Proximal operator

Remarks

The minimizer exists and is unique, since the function being minimized is lower semicontinuous, strongly convex and tends to $+\infty$ as $\norm{x} \to \infty$. Equivalently, $p = \operatorname{prox}_f(v)$ exactly when $v - p$ lies in the subdifferential $\partial f(p)$. Example: let $C \subseteq H$ be nonempty, closed and convex, and let $f$ be $0$ on $C$ and $+\infty$ outside. Then $f(x) + \tfrac12 \norm{x - v}^2$ is finite only on $C$, so $\operatorname{prox}_f(v)$ is the point of $C$ nearest to $v$, the projection of $v$ onto $C$. Example: for $f(x) = \lambda \abs{x}$ on $\R$ with $\lambda > 0$, one finds $\operatorname{prox}_f(v) = \sgn(v) \max(\abs{v} - \lambda, 0)$, the soft thresholding map. Non-example: let $f$ be $0$ on $\set{-1, 1}$ and $+\infty$ elsewhere on $\R$, so $f$ is not convex. At $v = 0$ both $-1$ and $1$ minimize $f(x) + \tfrac12 x^2$, so no single proximal point is defined.