For a proper lower-semicontinuous <convex function> $f$, its <proximal operator> is
$$
\operatorname{prox}_f(y)=
\arg\min_x\left\{f(x)+\frac12\lVert x-y\rVert_2^2\right\}.
$$
The squared norm is strongly convex, so the minimizer is unique. The <subdifferential sum rule> gives the necessary and sufficient condition
$$
x=\operatorname{prox}_f(y)
\quad\Longleftrightarrow\quad
0\in\partial f(x)+x-y
\quad\Longleftrightarrow\quad
y-x\in\partial f(x).
$$
More generally,
$$
x=\operatorname{prox}_{tf}(y)
\quad\Longleftrightarrow\quad
u:=\frac{y-x}{t}\in\partial f(x).
$$
The subgradient inversion rule for the <convex conjugate> says $u\in\partial f(x)$ exactly when $x\in\partial f^*(u)$. Hence
$$
\frac yt-u=\frac xt\in\frac1t\partial f^*(u)
=\partial(t^{-1}f^*)(u),
$$
which is precisely the proximal optimality condition
$$
u=\operatorname{prox}_{t^{-1}f^*}(y/t).
$$
Since $x=y-tu$, this proves the generalized <Moreau decomposition>
$$
\boxed{\operatorname{prox}_{tf}(y)=
y-t\operatorname{prox}_{t^{-1}f^*}(y/t)}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 339 / 2 / b / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-339/2/b/solution