Let $C \subset \mathbb R^n$ a nonempty subset. Let us define the indicator function of $C$ $$ I_C(x) = \begin{cases} 0 & x \in C \\ +\infty & x \notin C \end{cases}. $$
Let us consider, in particular, $A=\{(x,y) \in \mathbb R^2: y \ge x^2 \}$ and $B=\{(x,y) \in\mathbb R^2: y \le 0\}$.
Find the subdifferential of $I_A, I_B, I_A+I_B$ at the origin.
I show you what I have done so far; I have calculated the subdifferentials, they shold be
$$
\partial I_A(0) = \{s \in \mathbb R^2: \langle s,x \rangle \le 0, \forall x \in A\}
$$
$$
\partial I_B(0) = \{s \in \mathbb R^2: \langle s,x \rangle \le 0, \forall x \in B\}
$$
What about the third one? I know that
$$
\partial I_{A+B}(0) = \{s \in \mathbb R^2: \langle s,x \rangle \le 0, \forall x \in A+B\}
$$
but I'm stuck on subdifferentiating $I_A+I_B$. Indeed, what is $I_A+I_B$? Is it the function
$$
x \mapsto \begin{cases} 0 & x = 0 \\ +\infty & \text{elsewhere} \end{cases} = I_{\{0\}}?
$$
So who is its subdifferential?
Moreover is it possible to have a more explicit description of these sets? In particular, I would like to establish if $\partial (I_A+ I_B)(0) = \partial I_{A+B}(0) = \partial I_A(0)+\partial I_B(0)$.
What do you think? Thanks.
If $f:\mathbb{R}^n \to \mathbb{R} \cup \{+\infty \}$ is a proper convex function, then $g$ is a subgradient to $f$ at $x_0$ iff $f(x) \ge f(x_0) + \langle g, x-x_0 \rangle$ for all $x$. The subdifferential (or generalized gradient) of $f$ at $x_0$ is the set of all subgradients to $f$ at $x_0$.
You can see that the functional $\phi(x) = f(x_0) + \langle g, x-x_0 \rangle$ is equivalent to a hyperplane tangent to the epigraph of $f$ at $\binom{f(x_0)}{x}$, that is $\{(\alpha,x) \in \mathbb{R}^{n+1} | \langle \binom{-1}{g}, \binom{\alpha-f(x_0)}{x-x_0} \rangle = 0 \}$. So, one way of viewing a subgradient is as the lower component of a (non-vertical) hyperplane that passes through $\binom{f(x_0)}{x_0}$ and 'contains' the epigraph of $f$, that is, $\operatorname{epi} f \subset \{(\alpha,x) \in \mathbb{R}^{n+1} | \langle \binom{-1}{g}, \binom{\alpha-f(x_0)}{x-x_0} \rangle \le 0 \}$. The subdifferential is the set of all such subgradients. I find this perspective gives more intuition.
For $I_A$, we see that the indicator is zero on and above the parabola $A$, and infinity everywhere else. Hence the tangent hyperplanes passing through $(0,(0,0))$ are of the form $(-1, (0,g_2))$, where $g_2 \le 0$. That is, if we let $G_A = \{ (0, g_2) \}_{g_2 \le 0}$, then I claim $\partial I_A((0,0)) = G_A$.
Since $I_A$ is an indicator function, we have $\partial I_A((0,0)) = \{ (g_1,g_2) | x_1 g_1 + x_2 g_2 \le 0, \ \forall x_1,x_2 \text{ such that } x_2 \ge x_1^2 \}$. If $(x_1,x_2) \in A$, we have $x_2\ge 0$, hence we see that $G_A \subset \partial I_A((0,0))$. Now suppose $(g_1,g_2) \in \partial I_A((0,0))$. Then we have $t g_1+t^2 g_2 \le 0 $ for all $t$. It follows immediately that $g_1 = 0$ and $g_2 \le 0$, hence $(g_1,g_2) \in G_A$, and so the two sets are equal.
The epigraph of $I_B$ is simpler, it is zero whenever $x_2 \le 0$, and infinity otherwise. Hence the tangent hyperplanes passing through $(0,(0,0))$ are of the form $(-1, (0,g_2))$, where $g_2 \ge 0$. As above, if we let $G_B = \{ (0, g_2) \}_{g_2 \ge 0}$, then I claim $\partial I_B((0,0)) = G_B$.
We have $\partial I_B((0,0)) = \{ (g_1,g_2) | x_1 g_1 + x_2 g_2 \le 0, \ \forall x_1,x_2 \text{ such that } x_2 \le 0 \}$. It should be immediate that $G_B \subset \partial I_B((0,0))$. Now suppose $(g_1,g_2) \in \partial I_B((0,0))$. Then since $x_1 g_1+x_2 g_2 \le 0 $ for all $x_1$, $x_2 \le 0$, we see that $g_1 = 0$, and setting $x_2=-1$ shows that $g_2 \ge 0$, hence $(g_1,g_2) \in G_B$, and again the two sets are equal.
We see that $\partial I_A((0,0))+\partial I_B((0,0)) = \{ (0, y) \}_{y \in \mathbb{R}}$.
Finally, we see that $I_A(x)+I_B(x)$ is finite iff both $I_A(x),I_B(x)$ are finite iff $x \in A\cap B$. In other words, $I_A+I_B = I_{A\cap B}$.
In this case, $A\cap B = \{ (0,0) \}$. It should be clear either from the above considerations or direct calculations that $\partial (I_A+I_B)((0,0)) = \partial I_{A\cap B}((0,0)) = \mathbb{R}^2$.
In particular, it should be clear that $\partial I_A(x) + \partial I_B(x) \subsetneq \partial (I_A+I_B)(x)$, that is, they are not equal.
Comment: Regarding the last formula above, note that $A+B = \mathbb{R}^2$ and $\partial I_{A+B}(x) = \{0 \}$ for all $x$. Hence this can equal neither of $\partial I_A(x) + \partial I_B(x) $ or $ \partial (I_A+I_B)(x)$.