Does there exists a pseudomonotone operator $A: V \to V^*$, where $V$ is a (real reflexive, separable) Banach space such that $-A$ is also pseudomonotone?
An operator is called pseudomontone if $u_n \rightharpoonup u$ and $\limsup_{n \to \infty} \langle A u_n, u_n - u \rangle \le 0$ implies that $\langle A u, u - w \rangle \le \liminf_{n \to \infty} \langle A u_n, u_n - w \rangle$ holds for all $w \in V$.
Here's what I figured out so far: if such an operator exists then $u_n \rightharpoonup u$ and $\langle A u_n, u_n - u \rangle \to 0$ imply that $\langle A u_n, u_n - w \rangle \to \langle A u, u - w \rangle$ for all $w \in V$, since we have $$ \limsup_{n \to \infty} \langle -A u_n, u_n - u \rangle \le 0 \iff \liminf_{n \to \infty} \langle A u_n, u_n - u \rangle \ge 0 $$ and $$ \langle -A u, u - w \rangle \le \liminf_{n \to \infty} \langle -A u_n, u_n - w \rangle \iff \langle A u, u - w \rangle \ge \limsup_{n \to \infty} \langle A u_n, u_n - w \rangle. $$ Now if $\limsup_{n \to \infty} a_n \le b \le \liminf_{n \to \infty} a_n$ for any real sequence $(a_n)_n$ we have $a_n \to b$.
So in summary we have
- $u_n \rightharpoonup{} u$
- $\langle A u_n, u_n - u \rangle \to 0 = \langle A u, u - u \rangle$
- $\langle A u_n, u_n - w \rangle \to \langle A u, u - w \rangle$ for all $w \in V$.
Do such operators exist and can they be characterised in a good way?
Edit We also have the following alternative characterisation of pseudomonotone operators:
A bounded operator $A$ is pseudomonotone if and only if, from $u_n \rightharpoonup u$ in $V$ and $\limsup_{n \to \infty} \langle A u_n, u_n - u \rangle \le 0$ it follows that $A u_n \rightharpoonup A u$ and $\langle A u_n, u_n \rangle \to \langle A u, u \rangle$ hold.