conditions for norm of linear bounded operator to satisfy $\lvert T_x (y) \rvert = \lVert T_x \rVert$.

Question

Let $x = (x_n)_{n \in \mathbb N} \in l^\infty$ and let $T_x : l^1 \rightarrow \mathbb F$ be defined by $T_x (y) = \sum_{n=1}^\infty x_ny_n$. What condition on $x$ is needed so that there exists $y \in l^1$ such that $ \lVert y \rVert_1 = 1$ and $ \lvert T_x (y) \rvert = \lVert T_x \rVert$?

Solution:

$ \lvert T_x (y) \rvert = \lvert \sum_{n=1}^\infty x_ny_n \rvert \le \lVert x \rVert_\infty \cdot \lVert y \rVert_1$ for all $x \in l^\infty, y \in l^1$

For $ \lVert y \rVert_1 = 1$ we have $\lvert \sum_{n=1}^\infty x_ny_n \rvert \le \lVert x \rVert_\infty = sup_{n \ge 1} \lvert x_n \rvert$

I cannot see what condition on $x$ there is for $\lvert T_x (y) \rvert = \lVert T_x \rVert$.

Answer 1

For every $y\in \ell^1$ we have $$ |T_x(y)|\le \sum_{n=1}^\infty|x_n||y_n|\le \lVert x\rVert_\infty\sum_{n=1}^\infty|y_n|=\lVert x\rVert_\infty\lVert y\rVert_1, $$ and therefore $\lVert T_x\rVert \le \lVert x\rVert_\infty$. If we choose $x\in \ell^\infty$ such that $\lVert x\rVert_\infty=|x_k|$ for some positive integer $k$, then, denoting by $e_i$ the element $e_i=(\delta_{in})_{n\ge 1}$ of $\ell^1$ $$ |T_x(e_k)|=\sum_{n=1}^\infty x_n\delta_{kn}=|x_k|=\lVert x\rVert_\infty $$ and therefore $$ \lVert T_x\rVert=\max_{\lVert y\rVert_1=1}|T_x(y)|\ge |T_x(e_k)|=\lVert x\rVert_\infty, $$ i.e. $\lVert T_x\rVert=\lVert x\rVert_\infty$ provided $\sup_{n\ge 1}|x_n|=\max_{n\ge 1}|x_n|$. Hence, the condition on $x\in \ell^\infty$ is $\lVert x\rVert_\infty=|x_k|$ for some $k\in \mathbb{N}$.