A strongly pseudomonotone map that is not strongly monotone Let $K$ be a closed convex subset in $\mathbb{R}^n$ and $F: K\rightarrow \mathbb{R}^n$. We say that


*

*$F$ is strongly monotone on $K$ if there exists $\gamma>0$ such that
$$
\left<F(y)-F(x), y-x\right>\geq \gamma\|y-x\|^2, \quad \forall x,y\in K.
$$

*$F$ is strongly pseudomonotone on $K$ if there exists $\gamma>0$ such that
$$
\left<F(x), y-x\right>\geq 0 \Longrightarrow \left<F(y), y-x\right>\geq \gamma\|y-x\|^2
$$
for all $x,y\in K$.


It is easily to verify that strongly monotone implies strongly pseudomonotone.
The converse is not true in general. For example, in one-dimensional case 
$$
F(x)=(2-x), \quad K=[0,1],
$$ 
the mapping $F$ is strongly pseudomonotone but not strongly monotone on $K$.
Question: Can we find a mapping $F: K\rightarrow \mathbb{R}^n (n\geq 2)$ such that $F$ is strongly pseudomonotone but not strongly monotone on $K$. It is interesting to find an affine mapping as in the above example. 
 A: I give two examples: affine and non-affine. The non-affine is nice and easy to understand. The affine is a pain to write out in details, but I give the idea.
Non-affine example. I define it on the punctured unit ball $U=\{x\in \mathbb R^n : 0<\|x\|\le 1\}$; of course you can restrict it to a convex set. The definition is similar to yours: $F(x)=(2\|x\|^{-1}-1)x$. Clearly, this map is not monotone. To check strong pseudo-monotonicity, suppose  $\langle F(x),y-x\rangle \ge 0$. Then $\langle x,y-x\rangle \ge 0$. It follows that $$\langle F(y),y-x\rangle = (2\|y\|^{-1}-1)\langle y,y-x\rangle \ge (2\|y\|^{-1}-1)\langle y-x,y-x \rangle \ge \|y-x\|^2 $$
as required. 
Affine example. I give it in two dimensions: you can make it work in higher dimensions by adding a small multiple of identity to the planar map. In complex notation, $F(z)=1+z+\frac34(1-i)\bar z$. This is not monotone: in terms of $z=x+iy$ we have 
$$G(z):=\mathrm{Re}\,((F(z)-F(0))(\bar z-0))=|z|^2+\frac34 \mathrm{Re}\,((1-i)\bar z^2) = \frac74x^2+\frac14y^2-\frac32xy$$ 
which is a indefinite quadratic form. It is negative precisely in the narrow sectors between the lines $y=(3\pm \sqrt{2})x$. 
Let $U=\{z:|z|<r\}$ where $r>0$ is small enough for what follows. We must show that for all $z\in U$ and all $\zeta$ such that $\mathrm{Re}\,(F(z)\bar \zeta)\ge 0$ and $|\zeta|\le 2r$ the inequality $\mathrm{Re}\,(F(z)\bar \zeta)+G(\zeta)>0$ holds (we get $\ge \gamma |\zeta|^2$ by compactness and homogeneity). Clearly, we only have to worry about $\zeta$ in the aforementioned narrow sectors. 
Actually, there is nothing to worry about when $\zeta $ is in the 1st quadrant sector $(3-\sqrt{2})x\le y\le (3+\sqrt{2})x$, $x>0$, because here the linear term  $\mathrm{Re}\,(F(z)\bar \zeta)$ is positive and dominates everything quadratic. As for the opposite 3rd quadrant sector, $\zeta$ is forbidden from it by the condition $\mathrm{Re}\,(F(z)\bar \zeta)\ge 0$, because $F(z)\approx 1$ when  $r$ is small. 
