Pseudo Monotone Operator Suppose $X$ is a real Reflexive Banach space. Let $A:X\rightarrow X^{\star}$ be a Pseudo Monotone operator, i.e. if $u_{n}\rightharpoonup u$ and $\limsup\langle Au_{n},u_{n}-u\rangle\leq 0$, then $$\langle Au,u-w\rangle\leq\liminf\langle Au_{n},u_{n}-w\rangle,\ \forall\ w\in X$$
where $\rightharpoonup$ stands for weak convergence and $\langle\ \rangle$ represents duality.
I want to prove that $$Au_{n}\rightharpoonup Au$$ and $$\langle Au_{n},u_{n}\rangle\rightarrow \langle Au,u\rangle$$
What i have tried?
We can write $\langle Au_{n},u_{n}\rangle=\langle Au_{n}-Au,u_{n}-u\rangle-\langle Au,u\rangle+\langle Au_{n},u\rangle+\langle Au,u_{n}\rangle$. 
I showed that $\langle Au_{n}-Au,u_{n}-u\rangle\rightarrow 0$. Then if i can show that $Au_{n}\rightharpoonup Au$, the problem is solved because $X$ is reflexive (so $\langle Au_{n},u\rangle\rightarrow\langle Au,u\rangle$). 
Any idea? Thanks
 A: 1 - $\lim\langle Au_n,u_n-u\rangle=0$
Indeed, ofr $w=u$ we have 
\begin{eqnarray}
 0 &\leq& \liminf\langle Au_n,u_n-u\rangle      \nonumber \\
   &\leq& \limsup\langle Au_n,u_n-u\rangle \nonumber \\
   &\le& 0
\end{eqnarray}
Therefore, $\lim\langle Au_n,u_n-u\rangle=0$
2 - $\langle Au,u-w\rangle\leq\liminf\langle Au_n,u-w\rangle,\ \forall\ w\in X$
We have, (im using 1 here)
\begin{eqnarray}
 \langle Au,u-w\rangle &\leq&  \liminf\langle Au_n,u_n-w\rangle+\liminf\langle Au_n,-u_n+u\rangle     \nonumber \\
   &\leq& \liminf\langle Au_n,u_n-u_n+u-w\rangle \nonumber \\
   &=& \liminf\langle Au_n,u-w\rangle
\end{eqnarray}
3 - $\langle Au_n,v\rangle\rightarrow\langle Au,v\rangle,\ \forall\ v\in X$
Take $w=u-v$ in 2. Hence 
\begin{eqnarray}
 \langle Au,v\rangle &\leq& \liminf\langle Au_n,v\rangle      \nonumber \\
   &\leq& \limsup\langle Au_n,u_n-u_n+u-u+v\rangle \nonumber \\
   &\le& \limsup\langle Au_n,u_n-u\rangle+\limsup\langle Au_n,-u_n+u+v\rangle \\
   &=& \limsup-\langle Au_n,u_n-(u+v)\rangle \\
   &=& -\liminf\langle Au_n,u_n-(u+v)\rangle \\
   &\leq& -\langle Au,u-(u+v)\rangle \\
   &=& \langle Au,v\rangle
\end{eqnarray}
Because $X$ is reflexive, by using 3 we get that $Au_n\rightharpoonup Au$
4 - $\lim\langle Au_n-Au,u_n-u\rangle\rightarrow 0$
By using the hypothesis with $w=u$ we have that 
\begin{eqnarray}
0 &\leq& \liminf\langle Au_n,u_n-u\rangle       \nonumber \\
   &=& \liminf\langle Au_n,u_n-u\rangle+\liminf\langle Au,u_n-u\rangle \nonumber \\
   &\le& \liminf\langle Au_n-Au,u_n-u\rangle \\
   &\leq& \limsup\langle Au_n-Au,u_n-u\rangle \\
   &\leq& \limsup\langle Au_n,u_n-u\rangle \\
&\leq& 0
\end{eqnarray}
5 - $\langle Au_n,u_n\rangle\rightarrow\langle Au,u\rangle$
Note that $$\langle Au_n,u_n\rangle=\langle Au_n-Au,u_n-u\rangle-\langle Au,u\rangle+\langle Au_n,u\rangle+\langle Au,u_n\rangle$$
It follows from 3,5 and $u_n\rightharpoonup u$ that $$\lim\langle Au_n,u_n\rangle=\langle Au,u\rangle$$
Please verify if my proof is correct.
Id like to observe that the converse of this theorem is also true and much more easy to prove.
