Weak convergence and convergence of inner product implies strong convergence? Let $H$ be a real Hilbert space and assume that
$$ u_{n} \rightharpoonup u \textrm{ in } H, $$
$$ v_{n} \rightharpoonup v \textrm{ in } H $$
and
$$ \langle u_{n},v_{n}\rangle \rightarrow \langle u,v \rangle \textrm{ in } \mathbb{R}. $$ 
Is it then the case that $u_{n} \rightarrow u$ and $v_{n} \rightarrow v$ strongly in $H$?
EDIT Consider $\langle u, v \rangle > 0$
 A: No. For example, if $v_{n}=v=0$, then we always have $v_{n}\rightharpoonup v$
and $\langle u_{n},v_{n}\rangle\rightarrow\langle u,v\rangle=0$.
If your proposition is true, it implies that $u_{n}\rightarrow u$
whenever $u_{n}\rightharpoonup u$, which is obviously false.
A: No. Here is a counter-example. Consider the real Hilbert space $\mathcal{H}=l^{2}(\mathbb{N})$.
Let $a\in\mathcal{H}$ be defined by $a(n)=\frac{1}{n}$. Note that
$||a||^{2}=\sum_{n=1}^{\infty}\frac{1}{n^{2}}<\infty$ and $a\neq0$.
For each $n\in\mathbb{N}$, let $e_{n}\in\mathcal{H}$ be defined
by $e_{n}(k)=\delta_{nk}$. Let $x=y=a$, $x_{n}=e_{n}+a$, $y_{n}=e_{n+1}+a$.
We go to show that $x_{n}\rightarrow x$ weakly, $y_{n}\rightarrow y$
weakly, $\langle x_{n},y_{n}\rangle\rightarrow\langle x,y\rangle$,
$||x_{n}-x||\not\rightarrow0$ and $||y_{n}-y||\not\rightarrow0$,
$\langle x,y\rangle>0$.
Proof:


*

*Clearly $\langle x,y\rangle=||a||^{2}>0$.

*Let $u\in\mathbb{H}$ be arbitrary, then $\langle x_{n}-x,u\rangle=\langle e_{n},u\rangle\rightarrow0$
as $n\rightarrow\infty$. This shows that $x_{n}\rightarrow x$ weakly.
Similarly, we can show that $y_{n}\rightarrow y$ weakly.

*$\langle x_{n},y_{n}\rangle=||a||^{2}+\langle a,e_{n}\rangle+\langle a,e_{n+1}\rangle\rightarrow||a||^{2}=\langle x,y\rangle$.

*$||x_{n}-x||=||e_{n}||=1$, so $||x_{n}-x||\not\rightarrow0$.
Similarly, $||y_{n}-y||=||e_{n+1}||=1$, so $||y_{n}-y||\not\rightarrow0$.
