$\newcommand{\scp}[2]{\langle #1,#2\rangle}\newcommand{\id}{\mathrm{Id}}$
Let $f$ and $g$ be two proper, convex and lower semi-continuous functions (on a Hilbert space $X$ or $X=\mathbb{R}^n$) and let $g$ be continuously differentiable. Consequently, the subdifferential $\partial f$ and the gradient $\nabla g$ are monotone operators, i.e.
$$
\scp{\nabla g(x)-\nabla g(y)}{x-y}\geq 0
$$
for all $x,y$ and for $u\in\partial f(x)$ and $v\in\partial f(y)$
$$
\scp{u-v}{x-y}\geq 0.
$$
My question:



> 
> Is the operator $x\mapsto x - (\id + \gamma\partial f)^{-1}(x-\gamma\nabla g(x))$ monotone for $\gamma\geq 0$?
> 
> 
> 


Note that I ask for all values $\gamma\geq 0$ specifically for large values.


An equivalent question is



> 
> Does for $\gamma\geq 0$ it holds that
>  $$
> \scp{(\id + \gamma\partial f)^{-1}(x-\gamma\nabla g(x)) - (\id + \gamma\partial f)^{-1}(y-\gamma\nabla g(y))}{x-y}\leq \|x-y\|^2
> $$
> 
> 
> 


Thoughts:


For $f=0$ and $g=0$ it's clear (for $g=0$ this follows since the "proximal operator" $x\mapsto (\id + \gamma\partial f)^{-1}(x)$ is non-expansive, for $f=0$, the monotonicity of $\gamma\nabla g$ works in the right direction and does the trick). 


Also for small $\gamma$ (smaller that $2/L$ if $L$ is the Lipschitz constant of $\nabla g$, if is has one) the thing is clear as then the mapping $x\mapsto (\id + \gamma\partial f)^{-1}(x - \gamma\nabla g(x))$ is again non-expasive (and used as iteration in the so-called proximal gradient method). However, for large $\gamma$ I could not make use of any of these observation since using Cauchy-Schwarz for the inner product ruins the estimate then.


Moreover all examples I tried numerically (in various dimensions and for various functions $f$ and $g$) suggested that the claim holds.


Intuitively, all these together makes me think that the answer to my questions is yes, but I failed to prove it. Also all inequalities in Bauschke/Combettes "Convex analysis and Monotone Operator Theory in Hilbert spaces" I found were not helpful. 



Is the prox-residual monotone?
This inequality is not true. Here is a counterexample:


Let ${\bf x},{\bf y} \in \mathbb{R}^2$ with ${\bf y} = 0$ and ${\bf x} = (1,\epsilon)$ for $\epsilon$ small positive. Let $g(x,y) = Cy^2$ with $C(\epsilon)$ very large, chosen say so that
$${\bf x} - \nabla g({\bf x}) = (1, -M)$$
for some $M$ large.


We now construct $f$ convex so that $\nabla(|x|^2/2 + f)(2,-10) = (1,-M)$ and $\nabla f(0) = 0$, which would violate the desired inequality. We rewrite this as $\nabla f(2,-10) = (-1,10-M)$, $\nabla f(0) = 0$. Take
$$f(x,y) = \frac{1}{2}(C\_1y^2 + C\_2(x+y)^2)$$
for $C\_1,C\_2$ to be chosen momentarily. Then
$$\nabla f(2,-10) = (-8C\_2, -10C\_1 - 8C\_2), \quad \nabla f(0) = 0.$$
Taking $C\_2 = 1/8$ and $C\_1$ so that the second component is $10-M$ (if $M$ is large then $C\_1$ can be chosen positive, making $f$ convex), we are done.


The constants in this example are somewhat arbitrary; geometrically, the point is that we can make $g$ "very monotone" in one direction so that the non-expansivity of $(Id + \nabla f)^{-1}$ can't save us.


