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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, soThus even without assuming continuity initially, a weakly compact linear map is bounded.

The key bidual criterion isFor the forward implication, approximate $x^{\ast \ast }\in B_{X^{\ast \ast }}$ weak-star by points $J_X{x}_{\alpha }$ using the Goldstine theorem. Weak compactness supplies a subnet for which $T{x}_{\alpha }$ converges weakly to some $y\in Y$, while weak-star continuity of $T^{\ast \ast }$ makes $J_{Y}T{x}_{\alpha }$ converge to $T^{\ast \ast }{x}^{\ast \ast }$. Hence $T^{\ast \ast }{x}^{\ast \ast }=J_{Y}y$. Conversely, if the displayed inclusion holds, the weak-star compact set $T^{\ast \ast }(B_{X^{\ast \ast }})$ lies in $J_Y(Y)$, where the inherited weak-star topology is the weak topology of $Y$. It is a weakly compact set containing $J_{Y}T(B_X)$.

If $T$ is weakly compact and $y^{\ast \ast \ast }\in Y^{\ast \ast \ast }$, restriction of $y^{\ast \ast \ast }$ to $J_Y(Y)$ defines $y^{\ast }\in Y^{\ast }$. For $x^{\ast \ast }\in X^{\ast \ast }$,so $T^{\ast \ast \ast }{y}^{\ast \ast \ast }=J_{X^{\ast }}(T^{\ast }{y}^{\ast })$. The bidual criterion makes $T^{\ast }$ weakly compact. Conversely, if $T^{\ast }$ is weakly compact, then $T^{\ast \ast \ast }(Y^{\ast \ast \ast })\subseteq J_{X^{\ast }}(X^{\ast })$. Every $y^{\ast \ast \ast }$ annihilating $J_Y(Y)$ is sent to zero, so every $T^{\ast \ast }{x}^{\ast \ast }$ lies inThe criterion makes $T$ weakly compact. Hence $T$ is weakly compact exactly when $T^{\ast }$ 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 $T_n^{\ast \ast }{x}^{\ast \ast }\to T^{\ast \ast }{x}^{\ast \ast }$ 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 $(ATB{)}^{\ast \ast }=A^{\ast \ast }{T}^{\ast \ast }{B}^{\ast \ast }$ proves the ideal property. Thus the weakly compact operators form a norm-closed operator ideal.

The Krein-Smulian theorem states that the norm-closed convex hull $\overline{\mathrm{co}}K$ 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 $C(K{)}^{\ast }$ by the Banach-Alaoglu theorem. The barycentre mapis 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 $\mathrm{co}K$. The image of all probability measures is therefore the weak closure of $\mathrm{co}K$, 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. Thuswhenever 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 formuladefines a bounded operator $T:{\ell }^1\to Y$. Since $T{e}_n=y_n$, 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

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