Is the (pre)image of a convex set under a maximal monotone operator convex? $T:E\rightrightarrows E^*$ is a maximal monotone operator.
I know that for every $x\in E$, $T(x)$ is a convex set.
But is it true that $T(S)$ is a convex set for every convex subset $S\subset E$?
If not, does it hold under additional conditions?  
Similarly, is it true that $T^{-1}(K)$ is a convex set for every convex subset $K\subset E^*$?
 A: I see that the answer to the first question is no. There is a paper due to Jean-Pierre Gossez "On a convexity property of the range of a maximal monotone operator", where he exhibit a maximal monotone operator $T$ from $l^1$ to $l^\infty$ with domain $l^1$ such that its range (the image of the convex set $l^1$) has not convex closure, thus the set $T(l^1)$ cannot be convex either.
Concerning the second question, I have no idea which additional conditions ensure the property you want.
For the third question possed, the inverse of a maximal monotone operator is also a maximal monotone operator, and the preimage is the image of the inverse. Therefore, the answer is also negative.
Good bye
A: This is a variant on Yi-Hsuan Lin's answer.

*

*Rockafellar has in his book on Convex Analysis a proper lower semicontinuous convex function $f$ on $\mathbb{R}^2$ such that
$\operatorname{dom}\partial f$ is not convex. Set $T:= (\partial f)^{-1}$ and $S=\mathbb{R}^2$ to get also a counter-example which is much much simpler than the Gossez example.


*Sufficient conditions are $E=\mathbb{R}$ or $T$ is affine. The second condition is close to a necessary condition.


*See 1. Note that the inverse is not necessarily maximally monotone when $E$ is not reflexive.
