Image of double adjoint operator

Question

Let $X,\,Y$ be Banach spaces and $T:X\to Y$ a bounded operator. I want to determine whether $\text{Im }T^{\ast\ast}\subset J(Y)$, where $J$ is the canonical embedding from $Y$ into $Y^{\ast\ast}$, and moreover if for every $x\in X\subset X^{\ast\ast}$, $$T^{\ast\ast}(x) = J(Tx).$$

For a little bit of context, I'm reading a proof of the following theorem on Albiac and Kalton's Topics in Banach Space Theory:

Let $X$ be a Banach space and $\sum x_n$ a WUC (weakly unconditionally Cauchy) series on $X$. Then $\sum x_n$ is unconditionally convergent, i.e., the series $\sum x_{\pi(n)}$ converges for every permutation $\pi$ of $\mathbb{N}$, if and only if there is a compact operator $T:c_0\to X$ such that $Te_n = x_n$.

It is an earlier result that a series $\sum x_n$ is WUC if only if there is a continuous operator $T:c_0\to X$ such that $Te_n = x_n$.

The problem is that to prove the $(\Leftarrow)$ part of the theorem, the authors state that $$T^{\ast\ast}:\ell_\infty\to X\subset X^{\ast\ast}$$ and that $T(\sum e_n) = \sum x_n$, and I can't understand why these properties hold.

Answer 1

Let $T$ be the identity operator $T:c_0\to c_0$. Its adjoint is the identity operator on $\ell_1$. The second adjoint is the identity operator on $\ell_\infty$. So the range of $T^{**}$ need not be in $J(Y)$ for general operators.

However, in the context $T$ is a compact operator. Since $c_0$ has the approximation property, $T$ is the norm-limit of finite rank operators, so it suffices to verify the statement for the case of finite rank. Each such operator is a sum of rank-one operators, so we can assume $\operatorname{rank} T=1$; the general case follows by linearity.

Every rank-one operator $T:X\to Y$ can be written as $T(x) = f(x)y$ for some $f\in X^*$ and $y\in Y$. Then the adjoint is given by $T^*(g) = g(y)x$ (here $g\in Y^*$) and the second adjoint by $T^{**}(\phi) = \phi(f)y$ (here $\phi\in X^{**}$ so it acts on the elements of $X^*$). Clearly, the range of $T$ is in $J(Y)$, and the relation $T^{**}(x)=J(T(x))$ holds for $x\in X$.

Remarks

1. The relation $T^{**}\circ J_X = J_Y\circ T$ (where $J_X:X\to X^{**}$ and $J_Y:Y\to Y^{**}$ are canonical embeddings) holds for general bounded operators; this is just a matter of diagram chasing.

2. I don't know if $\operatorname{Im} T^{**} \subset J(Y)$ holds for compact operators in general (when $X$ does not have the approximation property).