# When is an operator on $\ell_1$ the dual of an operator on $c_0$?

Suppose $T:\ell_1\to\ell_1$ is a continuous linear operator. When can we say that $T$ is a dual, or adjoint, of an operator on $c_0$? In other words, under what conditions can we find a continuous linear $S:c_0\to c_0$ such that $T=S^*$?

Recall that $c_0$ is a subspace of $\ell_\infty=\ell_1^*$, and so we can talk about the restriction $T^*|_{c_0}$ of $T^*:\ell_\infty\to\ell_\infty$ to $c_0$.

Observation. If $T^*c_0\subseteq c_0$ then $T=S^*$ for some continuous linear $S:c_0\to c_0$.

Proof. Notice that $(T^*|_{c_0})^*$ is a continuous linear operator acting on $\ell_1$. We claim $T=(T^*|_{c_0})^*$. If not, then there is $z^*\in\ell_1$ so that $Tz^*\neq(T^*|_{c_0})^*z^*$. Hence, there is $y^{**}\in c_0\subset\ell_\infty$ such that $(Tz^*)(y^{**})\neq((T^*|_{c_0})^*z^*)(y^{**})$. However,

\begin{equation*}((T^*|_{c_0})^*z^*)(y^{**})=z^*(T^*y^{**})=(T^*y^{**})(z^*)=y^{**}(Tz^*)=(Tz^*)(y^{**}),\end{equation*}

a contradiction. $\square$

So, using the above, we could look for conditions to guarantee that $T^*c_0\subseteq c_0$. Perhaps it might help to try to adapt the above proof to work for certain kinds of transformations, say $U,V:\ell_\infty\to\ell_\infty$ with $VT^*Uc_0\subseteq c_0$.

Or, we could try an entirely different approach. Suggestions are welcome!

P.S. It would be fantastic if it turned out that if $T$ fails to be strictly singular (or have countable spectrum) then $T=S^*$ for some $S$. However, that seems too much to hope for.

• Some thoughts: In general, a bounded operator between dual spaces is an adjoint operator iff it is weak*-weak* continuous. Also of use may be the fact that operators from $\ell_1$ to a Banach space $X$ are characterized by the bounded sequences in $X$ (given a bounded sequence $(x_i)$ in $X$, map $e_i$ to $x_i$). So in order to be an adjoint operator, $(T e_i)$ must be weak* null. An example: define $T:\ell_1\rightarrow\ell_1$ via $Te_1=e_1$ and $Te_j=e_1+e_j$, $j>1$. $T$ is continuous, in fact an isomorphism, but not weak*-weak* continuous. – David Mitra Aug 24 '14 at 21:30
• If $(x_n^*)$ is weak* null in $\ell_1$ then $Se_i=x_i^*$ gives an operator from $\ell_1$ to $\ell_1$. One can define a bounded operator from $c_0$ to $c_0$ via $T x=(x_n^* x)$. One then has $T^*=S$, no? – David Mitra Aug 24 '14 at 22:00

We have the following characterization of adjoint operators:

Suppose that $X$ and $Y$ are normed spaces. If $T:X\rightarrow Y$ is a bounded linear operator, then $T^*$ is weak*-weak* continuous. Conversely, if $S$ is a weak*-weak* continuous linear operator from $Y^*$ to $X^*$, then there is a bounded linear operator $T:X\rightarrow Y$ such that $T^*=S$.

For a proof, see Robert E. Megginson's An Introduction to Banach Space Theory, Theorem 3.1.11.

The following two facts will also be useful here:

The bounded linear operators from $\ell_1$ to a Banach space $X$ correspond to the bounded sequences in $X$. The correspondence is given by $Te_i^*=x_i$ for the bounded sequence $(x_i)$, where $e_i^*$ is the $i^{\rm th}$ standard unit vector in $\ell_1$.

The bounded linear operators from a Banach space $X$ to $c_0$ correspond to the weak*-null sequences in $X^*$. The correspondence is given by $Tx=(x_i^* x)$ for the weak*-null sequence $(x_i^*)$.

From the first (and easy) part of the characterization of adjoint operators, one sees that if $T:\ell_1\rightarrow\ell_1$ is the adjoint of a bounded operator, then $(Te_i^*)$ is weak*-null.

Conversely, suppose $(x_i^*)$ is a weak*-null sequence in $\ell_1$. Then $(x_i^*)$ is bounded in $\ell_1$; so, the linear operator $T:\ell_1\rightarrow\ell_1$ defined by $Te_i^*=x_i^*$ is bounded. Also, the linear operator $S:c_0\rightarrow c_0$ defined by $Sx=(x_i^* x)$ is bounded. One can easily show $S^* e_i^* =x_i^*$ for each $i$:
$$(S^* e_i^*)e_j=e_i^*(Se_j)=e_i^*(x_n^* e_j)=x_i^*e_j,\text{ for all }j.$$ It follows that, $S^*=T$.

To summarize, an operator $T:\ell_1\rightarrow\ell_1$ is the adjoint of a bounded operator if and only if $(Te_i^*)$ is weak*-null.

• Wow, that's pretty sweet, thank you! I will try to work with this, to get what I need. – Ben W Aug 25 '14 at 1:48