To give some example where $x_n\rightarrow x$, $y_n\rightarrow y$ weakly but $(\langle x_n, y_n\rangle)_{n}$ is not convergent: Let us $X$ an vector space with inner product. To give some example where $x_n\rightarrow  x$, $y_n\rightarrow y$ weakly both but $(\langle x_n,y_n\rangle)_{n}$ is not convergent:
Well the excercise has a following hint: Consider $(e_1,e_2,\ldots)$ and $(e_2,e_2,e_4,e_4,\ldots)$ where $e_i$ are orthonormal.
Then for me, if we choose $x_n=(e_1,e_2,\ldots)$ and $y_n=(e_2,e_2,e_4,e_4,\ldots) $ we have $(\langle x_n,y_n\rangle)= (0,1,0,1, \ldots)$ so this sequence is not convergent.
My problem is understanding why the sequences above are weakly convergent, and which is their limit respectively.
For example $x_n\rightharpoonup x$ means that $f(x_n)\rightarrow f(x)$ for all $f\in X'$ (Dual of $X$-all linear continuous functions from $X\to \mathbb C$)
So, please i would like to know if i need to choose a particlar functional $f$ that show me what is the limit $x$?
I found that in $l^{\infty}$ the weak limit for $e_n$ is $0$ and their proof was by contradiction, but what is the tool that permit to find the weak limit of any sequence?
In my excercise do not say anything else, so i do not know who are the limit $x$ and $y$ respectively. Please, i want to know how to star when the problem say something as: Given the sequence $z_n$ calculate their weak limit? Please i will appreciate any hint. Thank you
 A: You can take for $x_n= \sin (nx)$ and for $y_n = \sin (nx)$ if $n$ is even and $y_n= \cos (nx)$ when $n$ is odd. Now, for any function $f$ in $L^2[0,2\pi]$ integral of product $fx_n$ or $fy_n$ tends to $0$, yet integral of $x_ny_n$ jumps between $0$ and $1$.
A: About your example: Let $H$ be a Hilbert space and $e_n$ an orthonormal sequence in $H$. Then $e_n$ must converge to zero weakly. This is due to Bessel's inequality, which states for every vector $x$ in $H$ and orthonormal sequence $e_n$, we have
$$\sum_{n=0}^{\infty} |(x|e_n)|^2 \leqslant \lVert x \rVert^2.$$
This implies that for each $x$, we must have $|(x|e_n)|^2 \to 0$, as the series couln't be convergent otherwise.
Proof of Bessel's inequality in prehilbert spaces:
Let $X$ be a prehilbert space and $e_n$ an orthonormal sequence in it. Let $x$ be an arbitrary vector in $X$. Let's define the following quantities:
$$\alpha_n := (x | e_n) \quad \text{and} \quad s_n := \sum_{k=0}^n \alpha_k e_k.$$
Then, using simple algebra:
$$\lVert x - s_n \rVert ^2 = \lVert x \rVert ^2 - (x | s_n) - (s_n|x) + \lVert s_n \rVert^2$$
From Pythagorean theorem, we also have that
$$\lVert s_n\rVert^2 = \sum_{k=0}^n |\alpha_k|^2,$$
and rewriting the "mixed" term
$$(x|s_n) = \sum_{k=0}^n \alpha^*_n(x|e_n) = \sum_{k=0}^n \alpha^*_n\alpha_n= \sum_{k=0}^n |\alpha_n|^2 \in \mathbb{R},$$
implying also $(x|s_n)=(s_n|x)$.
Putting it together,
$$\lVert x - s_n \rVert ^2 = \lVert x \rVert^2 - \sum_{k=0}^n |\alpha_k|^2 \geqslant 0,$$
proving the required inequality.
