= Solution
A linear map $T:X\to Y$ is a <weakly compact operator> when $T(B_X)$ is relatively weakly compact. Every weakly compact subset of a Banach space is norm bounded, so
$$
\sup_{\lVert x\rVert\leq1}\lVert Tx\rVert<\infty.
$$
Thus even without assuming continuity initially, a weakly compact linear map is bounded.
The key <Gantmacher theorem>[bidual criterion] is
$$
T\text{ is weakly compact}
\quad\Longleftrightarrow\quad
T^{**}(X^{**})\subseteq J_Y(Y).
$$
For the forward implication, approximate $x^{**}\in B_{X^{**}}$ weak-star by points $J_Xx_\alpha$ using the <Goldstine theorem>. Weak compactness supplies a subnet for which $Tx_\alpha$ converges weakly to some $y\in Y$, while weak-star continuity of $T^{**}$ makes $J_YTx_\alpha$ converge to $T^{**}x^{**}$. Hence $T^{**}x^{**}=J_Yy$. Conversely, if the displayed inclusion holds, the weak-star compact set $T^{**}(B_{X^{**}})$ lies in $J_Y(Y)$, where the inherited weak-star topology is the weak topology of $Y$. It is a weakly compact set containing $J_YT(B_X)$.
If $T$ is weakly compact and $y^{***}\in Y^{***}$, restriction of $y^{***}$ to $J_Y(Y)$ defines $y^*\in Y^*$. For $x^{**}\in X^{**}$,
$$
(T^{***}y^{***})(x^{**})
=y^{***}(T^{**}x^{**})
=x^{**}(T^*y^*),
$$
so $T^{***}y^{***}=J_{X^*}(T^*y^*)$. The bidual criterion makes $T^*$ weakly compact. Conversely, if $T^*$ is weakly compact, then $T^{***}(Y^{***})\subseteq J_{X^*}(X^*)$. Every $y^{***}$ annihilating $J_Y(Y)$ is sent to zero, so every $T^{**}x^{**}$ lies in
$$
(J_Y(Y)^\perp)_\perp=J_Y(Y).
$$
The criterion makes $T$ weakly compact. Hence $T$ is weakly compact exactly when $T^*$ is.
The bidual criterion also proves the structure of $\mathcal W(X,Y)$. It is closed under linear combinations. If $T_n\to T$ in operator norm and every $T_n$ is weakly compact, then $T_n^{**}x^{**}\to T^{**}x^{**}$ in norm; the canonical copy $J_Y(Y)$ is norm closed, so $T$ is weakly compact. For bounded composable maps $A$ and $B$, the image condition for $(ATB)^{**}=A^{**}T^{**}B^{**}$ proves the ideal property. Thus the weakly compact operators form a norm-closed <operator ideal of weakly compact operators>[operator ideal].
The <Krein-Smulian theorem> states that the norm-closed convex hull $\overline{\operatorname{co}}K$ of a weakly compact set $K$ is weakly compact. To prove it, view $K$ as a compact Hausdorff space in its weak topology. The probability measures on $K$ form a weak-star compact subset of $C(K)^*$ by the <Banach-Alaoglu theorem>. The barycentre map
$$
\mu\longmapsto b_\mu,
\qquad
x^*(b_\mu)=\int_Kx^*(x)\,d\mu(x)
$$
is weak-star-to-weak continuous; scalar integration and weak compactness ensure that $b_\mu\in X$. Finitely supported probability measures are weak-star dense and their barycentres are precisely $\operatorname{co}K$. The image of all probability measures is therefore the weak closure of $\operatorname{co}K$, and it is weakly compact. A convex set has the same weak and norm closures by the <Hahn-Banach separation theorem>, proving the claim.
If $X$ is reflexive, $B_X$ is weakly compact and every bounded $T:X\to Y$ is weak-to-weak continuous, so $T$ is weakly compact. If $Y$ is reflexive, every bounded subset of $Y$ is relatively weakly compact, with the same conclusion. Thus
$$
\mathcal B(X,Y)=\mathcal W(X,Y)
$$
whenever either space is reflexive.
If $X=\ell^1$ and $Y$ is nonreflexive, the <Eberlein-Smulian theorem> supplies a bounded sequence $(y_n)$ in $Y$ with no weakly convergent subsequence. The formula
$$
T(a)=\sum_{n=1}^\infty a_ny_n
$$
defines a bounded operator $T:\ell^1\to Y$. Since $Te_n=y_n$, it cannot be weakly compact. Therefore $\mathcal B(\ell^1,Y)\ne\mathcal W(\ell^1,Y)$.
This conclusion does not hold for every pair of nonreflexive spaces. Both $c_0$ and $\ell^1$ are nonreflexive, while the <Pitt theorem> says every bounded operator $c_0\to\ell^1$ is compact and hence weakly compact. Thus
$$
\mathcal B(c_0,\ell^1)=\mathcal W(c_0,\ell^1).
$$
Solved by gpt-5.6-sol high.
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