Show linear operator $f: \ell^1 \rightarrow (c_0)'$ is bounded

Question

I must be missing something very simple but may I quickly ask how to show that the linear operator $f: \ell^1 \rightarrow (c_0)'$, $f_y(x)= \sum \limits_{i=1}^\infty x_i*y_i\ $ from linear operator from $\ell^1$ to $(c_0)'$ is bounded?

Answer 1

Presumably $c_0$ is the space of sequences $\{x_n\}$ with $x_n \to 0$ with the $\|\cdot\|_\infty$ norm.

If $\phi \in (c_0)'$ then $\|\phi\|_{(c_0)'} = \displaystyle \sup_{\begin{array}{c} x \in c_0\\ \|x\|_\infty \le 1 \end{array}} \phi(x)$.

If $y \in \ell^1$ then $$ \|f_y\|_{(c_0)'} = \sup_{\begin{array}{c} x \in c_0\\ \|x\|_\infty \le 1 \end{array}} f_y(x) = \sup_{\begin{array}{c} x \in c_0\\ \|x\|_\infty \le 1 \end{array}} \sum_i |x_i y_i| \le \sum_i |y_i| = \|y\|_1$$

If you instead write $f(y)$ for $f_y$ you have that $f : \ell^1 \to (c_0)'$ and $\|f(y)\|_{(c_0)'} \le \|y\|_1$ for all $y \in \ell^1$.

Answer 2

$$|f_y(x)| = \left|\sum_i x_i y_i\right| \le \sum_i |x_i| |y_i| \le \sum_i (\sup_i |x_i|) |y_i| = (\sup_i |x_i|) \sum_i |y_i| = \|x \| \| y \|$$ which says $\| f_y\| \le \|y\|$.

Answer 3

For each $y\in\ell^1$ and $x\in c_0$ we have:

$|f_y(x)|=|\sum_{i=1}^\infty x_iy_i|\leq\sum_{i=1}^\infty |x_i||y_i|\leq ||x||_\infty\sum_{i=1}^\infty |y_i|\leq ||x||_\infty||y||_1$

So given $y\in\ell^1$ we have $\frac{|f_y(x)|}{||x||_\infty}\leq ||y||_1$ for all $0\ne x\in c_0$. Hence:

$||f(y)||=||f_y||=\sup_{x\ne 0}\frac{|f_y(x)|}{||x||_\infty}\leq ||y||_1$.

This means $f$ is bounded and its norm is at most $1$.