Proving equivalence of two definitions Hi all I am intersted in proving the equivalence of the following two definitions of pseudomontoncity:
Let $V$ be a reflexive Banach space and $K \subset V$ closed and convex.
Definition 1:
$A: V \rightarrow V^{*}$ is pseudomonotone if  $A \text{ is bounded}$ and $$u_{n} \rightharpoonup u \text{ and } \limsup\limits_{{n \rightarrow \infty}}\langle A(u_{n}), u_{n}-u\rangle \leq 0 ~ \Longrightarrow ~ \forall v \in V: \langle A(u), u-v\rangle \leq \liminf\limits_{{n \rightarrow \infty}}\langle A(u_{n}), u_{n}-v \rangle.$$
Defintion 2: $A$ is pseudomonotone if $A~~\text{is bounded}$ and $$ u_{n} \rightharpoonup u \text{ in } V,~~u_{n} \in K ~~\text{ and }~~ \limsup\limits_{n \rightarrow \infty}\langle A(u_{n}),u_{n}-u \rangle \leq 0$$ implies $$\lim\limits_{n \rightarrow \infty} \langle A(u_{n}), u_{n}-u \rangle = 0~~ \text{  and }~~ A(u_{n}) \rightharpoonup A(u) \text{  in  } V^{*}. $$
The implication Definition 2 $\implies$ Definition 1 is easy to show. I am interested in the other implication Definition 1 $\implies$ Definition 2. Does anyone have any idea how to prove this?
 A: Taking $v = u$ in definition 1, we find
$$\langle A(u), u-u\rangle = 0 \leqslant \liminf_{n\to\infty} \langle A(u_n), u_n - u\rangle \leqslant \limsup_{n\to\infty} \langle A(u_n), u_n - u\rangle \leqslant 0,$$
which shows the first part,
$$\lim_{n\to\infty} \langle A(u_n), u_n - u\rangle = 0,$$
of definition 2. Using that, we obtain that for all $v\in V$ we have
$$\begin{aligned}
\langle A(u), u-v\rangle &\leqslant \liminf_{n\to\infty} \langle A(u_n), u_n - v\rangle\\
&= \liminf_{n\to\infty} \bigl( \langle A(u_n), u_n-u\rangle + \langle A(u_n), u-v\rangle\bigr)\\
&= \lim_{n\to\infty} \langle A(u_n),u_n-u\rangle + \liminf_{n\to\infty} \langle A(u_n), u-v\rangle\\
&= \liminf_{n\to\infty} \langle A(u_n), u-v\rangle.
\end{aligned}$$
In other words, for all $w\in V$, we have - setting $v = u-w$ -
$$\langle A(u),w\rangle \leqslant \liminf_{n\to\infty} \langle A(u_n),w\rangle.\tag{1}$$
The same inequality for $-w$ yields
$$\begin{aligned}
-\langle A(u),w\rangle &= \langle A(u),-w\rangle\\
&\leqslant \liminf_{n\to\infty} \langle A(u_n),-w\rangle\\
&= \liminf_{n\to\infty} \bigl(-\langle A(u_n),w\rangle\bigr)\\
&= -\limsup_{n\to\infty} \langle A(u_n),w\rangle,
\end{aligned}$$
i.e.
$$\limsup_{n\to\infty} \langle A(u_n),w\rangle \leqslant \langle A(u),w\rangle.\tag{2}$$
Together, $(1)$ and $(2)$ yield
$$\langle A(u),w\rangle = \lim_{n\to\infty} \langle A(u_n),w\rangle\tag{3}$$
for all $w\in V$, i.e. $A(u_n) \rightharpoonup A(u)$ in $V^\ast$ (since $V$ is reflexive, hence the weak and weak$^\ast$ topologies coincide on $V^\ast$). That is the second part of definition 2, and we have shown that an operator that is pseudomonotone according to definition 1 is also pseudoomonotone according to definition 2.
