Question: 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

Answer: 
A more straightforward way to solve this would be to use that two nonzero complex numbers are equal if and only if their modules are equal, and their arguments are equal modulo $2\pi$ in the following way.

Write

$$z=re^{i\theta},$$

and note that your equation then reads

$$r^ne^{in\theta}=1.$$

Since $r^n=\lvert z^n\rvert=1$ by matching modulo, we have that $r=1$, and so $z=e^{i\theta}\ne0$. Furthermore, by matching arguments,

$$\exists k\in\mathbb{Z},\quad n\theta=2\pi k,$$

i.e.

$$\exists k\in\mathbb{Z},\quad\theta=\frac{2\pi k}n.$$

This gives the possible solutions as

$$z=e^{i\frac{2\pi k}{n}}$$

for $k\in\mathbb{Z}$. It is also easy to check (by plugging into the original equation) that these are indeed solutions for any choice of $k\in\mathbb{Z}$. Finally, notice that

$$e^{i\frac{2\pi (k+n)}{n}}=e^{i\frac{2\pi k}{n}}e^{i2\pi}=e^{i\frac{2\pi k}{n}},$$

and so there are only $n$ solutions, which can be found by taking, for example, $k=0,1,\dots, n-1$.
An easier way to solve would be:
$$
\begin{align\*}
z^{n} &= 1 \Rightarrow z^{n} = |z^{n}|\\
z &= |z| \cdot \mathrm{e}^{\arg(z) \cdot \mathrm{i}}
(|z| \cdot \mathrm{e}^{\arg(z) \cdot \mathrm{i}})^{n} &= 1\\
|z|^{n} \cdot (\mathrm{e}^{\arg(z) \cdot \mathrm{i}})^{n} &= 1\\
|z|^{n} \cdot \mathrm{e}^{n \cdot \arg(z) \cdot \mathrm{i}} &= 1\\
\mathrm{e}^{n \cdot \arg(z) \cdot \mathrm{i}} &= 1 \quad\mid\quad (\text{ 
})^{\frac{1}{2}}\\
\mathrm{e}^{\arg(z) \cdot \mathrm{i}} &= 1^{\frac{1}{2}}\\
\mathrm{e}^{\arg(z) \cdot \mathrm{i}} &= \pm \sqrt{1}\\
\mathrm{e}^{\arg(z) \cdot \mathrm{i}} &= \pm 1\\
&\Rightarrow \mathrm{e}^{\arg(z) \cdot \mathrm{i}} = \Re(\mathrm{e}^{\arg(z) \cdot \mathrm{i}})\\
&=\Rightarrow \arg{z} = 2 \cdot \frac{k\_{k \in \mathbb{Z}} + n}{n} \cdot \pi
\end{align\*}
$$

That means that there are n solutions with $$ k = 1, ~2, ~..., ~n - 1, $$
wich means that there are mure solutios for z then $$ z = \left\{1, ~i, ~-1, ~-i\right\}, $$ aka you should change this.
e.g.
$$
\begin{align\*}
z &= \frac{\mathrm{1}}{\sqrt{2}} + \frac{\mathrm{i}}{\sqrt{2}}\\
z^{4} &= (\frac{\mathrm{1}}{\sqrt{2}} + \frac{\mathrm{i}}{\sqrt{2}})^{4}\\
z^{4} &= (1^{\frac{1}{4}})^{4}\\
z^{4} &= 1
\end{align\*}
$$