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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 $$ \left\geq \gamma\|y-x\|^2, \quad \forall x,y\in K. $$
  • $F$ is strongly pseudomonotone on $K$ if there exists $\gamma>0$ such that $$ \left\geq 0 \Longrightarrow \left\geq \gamma\|y-x\|^2 $$ 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.

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    @Robert Israel: Dear Sir. I am looking forward to hearing your interesting comments2012-07-26
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    Notifications do not work when addressed to someone who has not commented, answered, or edited the question. In other words, Robert Israel will not find out about that comment unless he happens to come across this page on his own. There will be no "message notification".2012-07-26
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    @ArturoMagidin: Thank your for your comments.2012-07-26
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    Presumably you want $K$ to have nonempty interior, otherwise just transfer the one-dimensional example to a line segment in ${\mathbb R}^n$.2012-07-26
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    With that assumption, if $F$ is affine, i.e. $F(x) = c + Tx$ where $T$ is linear, then $F$ is strongly monotone iff there is no nonzero vector $v$ with $\langle T v, v \rangle \le 0$. The question then is whether $F$ can be strongly pseudomonotone if such a vector exists. I don't know the answer.2012-07-26
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    @Robert Israel: Dear Sir. I am confess that I would like to find an example in $\mathbb{R}^n (n\geq 2)$ and $\text{int}K\ne \emptyset$ (K has a nonempty interior). Do you have any comments on my problem. Thank you for your helping and your consideration on my work.2012-07-27

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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| 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.

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    Dir Sir. Thank you very much for your answer. I would like to discuss about your frist example (nonlinear case). Indeed, the mapping $F(u)=(2-\|u\|)u$ is **monotone** on the punctered unit ball $U$. Let $u,v\in U$, we note that $$2012-07-30
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    \begin{eqnarray*} \langle F(u)-F(v), u-v\rangle &=&\langle(2-\|u\|)u-(2-\|v\|)v, u-v\rangle\\ &=&\langle 2(u-v)-\|u\|(u-v)-v(\|u\|-\|v\|), u-v\rangle\\ &=&2\|u-v\|^2-\|u\|\|u-v\|^2-(\|u\|-\|v\|)\langle v, u-v\rangle\\ &\geq& 2\|u-v\|^2-\|u\|\|u-v\|^2-\left|\|u\|-\|v\|\right|\|v\|\|u-v\|\\ &\geq& 2\|u-v\|^2-\|u\|\|u-v\|^2-\|u-v\||\|v\|\|u-v\|\\ &\geq&0. \end{eqnarray*}2012-07-30
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    @blindman I fixed it now. The example is obtained by vectorizing the scalar function $y=2-x$, but I forgot that multiplying $2-\|x\|$ by the vector $x$ changes the magnitide, not just direction.2012-07-31
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    @user31373: I have an interesting question in the following http://math.stackexchange.com/questions/346937/on-the-weak-and-strong-convergence-of-an-iterative-sequence I would like to ask your comments?2013-04-12