# Weak to strong mapping

Let $H$ be a real Hilbert space. A mapping $F:H \rightarrow H$ is said to be

• strongly monotone if there exists $\alpha>0$ such that $$\langle F(u)-F(v), u-v\rangle\geq \alpha \|u-v\|^2, \quad \forall u,v\in H;$$

• inverse strongly monotone if there exists $\alpha>0$ such that $$\langle F(u)-F(v), u-v\rangle\geq \alpha \|F(u)-F(v)\|^2, \quad \forall u,v\in H;$$

• weak to strong if the following implication holds $$(u_n\rightharpoonup u_*, F(u_n)\rightarrow F(u_*))\; \Longrightarrow\;(u_n\rightarrow u_*)$$

• Lipschitz continuous if there exists $L>0$ such that $$\|F(u)-F(v)\|\leq L\|u-v\|, \quad \forall u,v\in H.$$

From the above definitions, we can check that

• If $F$ is Lipschitz continuous and strongly monotone then $F$ is inverse strongly monotone.

• If $F$ is strongly monotone than $F$ is weak to strong.

• If $F$ is inverse strongly monotone then $F$ is Lipschitz continuous.

I have some difficulties in the following question:

"Could we construct a class of inverse strongly monotone mappings in infinite dimensional real Hilbert space that are weak to strong but none of them is strongly monotone?"

I would like to thank for all kind comments and helping.

For a fixed $p>0$, define $$F(u)=\min(\|u\|^p,1)\, u$$ This map is a homeomorphism, with the inverse given by $$F^{-1}(u)=\max(\|u\|^{q},1)\, u, \quad q=-\frac{p}{p+1}$$ Therefore, $F(u_n)\to F(u)$ implies $u_n\to u$.
Observe that $F$ is Lipschitz but $F^{-1}$ is not. In particular, $F^{-1}$ is not inverse strongly monotone. This means that $F$ is not strongly monotone.
It remains to prove that $F$ is inverse strongly monotone. Equivalently, we can show that $F^{-1}$ is strongly monotone. One way to do it is to notice that $F^{-1}$ is the gradient of the convex function $V(x)=\max(\|u\|^2/2, \|u\|^{2+q}/(2+q))$. The computation is essentially finite-dimensional, so we can just work with the Hessian matrix of $V$. When $\|u\|>1$, the matrix is the identity. When $\|u\|<1$, it is a multiple of identity plus a positive semidefinite matrix. Either way, $V$ is strongly convex and therefore its gradient is strongly monotone.
• Dear Sir. When $\|u\|=0$ we can not find the value of $F^{-1}(u)$? – blindman Mar 27 '13 at 22:54