Weak convergence and convergence of inner product implies strong convergence?

Question

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 $\langle u, v \rangle > 0$

Answer 1

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:

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

2. 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.

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

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

Answer 2

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.