Scaling of a Proximal Operator - $\mathrm{Prox}_{f}(x)$ and $\mathrm{Prox}_{af}(x)$

Question

Let $a\in \mathbb{R}$, and $f$ is a convex function $f: \mathbb{R}^n\rightarrow \mathbb{R}$.

$\mathrm{Prox}_{f}(x)=y_1$ and

$\mathrm{Prox}_{af}(x')=y_2$.

Because I know $\mathrm{Prox}_{f}(x)$. And I do not want to calculate $\mathrm{Prox}_{af}(x)$.

So $y_1-x+\partial f(y_1)=0$ and $y_2-x'+ a \partial f(y_2)=0$

Then I have $ay_1-ax+a\partial f(y_1)=0$

So If I want to $\mathrm{Prox}_{af}(x')=y_2$. First I calculate $\mathrm{Prox}_{f}(\frac{x'}{a})=y'$.

Then I have $y'-\frac{x'}{a}+\partial f(y')=0$ By multiplying $a$ on both sides of the above eqution, I have $ay'-{x'}+a\partial f(y')=0$

So I think $ay'=\mathrm{Prox}_{af}(x')=a\mathrm{Prox}_{f}(\frac{x'}{a})$.

Am I right?

Answer 1

No, $y_2-x'+ a \partial f(y_2)=0$ and $ay'-{x'}+a\partial f(y')=0$ does not imply that $y_2=ay'$. It is true when $f$ is a norm, because then $\partial f(y)=\partial f(ay)$ for all $a>0$.

Here's a simple counterexample. Suppose $f$ is the indicator function of some closed convex set. Then the proximal operator is just the projection onto that set. That is, for some closed convex set $C$ $$ f(x) = \begin{cases} 0 & x \in C \\ +\infty & x \notin C \end{cases} $$ Then $\mathrm{Prox}_{af}(x)=\mathrm{Prox}_{f}(x) \in C$. Assume $C$ is not a cone. Then for some $x$ we must have $a \mathrm{Prox}_{f}(x/a) \notin C$, which means $\mathrm{Prox}_{af}(x) \ne a \mathrm{Prox}_{f}(x/a)$ .