# Operator attains norm

We have the following linear and well-defined mapping $T_{a}(b):=\sum_{j=1}^{\infty}a_{j}b_{j}$, with $\{a_{j}\}\in \ell^{1}$ and $\{b_{j}\} \in c_{0}$. Show that $\|T_{a}\|=\|a\|_{1}$.

My work:

We first prove that $\|T_{a}\|\leq \|a\|_{1}$. We can write

$$|T_{a}(b)| = \left|\sum_{j=1}^{\infty} a_{j}b_{j} \right| \leq \sum_{j=1}^{\infty} |a_{j}b_{j}| = \sum_{j=1}^{\infty} |a_{j}||b_{j}| \leq \|b\|_{\infty} \sum_{j=1}^{\infty} |a_{n}| = \|b\|_{\infty}\|a\|_{1}.$$

By definition of the operator norm we get:

$$\|T_{a}\|\leq \|a\|_{1}.$$

Now we want to prove that $\|T_{a}\|\geq \|a\|_{1}$, so we can conclude $\|T_{a}\|= \|a\|_{1}$. Given $\epsilon > 0$, there is an $J\in \mathbb{N}$ such that

$$\sum_{j=1}^{J}|a_{j}|\geq\|a\|_{1}-\epsilon.$$

Also for $z\in\mathbb{C}$ we define $$\mathrm{sign}(z) = \begin{cases} z/|z|, & \text{if } z\neq 0 \\ 0, & \text{otherwise}. \end{cases}$$

Note that $\overline{\mathrm{sign}(z)}z=|z|$ for all $z$. Further define $b\in c_{0}$ by

$$b_{j} = \begin{cases} \overline{\mathrm{sign}(a_{j})}, & \text{if } j\leq J \\ 0, & \text{otherwise}. \end{cases}$$

Then $b$ has norm at most one.

I think I am nearly there but am struggling with the last part.

Question: How do I proceed to get $\|T_{a}\|\geq\|a\|_{1}$.

First note that $\|b\|_\infty\leq 1$ and $$\|T_a\|\geq|T_a(b)|=\sum_{i=1}^Ja_ib_i=\sum_{i=1}^J|a_i| \geq \|a\|_1-\epsilon$$ It follows that $\|T_a\|+\epsilon \geq \|a\|_1$ for all $\epsilon>0$, hence $\|T_a\|\geq\|a\|_1$.
Yes, you are nearly there: assume that $a\neq 0$ (so that $\lVert b\rVert_\infty=1$ if $N$ is taken large enough); we have $$\lVert T_a\rVert=\sup_{\lVert x\rVert\leqslant 1}\frac{\lVert T_ax\rVert}{\lVert x\rVert}\geqslant |T_a(b)|=\sum_{j=1}^N|a_j|-\varepsilon.$$ As $\varepsilon$ was arbitrary, you can conclude.
In fact, I think you mean that the dual of $c_0$ is isometrically isomorphic to $\ell^1,$ so the norm of $T_a$ is $\lVert a\rVert_1$.