If $A$ is a bounded operator on $c_0$, then $\sum_{n=1}^\infty |(Ae_n)(m)|$ is bounded uniformly in $m$

Question

The following is Exercise 7,section 1, chapter 6 of Conway's A course in Functional Analysis.

Let $A\in {\cal B}(c_0)$ (${\cal B}(c_0)$ is linear bounded operators on $c_0$) and for $n\geq 1$, define $e_n \in c_0$ by $e_n(m)=\delta_{nm}$. Put $\alpha_{nm}=(Ae_n)(m)$ for $n,m\geq 1$. Prove $M = \sup_m\sum_{n=1}^\infty |\alpha_{nm}|< \infty.$

I think if I put $(Ae_n)(m)= \frac {(-1)^n}{nm} $, then $A\in B(c_0)$ but $\sum_{n=1}^\infty |\frac {(-1)^n}{nm}|=\infty$ . Where is my mistake? Please help me.

Answer 1

Your 'counterexample': With $x_n := (-1)^ne_n$ we have $Ax_n = \frac 1n\cdot(\frac 1m)_m$. For $n \in \mathbb N$ let $y_N = \sum_{n=1}^N x_n$, then $y_N \in c_0$ with $\|y_n\| = 1$. Extending your $A$ linearly to the span of $\{e_n\mid n \in \def\N{\mathbb N}\N\}$, we have $$ Ay_n = \sum_{n=1}^N Ax_n = \sum_{n=1}^N \frac 1n \cdot \left(\frac 1m\right)_m $$ Hence, as $\|y_N\| = 1$, $$ \|A\| \ge \|Ay_n\|= \sum_{n=1}^N \left\|\left(\frac 1m\right)_m\right\| = \sum_{n=1}^N \frac 1n \to \infty $$ So $A$ is unbounded.

Hint. As 900-sit-ups suggested, consider $A^*\colon \ell^1 \to \ell^1$. For $e_m \in \ell^1$, we have $$ (A^*e_m)_n = (A^*e_m)(e_n) = e_m(Ae_n) = \alpha_{nm} $$ So $\|A^*e_m\| = \sum_{n=1}^\infty |\alpha_{nm}|$. As $A$ is bounded, $A^*$ is.

Answer 2

Let $e_m^*$ be the $m$-th unit vector in $l_1$ (the dual of $c_0$). Let $\|A\|$ be the operator norm of $A$.

I have omitted a few details, but the general idea is much the same as that involved in showing that the induced $\infty$ matrix norm is the maximum row sum.

Let $x_N = \sum_{n=0}^N {\overline{e_m^*(A e_n)} \over | \overline{e_m^*(A e_n)}|}e_n$, then we see that $\|x_N\| \le 1$, hence $\|Ax_N\| \le \|A\|$, and in particular, $|e_m^*(Ax_N)| \le \|A\|$. However, since $e_m^*(Ax_N) = \sum_{n=0}^N |{e_m^*(A e_n)} | $, we see that $\sum_{n=0}^N |{e_m^*(A e_n)} | \le \|A\|$.