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The Hahn-Banach theorem says that if $Y$ is a linear subspace of a real normed space $X$ and $g:Y\to \mathbb{R}$ is bounded and linear, then there is a bounded linear extension $G:X\to \mathbb{R}$ with $\Vert G\Vert =\Vert g\Vert$.

For $x\in X$, the map $\hat{x}(f)=f(x)$ is linear on $X^{\prime }$ andso $\hat{x}\in X^{\prime \prime }$ and $\Vert \hat{x}\Vert \le \Vert x\Vert$. The assignment $Jx=\hat{x}$ is visibly linear. If $x\ne 0$, define $g$ on $\mathrm{span}\{x\}$ byIt has norm one, and Hahn--Banach extends it to an $f\in X^{\prime }$ of norm one with $f(x)=\Vert x\Vert$. ThereforeThus the canonical embedding into the bidual is an isometry and

On $C([0,1])$, point evaluation $\ell (f)=f(0)$ has norm one for the $L^{\infty }$ norm, because a continuous function's essential supremum equals its supremum. Hahn--Banach extends it to $\Lambda \in (L^{\infty }{)}^{\prime }$. If $\Lambda$ were represented by some $g\in L^1$, thenfor every continuous $f$. Choose continuous $0\le f_n\le 1$ with $f_n(0)=1$ and support in $[0,1/n]$. Absolute continuity of the integral makes the right side tend to zero, while the left side is always one, a contradiction. This singular functional on L infinity proves

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.

Part b(vi) and part c show that $\Vert T{\Vert }_1\le 1$ impliesConversely, choose a unit vector $x$ with $\Vert Sx\Vert$ arbitrarily close to $\Vert S\Vert$, put $y=Sx/\Vert Sx\Vert$, and take $T=x\otimes y$. Part c gives $\Vert T{\Vert }_1=1$ andTaking the supremum provesThus $S\mapsto [T\mapsto \mathrm{tr}(ST)]$ is an isometric embedding ${\mathcal{S}}_{\infty }\to {\mathcal{S}}_1^{\ast }$.

Suppose finite-rank operators are dense in ${\mathcal{S}}_1$ and let $L\in {\mathcal{S}}_1^{\ast }$. The sesquilinear formsatisfies $|b(x,y)|\le \Vert L\Vert \Vert x\Vert \Vert y\Vert$. The Riesz representation theorem gives $S\in {\mathcal{S}}_{\infty }$ with $b(x,y)=⟨Sx,y⟩$. Hence $L(x\otimes y)=\mathrm{tr}(S(x\otimes y))$, and linearity gives equality on every finite-rank operator. Density and continuity extend it to every $T\in {\mathcal{S}}_1$, proving surjectivity.

Conversely, if finite-rank operators were not dense, the Hahn-Banach theorem would give a nonzero $L\in {\mathcal{S}}_1^{\ast }$ vanishing on their closure. Surjectivity would represent it by some $S$, but thenfor all $x,y$, forcing $S=0$ and $L=0$, a contradiction. This proves the stated trace duality criterion.

For a real locally convex space $(X,\mathcal{P})$, the continuous dual space $X^{\ast }$ consists of all continuous real-linear maps $X\to \mathbb{R}$. If $g\in Y^{\ast }$ on a subspace $Y$, continuity gives seminorms $p_1,\dots ,p_k\in \mathcal{P}$ and $C>0$ such thatThe right side is a continuous sublinear functional on $X$. The dominated Hahn-Banach theorem extends $g$ to a linear $f:X\to \mathbb{R}$ satisfying the same bound, so $f\in X^{\ast }$.

If $Y$ is closed and $x_0\notin Y$, the Hausdorff locally convex quotient $X/Y$ has a continuous seminorm $p$ with $p(x_0+Y)>0$. On $Y+\mathbb{R}{x}_0$, define $g(y+t{x}_0)=t$ and rescale $p$ so that $|g|\le p$. Hahn--Banach extends it to $f\in X^{\ast }$ with $f{|}_Y=0$ and $f(x_0)=1$.

The separation of a point and an open convex set says that if $C\subseteq X$ is nonempty, open, and convex and $x_0\notin C$, there is $f\in X^{\ast }$ such thatTo prove it, choose $a\in C$, put $U=C-a$, and let $p_U$ be its Minkowski functional. For $z=x_0-a\notin U$, $p_U(z)\ge 1$. Define $h(tz)=t{p}_U(z)$ on $\mathbb{R}z$; then $h\le p_U$. Hahn--Banach extends $h$ to $f\le p_U$. Since $p_U(u)<1$ for $u\in U$, one has $f(c-a)<1\le f(x_0-a)$, proving the claim.

If $K$ is closed and convex and $x_0\notin K$, a Hahn-Banach separation theorem gives $f\in X^{\ast }$ and a real number separating $x_0$ from $K$. The corresponding inverse image of an open interval is a weak neighbourhood of $x_0$ disjoint from $K$, so $K$ is closed in $\sigma (X,X^{\ast })$.

The unit sphere $S_X$ of a normed space is norm closed. If $X$ is infinite-dimensional, every basic weak neighbourhood of an interior point $x\in B_X$ constrains only finitely many functionals $f_1,\dots ,f_n$. Their common kernel contains a nonzero $y$. Continuity of $t\mapsto \Vert x+ty\Vert$, together with its value below one at $t=0$ and divergence as $|t|\to \infty$, supplies $t$ with $\Vert x+ty\Vert =1$. Since all $f_i$ have the same values at $x+ty$ and $x$, every weak neighbourhood of $x$ meets $S_X$. Points outside $B_X$ are separated from it by Hahn--Banach, so the weak closure of the unit sphere is exactly $B_X$. In particular, $S_X$ is not weakly closed.

For $x$ in a unital complex algebra $A$, the spectrum of an element isFor a nonunital algebra one uses its unitization.

Now let $A$ be a Banach algebra. The invertible group is open, so the resolvent set is open and the spectrum is closed. If $|\lambda |>\Vert x\Vert$, the Neumann seriesconverges, so ${\sigma }_A(x)$ is contained in the closed disc of radius $\Vert x\Vert$ and is therefore compact.

If the spectrum were empty, $R(\lambda )=(\lambda 1-x{)}^{-1}$ would be an entire $A$-valued function. For each $f\in A^{\ast }$, the scalar function $f(R(\lambda ))$ is bounded: it tends to zero at infinity by the Neumann series and is bounded on every compact disc. The Liouville theorem makes it identically zero. Since the Hahn-Banach theorem separates points, this would give $R(\lambda )=0$, contradicting its invertibility. Hence the spectrum is nonempty.

LetChoose positive ${\epsilon }_n$ with ${\prod }_n(1+{\epsilon }_n)<\infty$. We construct $x_n\in K$ inductively. Once $E_n=\mathrm{span}\{x_1,\dots ,x_n\}$ has been chosen, compactness of its unit sphere gives finitely many members of $X^{\ast }$ that almost norm every vector of $E_n$. Because $0\in {\overline{K}}^w$, the next $x_{n+1}$ may be chosen so that all those functionals are as small on it as required. Choosing the error relative to $d$ givesfor all scalars $a_1,\dots ,a_{n+1}$. Indeed, if the last coefficient could threaten this estimate, $\Vert x_{n+1}\Vert \ge d$ first bounds that coefficient by a fixed multiple of the norm of the preceding sum; the selected norming functional then gives the displayed inequality. Iteration and the finite product bound satisfy the standard basis criterion, so $(x_n)$ is a basic sequence contained in $K$.

The same proof works for any Hausdorff locally convex vector topology $\tau$ weaker than the weak topology: on each finite-dimensional $E_n$, the $\tau$-continuous linear functionals still norm the space, and $0\in {\overline{K}}^{\tau }$ supplies the next point. A canonical strictly weaker example arises on $X^{\ast }$ when $X$ is not reflexive: the weak-star topology $\sigma (X^{\ast },X)$ is then strictly weaker than the weak topology $\sigma (X^{\ast },X^{\ast \ast })$.

We next prove the Eberlein-Smulian theorem. If the weak closure of a bounded set $K$ is weakly compact, take any sequence in $K$ and let $Y$ be its closed linear span. The relevant weak closure lies in the separable space $Y$. A countable weak-star dense subset of the dual unit ball separates points of this compact set, so its weak topology is metrizable. Compact metrizability gives a weakly convergent subsequence.

For the converse, suppose $K$ is not relatively weakly compact. In the canonical embedding into $X^{\ast \ast }$, chooseThe Hahn-Banach theorem gives $x^{\ast \ast \ast }\in X^{\ast \ast \ast }$ that vanishes on $J_X(X)$ but satisfies $x^{\ast \ast \ast }(x^{\ast \ast })=1$. Alternating Goldstine approximation with the fact that $x^{\ast \ast }$ lies in the weak-star closure of $J_{X}K$ constructs bounded $x_n^{\ast }\in X^{\ast }$ and $x_n\in K$ such that, up to errors tending to zero,If a subsequence $x_{n_j}$ converged weakly to $x$, then for each fixed $i$ the second relation would give $x_i^{\ast }(x)=1$. A weak-star cluster point $x^{\ast }$ of the bounded sequence $(x_i^{\ast })$ satisfies $x^{\ast }(x_{n_j})=0$ by the first relation, and hence $x^{\ast }(x)=0$ by weak convergence; but passing to the same cluster point in $x_i^{\ast }(x)=1$ gives $x^{\ast }(x)=1$. This contradiction produces a sequence in $K$ with no weakly convergent subsequence. Relative weak compactness is therefore equivalent to the subsequence condition.

Finally suppose $K$ is weakly sequentially compact and $x\in {\overline{K}}^w$. If $x$ lies in the norm closure of $K$, a norm-convergent sequence suffices. Otherwise apply the first part to $K-x$ and obtain a basic sequence $(y_n)$. A subsequence converges weakly by hypothesis. Its weak limit lies in the closed span of the basic sequence, while every coordinate functional is eventually zero on that subsequence. The limit is consequently zero, so the corresponding sequence from $K$ converges weakly to $x$.

This also shows that a weakly sequentially compact $K$ is weakly closed: any point of its weak closure is the limit of a sequence in $K$, and a weakly convergent subsequence has its limit in $K$. The relative form of the Eberlein-Smulian theorem then makes $K$ weakly compact. The reverse implication follows from the first direction of that theorem. Thus weak compactness and weak sequential compactness are equivalent.

A real positive linear functional on a vector subspace of $C(K)$ containing $1$ extends positively to all of $C(K)$. Its norm is $\varphi (1)$ by order bounds. If this is positive, apply the Hahn-Banach theorem to obtain an extension of the same norm, normalize it to value one at $1$, and use the unital contraction positivity criterion. If $\varphi (1)=0$, the functional is zero and its zero extension suffices.

If $C$ is a nonempty open convex subset of a real locally convex space and $x_0\notin C$, a continuous linear functional strictly separates them: after choosing a sign, $f(c)<f(x_0)$ for every $c\in C$. Apply the Hahn-Banach theorem to the Minkowski functional of a translate of $C$.

If $Y$ is a closed linear subspace of a Hausdorff locally convex space and $x_0\notin Y$, choose a continuous seminorm $q$ with $q(x_0-y)\ge 1$ for all $y\in Y$, using a basic neighborhood disjoint from $Y$. The linear functional $g(y+t{x}_0)=t$ is bounded by $q$. The Hahn-Banach theorem extends it to a continuous linear functional with the displayed values. This is the linear-subspace form of separation, sharper than merely distinguishing the two sets.

For $S={\sum }_i{a}_i{ϵ}_i$ and $F=\Vert S\Vert$, the even-function spectral gap on a hypercube gives $2\mathrm{Var}(F)\le -\mathbb{E}[FLF]$. A norming functional supplied by the Hahn-Banach theorem and convexity of the norm gives $LF\ge -F$, because $LS=-S$. Hence the Dirichlet form is at most $\mathbb{E}{F}^2$, proving the inequality. Two equal scalar coefficients attain equality, so the constant is sharp.

The sublinear functional $p(g)=\mu (\overline{g})$ has a linear supporting functional taking the value $p(f)$ at a specified continuous $f$, by the Hahn-Banach theorem. It is positive and normalized because $p$ has these values on constants. The Riesz-Markov-Kakutani representation theorem gives the probability $\nu$. Testing affine functions and their negatives shows that $\mu$ and $\nu$ have the same barycenter.

A bounded linear operator between Banach spaces is weakly compact exactly when its Banach-space adjoint is weakly compact. Equivalently, that adjoint is continuous from its domain's weak-star topology to its range's weak topology. For the reverse direction, the bidual condition applied to $T^{\ast }$ forces $T^{\ast \ast \ast }$ to annihilate every functional on $Y^{\ast \ast }$ vanishing on $J_{Y}Y$; the Hahn-Banach theorem then places $T^{\ast \ast }(X^{\ast \ast })$ inside $J_{Y}Y$.

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.