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A bounded linear operator $T:X\to Y$ between Banach spaces is a weakly compact operator exactly when its second adjoint takes values in the canonical copy of $Y$ inside $Y^{\ast \ast }$. The Goldstine theorem proves necessity; the Banach-Alaoglu theorem, continuity of the second adjoint, and the Mazur theorem prove sufficiency.

Use the canonical embedding into the bidual $J_{X}x(f)=f(x)$. It is an isometry by the Hahn-Banach theorem. The weak-star topology $\sigma (X^{\ast \ast },X^{\ast })$, restricted to $J_{X}X$, is exactly the weak topology $\sigma (X,X^{\ast })$: both are determined by the same scalar evaluations $f(x)$. Therefore if $K\subseteq X$ is a weakly compact set, $J_{X}K$ is compact in the Hausdorff weak-star topology of $X^{\ast \ast }$, and hence is closed for the weak-star topology. This explains the first assertion without treating the whole embedded space $J_{X}X$ as weak-star closed.

For a bounded linear operator $T:X\to Y$, its Banach-space adjoint isFor each $x\in X$, the scalar evaluation $y^{\ast }\mapsto (T^{\ast }{y}^{\ast })(x)=y^{\ast }(Tx)$ is weak-star continuous on $Y^{\ast }$. These are exactly the defining evaluations of the weak-star topology on $X^{\ast }$, so $T^{\ast }$ is weak-star-to-weak-star continuous. Equivalently, if $y_{\alpha }^{\ast }\stackrel{w^{\ast }}{\to }{y}^{\ast }$, then $(T^{\ast }{y}_{\alpha }^{\ast })(x)\to (T^{\ast }{y}^{\ast })(x)$ for every $x$.

The Goldstine theorem states that $J_X(B_X)$ is dense in $B_{X^{\ast \ast }}$ for $\sigma (X^{\ast \ast },X^{\ast })$, where the balls are closed unit balls. The Banach-Alaoglu theorem states that for any normed vector space $E$, the closed dual unit ball $B_{E^{\ast }}$ is compact for $\sigma (E^{\ast },E)$. Neither assertion claims norm compactness or, in general, sequential compactness.

For the weakly compact operator equivalences below, identify $X$ and $Y$ with their canonical images only when stated. In particular, the range condition means $T^{\ast \ast }(X^{\ast \ast })\subseteq J_{Y}Y$. We shall prove $(i)⟺(ii)$, $(ii)⟹(iii)⟹(iv)⟹(ii)$. The key identities, obtained by evaluating on $y^{\ast }\in Y^{\ast }$, are

Weak-star-to-weak continuity of the adjoint implies its weak compactness. By the Banach-Alaoglu theorem, $B_{Y^{\ast }}$ is compact for the weak-star topology. Under the assumed continuous map $T^{\ast }:(Y^{\ast },w^{\ast })\to (X^{\ast },w)$, its image $T^{\ast }(B_{Y^{\ast }})$ is weakly compact in $X^{\ast }$.

This image is closed for the weak topology, and hence for the norm topology, because the weak topology is Hausdorff and weaker than the norm topology. Thus its norm closure is the same weakly compact set. By the definition of a weakly compact operator, $T^{\ast }$ is weakly compact, proving $(iii)⟹(iv)$.

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-Šmulian 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-Šmulian 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