1
$\begingroup$

If $f$ and $g$ are respectively a differentiable function and a convex, lower semi-continuous function, then the algorithm defined by:

$$ x^{k+1} = \text{prox}_{\gamma{g}}[x^{k} - \gamma\nabla{f(x^{k}})]$$

converges to $\text{argmin}[f+g]$.

This is justified by the fact that if $x^{*}$ is a minimizer of $f+g$, then: $$ x^{*} = \text{prox}_{\gamma{g}}[x^{*} - \gamma\nabla{f(x^{*}})]$$

But I do not understand this relation. Why is it true?

That is, why $x^{*} = \text{argmin}[f+g] \Leftrightarrow x^{*} =\text{prox}_{\gamma{g}}[x^{*} - \gamma \nabla{f(x^{*}})] $ ?

$\endgroup$
1
$\begingroup$

I will show below that if $x^* = \text{prox}_{\gamma{g}}[x^{*} - \gamma \nabla{f(x^{*}})]$ then $x^* \in \text{argmin } f(x)+g(x) $.

Plugging in the definition of proximal operator, we have $$\text{prox}_{\gamma{g}}[x^{*} - \gamma \nabla{f(x^{*}})] = \text{argmin}_{x} \left\{\gamma g(x) + \frac{1}{2}\|x- (x^* - \gamma \nabla f(x^*))\|^2\right\}$$ Now since $x^* = \text{prox}_{\gamma{g}}[x^{*} - \gamma \nabla{f(x^{*}})]$, we have by Fermat's rule at $x=x^*$, the following $$0 \in \partial (\gamma g(x)) + (x-(x^*-\gamma\nabla f(x^*)))$$ Now just substitute $x=x^*$, we get $$0\in \gamma \partial g(x^*) + \gamma \nabla f(x^*)$$ This is equivalent to saying that $0 \in \partial F(x^*)$, where $F(x) = g(x)+f(x)$, so $x^*$ is the minimizer. The other direction of the proof is very similar.

Note: The last step $0 \in \partial F(x^*)$ need not always hold (check subdifferential properties).

$\endgroup$
1
$\begingroup$

Here's an explanation which assumes that we already understand the idea that the prox-operator of $g$ with parameter $t > 0$ is the operator $(I + t \partial g)^{-1}$, where $\partial g$ is the subdifferential of $g$.

I'll assume that $f$ is convex as well as differentiable, and that $g$ is convex and closed. Let $t >0$. A point $x$ is a minimizer of $f + g$ if and only if \begin{align} &0 \in \nabla f(x) + \partial g(x) \\ \iff &x \in x + t \nabla f(x) + t\partial g(x) \\ \iff &x - t \nabla f(x) \in (I + t \partial g)(x) \\ \iff &x = (I + t \partial g)^{-1}(x - t \nabla f(x)). \end{align} The final equation is another way of saying that $$ x = \text{prox}_{tg}(x - t \nabla f(x)). $$

We can then solve this equation using fixed point iteration, which yields the proximal gradient method.

By the way, if this derivation of the proximal gradient method doesn't seem intuitive, there are other ways to discover the proximal gradient method that are more obvious. The viewpoint given here has the advantage that it shows that the proximal gradient method is a fixed point iteration, which helps us with convergence proofs.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.