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A nontangential limit at $\xi \in \partial \mathbb{D}$ restricts approach to a fixed Stolz region ${\Gamma }_A(\xi )=\{z\in \mathbb{D}:|z-\xi |<A(1-|z|)\}$, where $A>1$. Radial approach is a special case. The Fatou theorem for bounded holomorphic functions says that every $f$ in H-infinity space has a finite boundary value $f^{\ast }(\xi )$ for almost every $\xi$, with convergence through every fixed Stolz region and $|f^{\ast }|\le \Vert f{\Vert }_{\infty }$ almost everywhere.

Here is a proof outline. Write $f=u+iv$ with bounded real harmonic functions $u,v$. Take a sequence $R_j\uparrow 1$. The bounded functions $u(R_j\xi )$ have a subsequence converging in the weak-star topology of $L^{\infty }(\partial \mathbb{D})$ by the Banach-Alaoglu theorem and weak-star metrizability of the dual ball; its $L^1$ predual is separable. Denote the limit by $U$. For $|z|<R_j$, the Poisson integral on the unit disk givesThe Poisson kernels converge uniformly for this fixed interior point and the boundary functions stay uniformly bounded, so the weak-star topology permits passage to the limit:Repeat for $v$, obtaining $V$. Thus $f$ is the Poisson integral of $F=U+iV\in L^{\infty }$.

The Lebesgue differentiation theorem makes almost every boundary point a Lebesgue point of $U$ and $V$. The lemma that a Poisson integral converges nontangentially at Lebesgue points then gives $f(z)\to F({\xi }_0)$ in every fixed Stolz region. To see why, put $\delta =1-|z|$ and let $d(\xi ,{\xi }_0)$ be the shorter angular distance. In ${\Gamma }_A({\xi }_0)$ the kernel satisfiesSplit the circle into a central arc of radius comparable to $\delta$ and annuli with radii $2^k\delta$. For a Lebesgue point, averages of $|F-F({\xi }_0)|$ on sufficiently small arcs are arbitrarily small. The annular contributions are bounded by these averages times a summable geometric sequence $C_A{2}^{-k}$. On the remaining fixed distant arc the kernel is $O(\delta )$, so its contribution vanishes. This proves the claimed nontangential limit. The same Lebesgue points work for every integer aperture $A\ge 2$, hence for all finite apertures.

These boundary values recover the interior by the Poisson integral, so they determine $f$ uniquely. They satisfy $\Vert f^{\ast }{\Vert }_{L^{\infty }}=\Vert f{\Vert }_{\infty }$: one inequality follows by taking limits of the bounded interior values, and the other from positivity and unit integral of the Poisson kernel. The dominated convergence theorem also gives $f(r\xi )\to f^{\ast }(\xi )$ in every finite $L^p$ norm. Its negative Fourier coefficients vanish, because the Taylor series of $f$ has only nonnegative powers. This explains the boundary realization of analytic Hardy spaces.

The exceptional set cannot generally be removed. For example, $f(z)=\exp (i\mathrm{Log}(1-z))$ is bounded and holomorphic on the unit disc, using the holomorphic logarithm on the right half-plane: $|f(z)|=e^{-\arg (1-z)}$ lies between $e^{-\pi /2}$ and $e^{\pi /2}$. At the boundary point one its radial values $e^{i\log (1-r)}$ oscillate without a limit. Nor does the theorem assert convergence along arbitrary tangential paths or a continuous boundary extension. Its strength is almost-everywhere control on every fixed approach cone.

True. Choose a norm-dense sequence $(x_n)$ in the unit ball of the separable normed vector space $E$. On the dual unit ball $B^{\prime }$, defineEach summand is at most $2^{1-n}$, so the series converges uniformly. Symmetry and the triangle inequality follow term by term. If $d(\varphi ,\psi )=0$, the two continuous functionals agree on the dense sequence, hence on the unit ball and then on all of $E$. Thus $d$ is a metric.

A relative weak-star neighbourhood controls finitely many evaluations. Conversely, to make $d$ small, first choose a tail whose sum is small and then control the finitely many evaluations at $x_1,\dots ,x_N$. This proves that each metric ball is weak-star open. For the other direction, let $x\in E$ and $ϵ>0$. Approximate $x/\Vert x\Vert$ by $x_n$ when $x\ne 0$. Since $\Vert \varphi -\psi \Vert \le 2$ on $B^{\prime }$,The second term can be made smaller than $ϵ/2$ by choosing $n$; the first is smaller than $ϵ/2$ whenever $d(\varphi ,\psi )$ is sufficiently small, because $|\varphi (x_n)-\psi (x_n)|\le 2^{n}d(\varphi ,\psi )$. Every evaluation is therefore metric-continuous. These two inclusions show that $d$ induces exactly the weak-star topology, proving weak-star metrizability of the dual ball without a completeness assumption on $E$.

True. Choose a countable dense sequence $(x_j)$ in the separable Banach space $E$. For $\varphi ,\psi$ in its closed dual unit ball, putThis is a metric: positivity follows because continuous functionals agreeing on a dense subset agree everywhere; symmetry is immediate; and the triangle inequality follows from subadditivity of $t/(1+t)$ for $t\ge 0$. The tail of the series is uniformly small, so convergence on these coordinates implies convergence for this metric, and conversely metric convergence controls each individual coordinate.

The uniform bound $\Vert \varphi -\psi \Vert \le 2$ upgrades coordinate convergence to pointwise convergence on all of $E$. For any $x\in E$, choose $x_j$ close to $x$ and useFor neighbourhoods involving finitely many $x$, choose finitely many such approximations. This proves equality of the induced topologies, including for arbitrary nets. Thus the weak-star metrizability of the dual ball gives

For a convex subset of a Banach space, weak closure equals norm closure. The Mazur theorem states that for every convex set $C\subseteq X$,Since every bounded linear functional is norm-continuous, the weak topology is weaker than the norm topology, giving ${\overline{C}}^{\Vert \cdot \Vert }\subseteq {\overline{C}}^w$.

For the reverse inclusion, use this form of the Hahn-Banach separation theorem: if $D$ is a nonempty norm-closed convex set in a real normed vector space and $x\notin D$, there is a continuous real-linear functional $F$ such thatApply it to $D={\overline{C}}^{\Vert \cdot \Vert }$. In a complex Banach space, use the underlying real space; every continuous real-linear $F$ is the real part of the complex bounded linear functional $f(v)=F(v)-iF(iv)$. The separating strict inequality gives a neighbourhood open for the weak topology of $x$ disjoint from $D$, so $x\notin {\overline{C}}^w$. The empty set case is immediate. This proves the Mazur theorem.

A weakly null sequence has disjoint convex blocks converging to zero in norm. If $x_n\stackrel{w}{\to }0$, then for every starting index $m$,The Mazur theorem places $0$ in the norm closure of that tail convex hull. Thus a finite convex combination of vectors from any tail can have norm less than any prescribed $\epsilon >0$.

Choose the convex blocks recursively. After selecting the previous terminal index $q_{n-1}$, let $p_n>q_{n-1}$, and choose a finite convex combination from $\{x_i:i\ge p_n\}$ with norm less than $2^{-n}$. Choose $q_n$ beyond its largest used index, padding the intervening coefficients and the final coefficient with zero. This ensures the printed strict condition $p_n<q_n$ as well as $q_n<p_{n+1}$. Setting unused coefficients to zero yieldsTherefore $u_n\to 0$ in norm. The recursive tail selection is what makes these convex blocks, rather than merely unrelated convex combinations.

Next suppose $X^{\ast }$ is separable and $(x_n)$ is bounded. Use the canonical embedding into the bidual $J_X:X\to X^{\ast \ast }$, where $J_{X}x(f)=f(x)$. If $\Vert x_n\Vert \le M$, then $J_X{x}_n\in M{B}_{X^{\ast \ast }}$. The Banach-Alaoglu theorem makes this ball compact in $\sigma (X^{\ast \ast },X^{\ast })$, and weak-star metrizability of the dual ball makes it metrizable because its predual $X^{\ast }$ is separable. A compact space that is a metric space is sequentially compact, so some subsequence $(y_n)$ satisfiesThe limit need not belong to $J_{X}X$; it is precisely the use of the bidual space that supplies compactness without assuming reflexivity.

For a convex block $u_n={\sum }_{i=p_n}^{q_n}{a}_i{y}_i$, every $f\in X^{\ast }$ satisfiesBoth $f(y_n)$ and $f(u_n)$ tend to $\varphi (f)$, soThis is the useful convex-block cancellation of a weak-star limit; it does not require the limit to be a vector in $X$.

The quotient sequence has approximate lifts in $3B_X$ that are weakly null after passing to a subsequence. Let $q:X\to Z=X/Y$ be the quotient map, with the quotient norm. Since $\Vert z_n\Vert \le 1$, choose $v_n\in X$ such thatThis uses the infimum defining the quotient norm and does not assume that it is attained. By the preceding compactness argument, pass to a subsequence $y_n=v_{k_n}$ with a common weak-star limit in $X^{\ast \ast }$. Its quotient images $q(y_n)=z_{k_n}$ form a weakly null sequence.

Apply the convex block construction to $(q(y_n))$ in the quotient Banach space. Using the same coefficients on $(y_n)$ gives convex blocksSet $x_n=y_n-u_n$. The convex-block cancellation of a weak-star limit proves $x_n\stackrel{w}{\to }0$, whileThus the approximate weakly null lifting through a quotient is achieved with the required bound:

The weak-star topology $\sigma (X^{\ast },X)$ is the coarsest topology making $f\mapsto f(x)$ continuous for every $x\in X$. Its basic neighbourhoods of zero impose finitely many inequalities $|f(x_j)|<ϵ$.

Suppose first that zero had a countable neighbourhood base $(U_n)$ in the entire dual. Choose a basic neighbourhood $V_n\subset U_n$ controlled by a finite set $F_n\subset X$, and let $E_n=\mathrm{span}(F_1\cup \cdots \cup F_n)$. Given $x\in X$, the neighbourhood $W_x=\{f:|f(x)|<1\}$ contains some $U_n$, hence $V_n$. If $x\notin \mathrm{span}{F}_n$, the Hahn-Banach theorem supplies a bounded linear functional $h$ which vanishes on $F_n$ but has $h(x)\ne 0$. Indeed, the finite-dimensional span is closed, and the functional taking $e+tx$ to $t$ on its sum with $\mathbb{F}x$ is bounded because $\mathrm{dist}(x,\mathrm{span}{F}_n)>0$. Every scalar multiple of $h$ lies in $V_n$, contradicting $V_n\subset W_x$. Thus $x\in E_n$, and $X={\bigcup }_n{E}_n$ has countable Hamel dimension.

If $X$ is infinite-dimensional, every finite-dimensional $E_n$ is closed and has empty interior. Completeness rules out their union. For clarity, the Baire category theorem needed here has a direct nested-ball proof: inside a starting open ball choose a closed ball missing $E_1$, then inside its interior choose a closed ball missing $E_2$, and continue with positive radii tending to zero. The centres are Cauchy; their limit belongs to every ball and to none of the $E_n$, a contradiction. A metric topology would have a countable ball base at zero. HenceThis is the weak-star topology on an entire infinite-dimensional Banach dual is not metrizable result. It does not contradict weak-star metrizability of the dual ball for a separable predual: that assertion concerns a bounded subset.

The Banach-Alaoglu theorem states that $B_{X^{\ast }}$ is compact for $\sigma (X^{\ast },X)$ for every normed vector space $X$, whether complete or not. Embed it in the product ${\prod }_{x\in X}\{a\in \mathbb{F}:|a|\le \Vert x\Vert \}$ by its evaluations. Each factor is compact, so Tychonoff theorem makes the product compact Hausdorff. The equations expressing additivity and scalar homogeneity define a closed subset of the product. Its elements are exactly the bounded linear functionals of norm at most one, since the coordinate bounds give $|f(x)|\le \Vert x\Vert$. The induced product topology is precisely weak-star, proving the theorem.

The theorem printed as “Goldstein” is the standard Goldstine theorem: $J(B_X)$ is weak-star dense in $B_{X^{\ast \ast }}$ for the canonical embedding into the bidual. To prove it, fix $\Phi \in B_{X^{\ast \ast }}$ and $f_1,\dots ,f_m\in X^{\ast }$. The set $S=\{(f_1(x),\dots ,f_m(x)):x\in B_X\}$ is convex. If $b=(\Phi (f_1),\dots ,\Phi (f_m))$ were outside its closure, finite-dimensional Hahn-Banach separation theorem would supply coefficients $a_j$ such that, for $f={\sum }_j{a}_j{f}_j$,The real case omits real parts; in the complex case every real-linear separating functional on ${\mathbb{C}}^m$ has the displayed form. The supremum equals the dual norm by multiplying vectors by signs or unimodular scalars. The strict inequality contradicts $\Vert \Phi \Vert \le 1$. Thus $b\in \overline{S}$, which is exactly approximation on every prescribed finite family of weak-star coordinates.

Now take $K=B_{X^{\ast }}$ with its compact Hausdorff weak-star topology. For each $x\in X$, the evaluation $\hat{x}:f\mapsto f(x)$ is continuous on $K$, and Hahn-Banach theorem gives $\Vert \hat{x}{\Vert }_{\infty }=\Vert x\Vert$. Thereforeis the isometric evaluation embedding into continuous functions on a compact dual ball, over the same scalar field as $X$.

For the final sequential assertions, choose a norm-dense sequence $(f_j)$ in the separable dual $X^{\ast }$ and suppose $\Vert x_n\Vert \le M$. Each bounded scalar sequence $f_j(x_n)$ has a convergent subsequence: divide a bounding interval or square into finitely many smaller closed pieces, repeatedly retain one containing infinitely many terms, and let the diameters tend to zero. Successive extraction for $j=1,2,\dots$, followed by a diagonal subsequence $(y_n)$, makes every $f_j(y_n)$ converge.

If $M=0$ the conclusion is immediate. Otherwise, for any $f\in X^{\ast }$ and $ϵ>0$, choose $j$ with $\Vert f-f_j\Vert <ϵ/(4M)$. For sufficiently large $n,m$,Thus $\phi (f)={\lim }_{n}f(y_n)$ exists for every $f$. Taking limits proves linearity, and $|\phi (f)|\le M\Vert f\Vert$ provesThis diagonal argument includes the continuity estimate rather than appealing to a sequential compactness theorem.

If $X$ is not reflexive, choose $\Phi \in B_{X^{\ast \ast }}\setminus JX$, scaling a bidual element outside $JX$ if necessary. The Goldstine theorem proved above supplies $x_n\in B_X$ with $|f_j(x_n)-\Phi (f_j)|<1/n$ for $j\le n$. Norm density and the bound $\Vert J{x}_n-\Phi \Vert \le 2$ extend this convergence to every $f\in X^{\ast }$. Hence $J{x}_n\to \Phi$ weak-star. Every subsequence has the same coordinate limits, so a weakly convergent subsequence with limit $x\in X$ would give $Jx=\Phi$, a contradiction. This sequential Goldstine approximation for a separable dual provesThe construction uses the Goldstine proof already given, not an unproved weak sequential compactness result.

If $X$ is reflexive, then $X^{\ast }$ is reflexive. The weak topology $\sigma (X^{\ast },X^{\ast \ast })$ and weak-star topology $\sigma (X^{\ast },X)$ therefore coincide under the canonical identification $X^{\ast \ast }=X$. Thus every weak-star convergent sequence in $X^{\ast }$ is weakly convergent, so $X$ is a Grothendieck space.

Conversely, suppose that $X$ is separable and Grothendieck. The Banach-Alaoglu theorem and weak-star metrizability of the dual ball make $B_{X^{\ast }}$ weak-star compact and metrizable, hence weak-star sequentially compact. Every convergent subsequence is weakly convergent by the Grothendieck property. Thus $B_{X^{\ast }}$ is weakly sequentially compact. The stated converse to part (a), equivalently the other direction of the Eberlein-Šmulian theorem, makes $B_{X^{\ast }}$ weakly compact. Hence $X^{\ast }$ is reflexive, and therefore so is $X$.

Finally let $T:X\to Y$ be bounded and onto, with $X$ Grothendieck, and suppose $y_n^{\ast }\to y^{\ast }$ weak-star in $Y^{\ast }$. Thenweak-star in $X^{\ast }$, hence weakly. The open mapping theorem implies that $T^{\ast }:Y^{\ast }\to X^{\ast }$ is an isomorphism onto its closed range. Given $y^{\ast \ast }\in Y^{\ast \ast }$, the functionalis bounded on $T^{\ast }(Y^{\ast })$ and extends by the Hahn-Banach theorem to some $x^{\ast \ast }\in X^{\ast \ast }$. ThereforeThis is weak convergence in $Y^{\ast }$, so $Y$ is Grothendieck.

Choose a norm-dense sequence $(f_n)$ in the unit sphere of the separable dual $X^{\ast }$. For every $n$, choose $x_n\in B_X$ with $|f_n(x_n)|>1/2$. If a functional $f\in X^{\ast }$ vanished on the closed linear span of the $x_n$, normalize it and choose $f_n$ with $\Vert f-f_n\Vert <1/4$; then $|f(x_n)|>1/4$, a contradiction. The Hahn-Banach theorem therefore shows that the span of $(x_n)$ is dense, so $X$ is separable.

Choose a norm-dense sequence $(u_n)$ in $B_X$. On $B_{X^{\ast }}$ defineWeak-star convergence implies convergence in this metric. Conversely, metric convergence gives convergence on the dense set $(u_n)$, and the uniform norm bound on the dual ball extends it to every $u\in X$. Thus $d_{\ast }$ metrizes the weak-star topology, proving weak-star metrizability of the dual ball.

Similarly, for a norm-dense sequence $(f_n)$ in $B_{X^{\ast }}$,metrizes the weak topology on $B_X$. Indeed, convergence against the dense functionals extends to every $f\in X^{\ast }$ because $x-y$ remains norm bounded.

The continuous dual of a separable Banach space is separable in its weak-star topology, even when it is not norm separable. Banach-Alaoglu theorem and weak-star metrizability of the dual ball make the dual unit ball a separable compact metric space. The countable union of positive integer multiples of a countable dense subset of that ball is dense in the whole dual. This does not imply global weak-star metrizability.

A countable local base at zero would give countably many finite sets of evaluation vectors. Any other evaluation must lie in the span of one of those finite sets: otherwise a bounded functional vanishing on that set but not on the evaluation vector, supplied by Hahn-Banach theorem, contradicts inclusion of neighbourhoods after scalar rescaling. Thus the predual has countable Hamel dimension, impossible for an infinite-dimensional Banach space by the Baire category theorem. This concerns the entire dual, unlike weak-star metrizability of the dual ball.