$f:l^1 \to l^1$ be a linear operator given by $f((y_1,y_2, ...))=(x_1y_1,x_2y_2, ...), \forall y=(y_1,y_2,....) \in l^1$ , where $x=(x_1,x_2,....) \in l^{\infty}$ is a fixed element. We have to show that $||f|| = ||x||_{\infty}$. I have shown that -
$||f(y)|| = \sum_{i=1}^{\infty} |x_iy_i| \leq ||x||_{\infty} \sum_{i=1}^{\infty} |y_i| = ||x||_{\infty} ||y||_1$, $\forall y \in l^1$ $\Rightarrow ||f|| \leq ||x||_{\infty}$
But I am unable to find any suitable element in $l^1$ for proving the other part of the inequality. Please give some hints. Thank you.
$f(e_n)=x_ne_n$ so $\|f(e_n)\|=\|x_n\|$ By definition of norm of an operator this implies $\|f\| \geq |x_n|$. This is true for all $n$, so $\|f\| \geq \|x\|_{\infty}$. [$e_n$ is the sequence with $1$ in the $n-$th place and $0$ elsewhere].