(1) Conside the convex set $C:=\left{(x, y) \in \mathbb{R}^2 \mid x^2+y^2 \leq 1\right.$ and $\left.x, y \geq 0\right}$, the unit disc intersect the first quadrant, and let $\delta_C(x, y)$ be its indicator function. Find $\partial \delta_C\left(x_0, y_0\right)$ where $\left(x_0, y_0\right)$ is a boundary point of $C$. Hint: Draw a picture. There should be 6 cases to consider. Note that $C=C_1 \cap C_2 \cap C_3$ where $C_1=\left{(x, y) \in \mathbb{R}^2 \mid x^2+y^2 \leq 1\right}$, $C_2=\left{(x, y) \in \mathbb{R}^2 \mid-x \leq 0\right}$, and $C_3=\left{(x, y) \in \mathbb{R}^2 \mid-y \leq 0\right}$.
To find $\partial \delta_C(x_0, y_0)$, we need to find the subdifferential of the indicator function $\delta_C$ at the point $(x_0, y_0)$. Since $C$ is a convex set, the subdifferential of $\delta_C$ at $(x_0, y_0)$ is the set of all vectors $v \in \mathbb{R}^2$ such that $$\delta_C(x,y) \geq \delta_C(x_0,y_0) + \langle v, (x-x_0, y-y_0)\rangle$$ for all $(x,y)\in C$.
Since $C$ is defined as the intersection of three sets $C_1$, $C_2$, and $C_3$, we can find the subdifferential of $\delta_C$ at $(x_0, y_0)$ by finding the subdifferentials of $\delta_{C_1}$, $\delta_{C_2}$, and $\delta_{C_3}$ at $(x_0, y_0)$ and taking their intersection.
We start by finding the subdifferential of $\delta_{C_1}$ at $(x_0, y_0)$. Since $C_1$ is a closed convex set, its subdifferential at any point is the set of all vectors pointing inwards from the boundary of $C_1$ at that point. Thus, if $(x_0, y_0)$ is a point on the boundary of $C_1$, the subdifferential of $\delta_{C_1}$ at $(x_0, y_0)$ is the set of all vectors pointing inwards from the boundary of the unit disc at $(x_0, y_0)$.
Next, we find the subdifferential of $\delta_{C_2}$ at $(x_0, y_0)$. Since $C_2$ is defined by the inequality $-x \leq 0$, its indicator function $\delta_{C_2}(x,y)$ is equal to $0$ if $x \geq 0$ and $-\infty$ if $x < 0$. Therefore, if $(x_0, y_0)$ is a point on the boundary of $C_2$, its subdifferential is the set of all vectors $(v_1,v_2)$ satisfying $$v_1 \geq 0, v_1(x-x_0) + v_2(y-y_0) \leq 0 \text{ for } x < x_0.$$
Finally, we find the subdifferential of $\delta_{C_3}$ at $(x_0, y_0)$. Since $C_3$ is defined by the inequality $-y \leq 0$, its indicator function $\delta_{C_3}(x,y)$ is equal to $0$ if $y \geq 0$ and $-\infty$ if $y < 0$. Therefore, if $(x_0, y_0)$ is a point on the boundary of $C_3$, its subdifferential is the set of all vectors $(v_1,v_2)$ satisfying $$v_2 \geq 0, v_1(x-x_0) + v_2(y-y_0) \leq 0 \text{ for } y < y_0.$$
(2) Let $F(x, y)=f(x)+g(y)$ where $f, g: \mathbb{R}^1 \rightarrow \mathbb{R}$.
(a) Prove that $\partial F\left(x_0, y_0\right)=\partial f\left(x_0\right) \times \partial g\left(y_0\right)$. Recall that the product of two sets $A \times B={(a, b) \mid a \in A, b \in B}$.
(b) Use the formula in part (a) to compute $\partial F\left(x_0, y_0\right)$ for the function $F(x, y)=|x|+y^2$.
(a) To prove that $\partial F(x_0,y_0)=\partial f(x_0)\times \partial g(y_0)$, we need to show that for any $(p,q) \in \partial f(x_0) \times \partial g(y_0)$, we have $(p,q) \in \partial F(x_0,y_0)$, and conversely, for any $(u,v) \in \partial F(x_0,y_0)$, we have $u \in \partial f(x_0)$ and $v \in \partial g(y_0)$.
First, let $(p,q) \in \partial f(x_0) \times \partial g(y_0)$. This means that there exist $x_1 \in \mathbb{R}$ and $y_1 \in \mathbb{R}$ such that $$f(x) \geq f(x_0) + p(x-x_0) \quad \text{for all } x \in \mathbb{R}$$ and $$g(y) \geq g(y_0) + q(y-y_0) \quad \text{for all } y \in \mathbb{R}.$$ Adding these two inequalities, we get $$F(x,y) = f(x) + g(y) \geq f(x_0) + p(x-x_0) + g(y_0) + q(y-y_0) = F(x_0,y_0) + p(x-x_0) + q(y-y_0)$$ for all $(x,y) \in \mathbb{R}^2$. This shows that $(p,q) \in \partial F(x_0,y_0)$.
Conversely, let $(u,v) \in \partial F(x_0,y_0)$. This means that there exists $(x_1,y_1) \in \mathbb{R}^2$ such that $$F(x,y) \geq F(x_0,y_0) + u(x-x_0) + v(y-y_0) \quad \text{for all } (x,y) \in \mathbb{R}^2.$$ Setting $y = y_0$ in this inequality and using the fact that $F(x,y) = f(x) + g(y)$, we get $$f(x) + g(y_0) \geq f(x_0) + u(x-x_0) + g(y_0) + v(y-y_0) \quad \text{for all } x \in \mathbb{R}.$$ Rearranging, we get $$f(x) \geq f(x_0) + u(x-x_0) \quad \text{for all } x \in \mathbb{R}.$$ This shows that $u \in \partial f(x_0)$. Similarly, setting $x = x_0$ in the inequality $F(x,y) \geq F(x_0,y_0) + u(x-x_0) + v(y-y_0)$ and using the fact that $F(x,y) = f(x) + g(y)$, we get $v \in \partial g(y_0)$. Therefore, we have shown that $(u,v) \in \partial f(x_0) \times \partial g(y_0)$.
Combining the two parts, we conclude that $\partial F(x_0,y_0) = \partial f(x_0) \times \partial
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