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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

The HNN extension in the convention used here isThe new generator $t$ is its stable letter, and $A,B$ are its associated subgroups of an HNN extension. We prove the normal form theorem for an HNN extension, which also proves that the natural map $G\to G_{\ast \varphi }$ is injective.

Set $C_{+}=A$, $C_{-}=B$, ${\psi }_{+}=\varphi$, and ${\psi }_{-}={\varphi }^{-1}$. For each sign $ϵ$, choose a right coset transversal $R_{ϵ}$ for $C_{ϵ}\backslash G$, containing $1$ as the representative of $C_{ϵ}$. Thus every $g\in G$ has a unique decomposition $g=cr$, with $c\in C_{ϵ}$ and $r\in R_{ϵ}$. A normal form iswhere $g_0\in G$ is unrestricted and the case $k=0$ is just $g_0$.

Here is a constructive proof. Let $\mathcal{N}$ be the set of these formal sequences. Left multiplication by $h\in G$ changes only $g_0$ to $h{g}_0$. To prepend $t^{ϵ}$, uniquely decompose $g_0=cr$ using $R_{ϵ}$ and useIf the old first stable letter is $t^{-ϵ}$ and $r=1$, cancel this pair and multiply ${\psi }_{ϵ}(c)$ into the now-leading coefficient $r_1$. Otherwise retain the new first stable letter. Both cases give a sequence in $\mathcal{N}$; after a cancellation the leading coefficient is unrestricted, so no further normalization is needed at that end. Prepending the letters of a word from right to left proves existence.

These operations define permutations of $\mathcal{N}$. The base-group operations satisfy $L_h{L}_{h^{\prime }}=L_{h{h}^{\prime }}$. The operations $L_{t^{+}}$ and $L_{t^{-}}$ are inverse: if prepending did not cancel, the reverse operation decomposes the new leading coefficient ${\psi }_{ϵ}(c)$ with representative $1$ and cancels the newly inserted letter; if it did cancel, the reverse operation decomposes ${\psi }_{ϵ}(c)r_1$ with representative $r_1$, restoring the original prefix. The normal-form restriction rules out an unwanted second cancellation in the latter case.

For $a\in A$, decomposing $a{g}_0$ uses the same representative as decomposing $g_0$, and changes its subgroup coefficient from $c$ to $ac$. Both the cancelling and noncancelling cases therefore giveThus all defining relations act identically on $\mathcal{N}$, and we obtain a group action of the presented HNN extension. Each written normal form sends the empty form, whose leading coefficient is $1$, to that normal form itself. Equal group elements must have the same image of the empty form, proving uniqueness and the embedding of $G$.

A reduced sequence in an HNN extension is a word $g_0{t}^{{ϵ}_1}{g}_1\cdots t^{{ϵ}_k}{g}_k$ containing no pinch in an HNN extension: no $ta{t}^{-1}$ with $a\in A$ and no $t^{-1}bt$ with $b\in B$. Equivalently, whenever ${ϵ}_{i+1}=-{ϵ}_i$, one requires $g_i\notin C_{{ϵ}_i}$.

Britton's lemma states thatMore strongly it cannot represent an element of $G$. To prove this from the normal form theorem for an HNN extension, normalize the coefficients from right to left. Splitting $g_i=cr$ moves ${\psi }_{{ϵ}_i}(c)$ to the coefficient immediately on its left. If the preceding stable letter has the opposite sign, this transported factor belongs to the subgroup relevant to that inverse pair; multiplying by it cannot turn a coefficient outside that subgroup into one inside it. If the signs agree, cancellation is impossible anyway. Hence no stable letter disappears during normalization of a reduced sequence. Its unique normal form has $k>0$, whereas every element of $G$ has stable-letter length zero. This proves Britton's lemma. The same proof works with finitely many stable letters, with pinches requiring the same letter and its inverse.

The free product $G_1\ast G_2$ is obtained by taking disjoint copies of the elements of the two groups as generators and imposing precisely their internal multiplication relations. Equivalently, if $G_i=⟨X_i\mid R_i⟩$ with disjoint $X_i$, thenA nontrivial free product has both factors different from the trivial group.

The normal form theorem for a free product says that every element has a unique expressionwhere the empty expression represents the identity. The $g_j$ are called syllables in a free product. Thus nonempty alternating products cannot be the identity, and the natural maps of the factors into the free product are injective.

For existence, multiply adjacent syllables from the same factor and delete identity syllables until the word alternates. Each change uses a defining relation and decreases the number of syllables. To establish uniqueness without assuming that these reductions are confluent, let $\mathcal{W}$ be the set of alternating nonidentity syllable sequences, including the empty sequence. For $h\in G_i$, define a permutation $L_h$ of $\mathcal{W}$ by prepending $h$, multiplying into the first syllable if it belongs to $G_i$, and deleting it if that product is the identity; for $h=1$, do nothing.

The multiplication law inside $G_i$ gives $L_h{L}_k=L_{hk}$ for $h,k\in G_i$, and $L_h^{-1}=L_{h^{-1}}$. This can be checked at the first syllable: if a product deletes it, the next syllable belongs to the other factor and is treated as a new first syllable. Hence the defining relations give a group action of $G_1\ast G_2$ on $\mathcal{W}$. An alternating word $g_1\cdots g_r$ sends the empty sequence to its own syllable sequence. Two such words representing the same group element induce the same permutation, and therefore have identical sequences. This proves the normal form theorem for a free product.

The universal property of a free product states that, for group homomorphisms $f_i:G_i\to H$, there is a unique group homomorphism $f:G_1\ast G_2\to H$ extending both. Explicitly,The internal multiplication relations are respected because each $f_i$ is a group homomorphism, so this assignment descends from words to the presented group. Alternatively, multiplication of two normal forms is concatenation followed by precisely those internal relations, which do not change its value in $H$. Uniqueness holds because the two factors generate the free product.