Proof of Banach–Alaoglu Theorem for a Separable Normed Space: the Closed Balls of the Dual Are Weak-Star Sequentially Compact and Compact for a Metric of Weak-Star Convergence

Throughout, $U=\{x\in X:\|x\|\le1\}$ is the closed unit ball of $X$, and every $F\in B^{*}_{M}$ satisfies

$$|F(x)|\le\|F\|_{*}\,\|x\|\le M\|x\|\qquad(x\in X).$$

The proofs of (i) and (ii) use no choice principle; the proof of (iii) uses the axiom of dependent choice, a standing hypothesis by Radon Measures on Locally Compact Metric Spaces: Standing Notation (iv), both directly and through the compactness lemma it cites.

Proof of (i): such a sequence exists. The proof of (i) needs a sequence $(x_{k})$ in $U$ whose range is dense in $U$; we construct one. By Separable Metric Space and Separable Real Normed Space (ii) and Separable Metric Space and Separable Real Normed Space (i) there is a countable subset $D\subseteq X$ that is dense in $(X,d_{X})$, $d_{X}$ the norm metric. It is nonempty, since the open ball $B(0,1)$ meets it, so by Countable Set (i) there is a surjective function $\mathbb{N}\to D$, $k\mapsto y_{k}$, whose range $D$ is dense in $(X,d_{X})$. Let

$$x_{k}=\frac{1}{\max\{1,\|y_{k}\|\}}\,y_{k}\qquad(k\in\mathbb{N}).$$

Then $\|x_{k}\|=\|y_{k}\|/\max\{1,\|y_{k}\|\}\le1$, so $x_{k}\in U$.

The range of $(x_{k})$ is dense in $U$. Let $x\in U$ and $\delta>0$; we find $k$ with $\|x_{k}-x\|<\delta$. By density choose $k$ with $\|y_{k}-x\|<\delta/2$. If $\|y_{k}\|\le1$, then $x_{k}=y_{k}$. Otherwise $x_{k}=y_{k}/\|y_{k}\|$ and

$$\|y_{k}-x_{k}\|=\Bigl(1-\frac{1}{\|y_{k}\|}\Bigr)\|y_{k}\|=\|y_{k}\|-1\le\|y_{k}\|-\|x\|\le\|y_{k}-x\|<\frac{\delta}{2}.$$

In both cases $\|x_{k}-x\|\le\|x_{k}-y_{k}\|+\|y_{k}-x\|<\delta$. So $(x_{k})$ has the properties the statement requires of the sequence in (ii); the proof of (i) uses this sequence, and the proof of (ii) the one of the statement.

An approximation estimate. Let $(x_{k})$ be a sequence in $U$ whose range is dense in $U$. For $F,G\in B^{*}_{M}$, $x\in U$ and $k\in\mathbb{N}$, linearity gives $F(x)-G(x)=\bigl(F(x_{k})-G(x_{k})\bigr)+F(x-x_{k})-G(x-x_{k})$, and hence

$$|F(x)-G(x)|\le|F(x_{k})-G(x_{k})|+2M\|x-x_{k}\|.\tag{1}$$

Moreover, for $x\in X$ with $x\neq0$ the vector $x/\|x\|$ lies in $U$ and $F(x)=\|x\|\,F(x/\|x\|)$; so a statement about the values at the points of $U$ that is preserved under multiplication by a constant passes to all of $X$.

Proof of (i): the diagonal subsequence. Let $(x_{k})$ be the sequence constructed above, let $(F_{m})$ be a sequence in $B^{*}_{M}$, and put $a_{m,k}=F_{m}(x_{k})$. Since $|a_{m,k}|\le M\|x_{k}\|\le M$, every sequence $(a_{m,k})_{m\in\mathbb{N}}$ is bounded, and The Diagonal Subsequence Lemma for Bounded Real Arrays (i) gives a strictly increasing sequence $(m_{j})_{j\in\mathbb{N}}$ in $\mathbb{N}$ such that $(F_{m_{j}}(x_{k}))_{j\in\mathbb{N}}$ converges for every $k$. Write $G_{j}=F_{m_{j}}$; then $(G_{j})$ is a subsequence of $(F_{m})$, and each $(G_{j}(x_{k}))_{j}$ is a Cauchy sequence by Completeness of the Real Numbers for Sequences: Monotone Convergence, the Bolzano-Weierstrass Theorem and Cauchy Sequences (v).

Convergence at every point. Let $x\in U$ and $\varepsilon>0$. First choose $k$ with $\|x-x_{k}\|<\varepsilon/(4M+1)$, so that $2M\|x-x_{k}\|<\varepsilon/2$; then, with this $k$ fixed, choose $N\in\mathbb{N}$ with $|G_{j}(x_{k})-G_{l}(x_{k})|<\varepsilon/2$ for all $j,l\ge N$. By (1), $|G_{j}(x)-G_{l}(x)|<\varepsilon$ for all $j,l\ge N$. Thus $(G_{j}(x))_{j}$ is a Cauchy sequence, and so convergent by Completeness of the Real Numbers for Sequences: Monotone Convergence, the Bolzano-Weierstrass Theorem and Cauchy Sequences (v), for every $x\in U$; for $x\notin U$ the sequence $G_{j}(x)=\|x\|\,G_{j}(x/\|x\|)$ converges as well, by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (vi).

The limit functional. Define $F:X\to\mathbb{R}$ by $F(x)=\lim_{j\to\infty}G_{j}(x)$. Passing to the limit in $G_{j}(u+v)=G_{j}(u)+G_{j}(v)$ and $G_{j}(\lambda u)=\lambda G_{j}(u)$ shows, by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (vi) and Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (ii), that $F$ is a linear functional on $X$; passing to the limit in $|G_{j}(x)|\le M\|x\|$ gives $|F(x)|\le M\|x\|$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (viii) and Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (x). So $F\in X^{*}$ and, $\|F\|_{*}$ being the infimum of the bounds of $F$ by The Dual Norm (i), $\|F\|_{*}\le M$. Hence $F\in B^{*}_{M}$, and $G_{j}(x)\to F(x)$ for every $x\in X$, which is $G_{j}\overset{*}{\rightharpoonup}F$ by Weak-Star Convergence in the Dual of a Real Normed Space (i).

Proof of (ii): $d$ is a metric. From here on, $(x_{k})$ and $d$ are as in the statement. Let $F,G,H\in B^{*}_{M}$. For every $k$, $|F(x_{k})-G(x_{k})|\le2M\|x_{k}\|\le2M$ and $2^{-k}\le1/2$, so the numbers $2^{-k}|F(x_{k})-G(x_{k})|$ form a nonempty subset of $[0,M]$; its supremum $d(F,G)$ exists and lies in $[0,M]$. Clearly $d(F,G)=d(G,F)$ and $d(F,F)=0$. For every $k$,

$$2^{-k}|F(x_{k})-H(x_{k})|\le2^{-k}|F(x_{k})-G(x_{k})|+2^{-k}|G(x_{k})-H(x_{k})|\le d(F,G)+d(G,H),$$

and taking the supremum over $k$ gives the triangle inequality. Finally suppose $d(F,G)=0$, so that $F(x_{k})=G(x_{k})$ for every $k$. For $x\in U$ and $\varepsilon>0$ choose $k$ with $2M\|x-x_{k}\|<\varepsilon$; then (1) gives $|F(x)-G(x)|<\varepsilon$. Hence $F(x)=G(x)$ for $x\in U$, and by homogeneity for every $x\in X$, so $F=G$. By Metric Space (i), $d$ is a metric on $B^{*}_{M}$.

Convergence for $d$ implies weak-star convergence. Let $(G_{m})$ be a sequence in $B^{*}_{M}$ and $G\in B^{*}_{M}$, and suppose $G_{m}\to G$ in $(B^{*}_{M},d)$. For every $k$, $|G_{m}(x_{k})-G(x_{k})|\le2^{k}d(G_{m},G)$, so $G_{m}(x_{k})\to G(x_{k})$. Let $x\in U$ and $\varepsilon>0$. First choose $k$ with $2M\|x-x_{k}\|<\varepsilon/2$; then, with $k$ fixed, choose $N$ with $|G_{m}(x_{k})-G(x_{k})|<\varepsilon/2$ for every $m\ge N$. By (1), $|G_{m}(x)-G(x)|<\varepsilon$ for every $m\ge N$. So $G_{m}(x)\to G(x)$ for every $x\in U$, hence, by homogeneity and Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (vi), for every $x\in X$: $G_{m}\overset{*}{\rightharpoonup}G$.

Weak-star convergence implies convergence for $d$. Conversely, suppose $G_{m}\overset{*}{\rightharpoonup}G$, and let $\varepsilon>0$. First choose $K\in\mathbb{N}$ with $2^{-K}(2M+1)<\varepsilon/2$, which is possible since $(1/2)^{n}\to0$ by Completeness of the Real Numbers for Sequences: Monotone Convergence, the Bolzano-Weierstrass Theorem and Cauchy Sequences (ii). Then, with $K$ fixed, choose for each $k\le K$ an index $N_{k}$ with $|G_{m}(x_{k})-G(x_{k})|<\varepsilon/2$ for every $m\ge N_{k}$, and let $N=\max\{N_{1},\dots,N_{K}\}$. For $m\ge N$ and every $k$: if $k\le K$, then $2^{-k}|G_{m}(x_{k})-G(x_{k})|<\varepsilon/2$; if $k>K$, then $2^{-k}|G_{m}(x_{k})-G(x_{k})|\le2^{-K}\cdot2M<\varepsilon/2$. Hence $d(G_{m},G)\le\varepsilon/2<\varepsilon$ for every $m\ge N$, that is, $G_{m}\to G$ in $(B^{*}_{M},d)$ by Convergent Sequence in a Metric Space (i).

Proof of (iii): sequential compactness. Every sequence in $B^{*}_{M}$ has, by (i), a subsequence converging weak-star to some $G\in B^{*}_{M}$, and by (ii) this subsequence converges to $G$ in $(B^{*}_{M},d)$. So $B^{*}_{M}$ is sequentially compact in $(B^{*}_{M},d)$.

Completeness. Let $(G_{m})$ be a Cauchy sequence in $(B^{*}_{M},d)$. By sequential compactness it has a subsequence $(G_{\sigma(j)})_{j}$ converging to some $G\in B^{*}_{M}$. Given $\varepsilon>0$, choose $N$ with $d(G_{m},G_{l})<\varepsilon/2$ for all $m,l\ge N$, and then $j\ge N$ with $d(G_{\sigma(j)},G)<\varepsilon/2$; since $\sigma$ is strictly increasing, $\sigma(j)\ge j\ge N$. For every $m\ge N$,

$$d(G_{m},G)\le d(G_{m},G_{\sigma(j)})+d(G_{\sigma(j)},G)<\varepsilon .$$

So $(G_{m})$ converges to $G$, and $(B^{*}_{M},d)$ is complete.

Total boundedness. Suppose that $B^{*}_{M}$ is not totally bounded in $(B^{*}_{M},d)$. Then there is $\varepsilon>0$ such that for every finite subset $S\subseteq B^{*}_{M}$ some $G\in B^{*}_{M}$ satisfies $d(G,H)\ge\varepsilon$ for every $H\in S$ (for $S=\emptyset$, any $G$, such as $0\in B^{*}_{M}$). Applying dependent choice to the relation that extends a finite sequence $(G_{1},\dots,G_{n})$ in $B^{*}_{M}$ by a $G_{n+1}\in B^{*}_{M}$ with $d(G_{n+1},G_{l})\ge\varepsilon$ for every $l\le n$, we obtain a sequence $(G_{m})$ in $B^{*}_{M}$ with $d(G_{m},G_{l})\ge\varepsilon$ whenever $l<m$. No subsequence of $(G_{m})$ converges in $(B^{*}_{M},d)$: if $G_{\sigma(j)}\to G$, then $d(G_{\sigma(j)},G_{\sigma(i)})\le d(G_{\sigma(j)},G)+d(G,G_{\sigma(i)})<\varepsilon$ for all large $i<j$, a contradiction. This contradicts sequential compactness, so $B^{*}_{M}$ is totally bounded in $(B^{*}_{M},d)$.

Compactness. By The Open Subsets of a Metric Space Form a Topology (i), $\mathcal{T}_{d}$ is a topology on $B^{*}_{M}$, so $\emptyset=B^{*}_{M}\setminus B^{*}_{M}$ lies in $\mathcal{T}_{d}$ and $B^{*}_{M}$ is closed in $(B^{*}_{M},\mathcal{T}_{d})$. Thus $B^{*}_{M}$ is a closed, totally bounded subset of the complete metric space $(B^{*}_{M},d)$, and A Closed Totally Bounded Subset of a Complete Metric Space is Compact (ii) shows that it is compact in $(B^{*}_{M},\mathcal{T}_{d})$.