Finding the norm of $M_x(y)=(x_n y_n)$ I would like to find the norm of the operator: 
Given $x=(x_n)\in l_{\infty}$,
\begin{align} M_x: l_1 &\to l_1 \\ y=(y_n)&\mapsto M_x(y)=(x_n y_n) \end{align}
I proved that $M_x$ is continuous:
$$\|M_x(y)\|_1=\|x_ny_n\|_1=\sum_{n=1}^{\infty}|x_ny_n|\leq \sum_{n=1}^{\infty}\|x\|_\infty|y_n|=\|x\|_{\infty}\|y_n\|_1$$
So, I proved too that $\|M_x\|\leq \|x\|_{\infty}$. To find the norm, iIneed to take some $y\in l_1$ that $\|M_x\|\geq \|x\|_{\infty}$. What can I take?
 A: The second condition you state is not correct;  you have shown already $\lVert M_x y \rVert \leqslant \lVert x \rVert_\infty$ when $\lVert y \rVert = 1$. There may not be a single $y$ of unit $\ell_1$ norm that satisfies your second inequality. 
Rather, what you do need to show is that there exist vectors $y \in \ell_1$ with $\lVert y \rVert = 1$ and such that $ \lVert M_x y \rVert $ is arbitrarily close to $\lVert x \rVert_\infty $.  
By definition of the $\ell_\infty$ norm, $\lVert x \rVert_\infty = \sup_n\{\lvert x_n\rvert  \}$, so there is a sequence of indices, $n_1, n_2, \cdots $ for which $\lvert x_{n_i} \rvert \to \lVert x \rVert_\infty$ as $i \to \infty$.  Define,
$$ y_i =\left\{ \array{ 1 & \text{if } i = n_i \\ 0 &\text{otherwise.}}\right.$$
The $\lVert y_i \rVert_1 = 1$ for all $i$ and $\lVert M_x y_i \rVert_1 = \lvert x_{n_i} \rvert$.  But, by the selection of the $n_i$, we already have $\lvert x_{n_i} \rvert \to \lVert x \rVert_\infty$ as $i \to \infty$,and therefore $\lVert M_xy\rVert \to \lVert x \rVert_\infty$.  Together with the inequality you have already established it now follows $\lVert M_x \rVert = \lVert x \rVert_\infty$ as required.
