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

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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.

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