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Could anyone please help me to prove that a proximal operator of squared Euclidean distance to a set $\mathcal{C}$ is

$$\operatorname{prox}_{t\operatorname{dist}^2}\left(x\right) = \frac{1}{(t+1)} \left( x + t \operatorname{proj}_{\mathcal{C}}\left( x \right) \right)$$ cf. slide#6.21 in [1]?

The squared distance function is (with $1/2$ scaling):

\begin{align} \operatorname{dist}^2\left(x\right) &= \frac{1}{2}\inf_{y \in \mathcal{C}} \| x - y \|^2 \equiv \frac{1}{2}\|x - \operatorname{proj}_{\mathcal{C}}\left( x \right) \|_2^2. \end{align}

Attempt:

Definition: The proximal mapping of closed convex function $f$ is $$\operatorname{prox}_f\left(x\right) = \arg\min_{z} f(z) + \frac{1}{2t} \| z - x\|_2^2 $$

Let $f(z) := \operatorname{dist}^2\left(z\right) = \frac{1}{2}\|z - \operatorname{proj}_{\mathcal{C}}\left( z \right) \|_2^2$.

To this end, we can plugin the definition of the considered function to the proximal mapping such that \begin{align} &0 = \nabla f(z) + \frac{1}{t}\left(z - x\right) \end{align}

Now, I am unsure about the gradient of squared distance function unless I assume that $\operatorname{proj}_{\mathcal{C}}\left( x \right) \leftarrow\operatorname{proj}_{\mathcal{C}}\left( z \right) $, then $\operatorname{proj}_{\mathcal{C}}\left( x\right)$ will be seen as a constant and straightfoward to compute the gradient of $f(z)$, i.e., the gradient is $\nabla f(z) = z - \operatorname{proj}_{\mathcal{C}}\left( x \right)$, otherwise I don't think it is possible to find the gradient in general, correct? Then, we can find the proximal operator $$z = \frac{1}{(t+1)}\left(x + t\operatorname{proj}_{\mathcal{C}}\left( x \right) \right)$$

I think the problem in my understanding is the gradient of such a squared distance to a set. I will raised another question instead for the clarity.

user550103
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  • The gradient of the squared distance function is in fact given by the identity minus projection ($\nabla f(z) = z - \operatorname{proj}C(z)$. This can be looked up. In order to have the same solution you should note that the definition of the proximal operator is usually written with a parameter, i.e. $\operatorname{prox}{t f}(x) = \arg\min_u{f(u) + \frac{1}{2 t}|| u-x||^2}$. – xel Mar 23 '21 at 11:34
  • However, if I remember correctly there should be a way to solve this problem by pure algebra (manipulating the squares inside the argmin and noting that the proximal operator of the 2 norm squared is given by $\operatorname{prox}_{t ||\cdot||^2}(x) = \frac{1}{1+t}x$, which looks very similar to the desired form of your prox. – xel Mar 23 '21 at 11:36
  • @xel Thank you. Yes, I missed that $t$. So, I can match my derivations, but I am not sure whether my understanding of the gradient of a distance function is correct. I am not aware of any good reference where they have proved the gradient of a distance function. Are you ware of it? Thank you in advance – user550103 Mar 23 '21 at 15:00
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    for example: Bauschke, Heinz H., and Patrick L. Combettes. Convex analysis and monotone operator theory in Hilbert spaces. Vol. 408. New York: Springer, 2011. Corollary 12.30. But as I mentioned, there is most likely also an elementary way to show it (which I assume is the way to go if this some kind of homework/exercise. – xel Mar 23 '21 at 16:45
  • @xel Thank you for the pointer. I really appreciate it. – user550103 Mar 23 '21 at 20:03

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