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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 $$ \left<F(y)-F(x), y-x\right>\geq \gamma\|y-x\|^2, \quad \forall x,y\in K. $$
  • $F$ 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.

blindman
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  • Presumably you want $K$ to have nonempty interior, otherwise just transfer the one-dimensional example to a line segment in ${\mathbb R}^n$. – Robert Israel Jul 26 '12 at 19:37
  • 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. – Robert Israel Jul 26 '12 at 19:44
  • @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. – blindman Jul 27 '12 at 00:02

1 Answers1

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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 $U=\{z:|z|<r\}$ 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.

  • 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 $$ – blindman Jul 30 '12 at 22:47
  • \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} – blindman Jul 30 '12 at 23:11
  • @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. –  Jul 31 '12 at 01:48