Q: Sum of Coordinate Wise Proximal Operators Consider a function $f(x)$ with $x \in \mathbb{R}^d$. The proximal operator is defined as $\text{prox}(x) = \text{argmin}_{y} \{ f(y) + \frac{1}{2}\|x - y\|_2^2 \}$. Now, we can similarly define for each direction $i = 1, 2, \dots, d$ a proximal operator $\text{prox}_i$ such that one is only allowed to change the $i^{\text{th}}$ coordinate. Formally speaking, let $z = \text{argmin}_{y_i} \{ f((x_1, x_2, \dots,x_{i-1}, y_i, x_{i+1}, \dots x_d)) + \frac{1}{2}|x_i - y_i|_2^2 \}$, then $\text{prox}_i(x) = (x_1, x_2, \dots,x_{i-1}, z, x_{i+1}, \dots x_d)$
In the scenario I am considering $f$ is $\sigma$-strongly convex functions and so are all the functions obtained by keeping all but one variables fixed. I am interested in proving that $\|x - \text{prox}_i(x)\|^2 \leq \| x - \text{prox}(x) \|^2 $ for all $i$ and also $\sum_{i = 1}^d \|x - \text{prox}_i(x)\|^2 \geq \| x - \text{prox}(x) \|^2$.
They seem to be intuitively correct to me and also kind of obvious. However, I am unable to formalise the argument. It seems like a simple triangle inequality or Cauchy Schwarz kind of argument, however, I am unable to prove the same.
Any help would be highly appreciated. Thank you!
