$\newcommand{\prox}{\operatorname{prox}}$
Probably the most remarkable property of the proximal operator is the **fixed point** property:

*The point $x^*$ minimizes $f$ if and only if $x^* = \prox_f(x^*) $*

So, indeed, $f$ can be minimized by find a fixed point of its proximal operator. See Proximal Algorithms by Neal Parikh and Stephen Boyd.

**Question 1**. In the paper given above, author is saying:

*If $\prox_f$ were a contraction, i.e., Lipschitz continuous with
constant less than $1$, repeatedly applying $\prox_f$ would find a (here,
unique) fixed point*

Why the bound on the first-order derivative guarantees finding a fixed point by repeatedly applying proximal operator?

**Question 2**. Let me quote a paragraph from the same paper:

*It turns out that while $\prox_f$ need not be a contraction (unless $f$ is strongly convex), it does have a different property, firm nonexpansiveness, sufficient for fixed point iteration:*

*$\|\prox_f(x) - \prox_f(y)\|^2_2 \le (x-y)^T (\prox_f(x) - \prox_f(y))$*

$\forall x,y \in \mathbb{R}^n$

*Firmly nonexpansive operators are special cases of nonexpansive operators (those that are Lipschitz continuous with constant 1). Iteration of a general nonexpansive operator need not converge to a fixed point: consider operators like $-I$ or rotations. However, it tunrs out that if $N$ is nonexpansive, then the operator $T = (1-\alpha)I + \alpha N$, where $\alpha \in (0,1)$ has the same fixed points as $N$ and simple iteration of $T$ will converge to a fixed point of $T$ (and thus of $N$), i.e. the sequence:*

$x^{k+1} := (1-\alpha)x^k +\alpha N(x^k)$

*will converge to a fixed point of $N$. Put differently, damped iteration of
a nonexpansive operator will converge to one of its fixed points.*

*Operators in the form $(1-\alpha)I + \alpha N$, where $N$ is non-expansive and $\alpha \in (0,1)$, are called $\alpha$-averaged operators.*

*Firmly nonexpansive operators are averaged: indeed, they are precisely the (1/2)-averaged
operators.*

Why "unless $f$ is strongly convex"?

What is the intuition behind the given expression for firm nonexpansiveness?

How can you show that firm nonexpansive operators are $\alpha$-averaged with $\alpha = \frac{1}{2}$?

Is anyone aware of the proof of why proximal map is firm nonexpansive?