Question

4. Prove the following proposition: Proposition 13.8 (Separable Sum) Let $f_i$ be convex and proper on $X_i$, for $i \in \{1,...,m\}$, and define $f(x_1,...,x_m) := f_1(x_1) + ... + f_m(x_m)$. Then the subdifferential of $f$ at $(x_1,...,x_m)$ is given by the Cartesian product of the subdif- ferentials of each $f_i$ at $x_i$, i.e., $\partial f(x_1,...,x_m) = \partial f_1(x_1) \times ... \times \partial f_m(x_m)$. Compute the subdifferential of (a) for $\tau > 0$, $f(x_1,...,x_m) = \sum_{i=1}^m ||x_i||$ where the norm $||.||$ is an Euclidean norm on $X_i$, for $i \in \{1,...,m\}$. (b) $f(x) = ||x||_1$ on $\mathbb{R}^n$.

          4. Prove the following proposition:
Proposition 13.8 (Separable Sum) Let $f_i$ be convex and proper on $X_i$, for $i \in \{1,...,m\}$,
and define
$f(x_1,...,x_m) := f_1(x_1) + ... + f_m(x_m)$.
Then the subdifferential of $f$ at $(x_1,...,x_m)$ is given by the Cartesian product of the subdif-
ferentials of each $f_i$ at $x_i$, i.e.,
$\partial f(x_1,...,x_m) = \partial f_1(x_1) \times ... \times \partial f_m(x_m)$.
Compute the subdifferential of
(a) for $\tau > 0$, $f(x_1,...,x_m) = \sum_{i=1}^m ||x_i||$ where the norm $||.||$ is an Euclidean norm on $X_i$,
for $i \in \{1,...,m\}$.
(b) $f(x) = ||x||_1$ on $\mathbb{R}^n$.

$4. Prove the following proposition: Proposition 13.8 (Separable Sum) Let fi be convex and proper on Xi, for i ∈{1,...,m}, and define f(x1,...,xm) := f1(x1) + ... + fm(xm). Then the subdifferential of f at (x1,...,xm) is given by the Cartesian product of the subdif- ferentials of each fi at xi, i.e., ∂ f(x1,...,xm) = ∂ f1(x1) × ... ×∂ fm(xm). Compute the subdifferential of (a) for τ > 0, f(x1,...,xm) = ∑i=1^m ||xi|| where the norm ||.|| is an Euclidean norm on Xi, for i ∈{1,...,m}. (b) f(x) = ||x||1 on ℝ^n.$

Added by Elena P.

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