball of radius $1$ centred at $(-1,0)$ and $B=$ ball of radius $1$ centred at $(1,0)$}.$$ Note that $A\cap B = \{(0,0)\}$. Set $z := (0,0)\in\mathbb{R}^2$. Now define $$ f(x) = \begin{cases} 0, &\text{if $x\in A$;}\\ +\infty, &\text{else} \end{cases} \quad\text{and}\quad g(x)= \begin{cases} 0, &\text{if $x\in B$;}\\ +\infty, &\text{else.} \end{cases}$$ Note that $\partial f(z)=\mathbb{R}_+(1,0) = \{(\rho,0)\,|\,\rho\geq 0\}$ and $\partial g(z)=\mathbb{R}_+(-1,0) = \{(\rho,0)\,|\,\rho \leq 0\}$; thus, $$\partial f(z)+g(z) = \mathbb{R}_+(1,0) + \mathbb{R}_-(1,0) = \mathbb{R}\times\{0\}.$$ Moreover, $$ (f+g)(x) = f(x)+g(x) = \begin{cases} 0, &\text{if $x\in A\cap B=\{(0,0)\}$;}\\ +\infty, &\text{else.} \end{cases} $$ Hence $$\partial(f+g)(z) = \mathbb{R}^2$$ which is a strict superset of $\mathbb{R}\times\{0\}$.

Comments.
(1) The sum rule is true when the underlying space is just $\mathbb{R}$ as long as the domains of the subdifferential operators intersect. It fails for merely having the domains intersect, see example by @gerw below.
(2) The sum rule can fail as shown above even when the domains do intersect nontrivially.
(3) Finally, the sum rule holds when the domains overlap sufficiently (see Rockafellar's book for details involving say the relative interior or polyhedral functions).

Answer 2

Here is an example in $\mathbb R^1$: Define $$ f(x) = \begin{cases} -\sqrt{x} & \text{for } x \ge 0, \\ +\infty & \text{for } x < 0, \end{cases} $$ and $$ g(x) = \begin{cases} 0 & \text{for } x \le 0, \\ +\infty & \text{for } x > 0. \end{cases} $$ Thus, $f+g$ is the indicator function of $\{0\}$. Then, \begin{align*} \partial f(0) &= \emptyset, \\ \partial g(0) &= [0,\infty), \\ \partial f(0) + \partial g(0) &= \emptyset, \\ \partial (f+g)(0) &= \mathbb R. \end{align*}