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A functional $p:X\to \mathbb{R}$ is sublinear when

Any extension to $\tilde{M}=M+\mathbb{R}x$ must have the formChoose $c$ satisfyingThis interval is nonempty: for $y,z\in M$,which rearranges to the required lower bound being at most the upper bound. For $t>0$, positive homogeneity reduces
$\tilde{\ell }(y+tx)\le p(y+tx)$ to the upper inequality after replacing $y$ by $y/t$; for $t<0$ it reduces to the lower inequality. The case $t=0$ is the hypothesis on $\ell$. Thus this choice gives the required dominated linear extension.

The dominated form of the Hahn-Banach theorem states that a linear functional on a subspace, bounded above by a sublinear functional, extends linearly to the whole real vector space while retaining that bound.

Let $M=\mathrm{span}\{z_1,\dots ,z_n\}$. Its coordinate mapsare continuous because every linear map on a finite-dimensional normed space is continuous. Hahn--Banach extends each to
${\ell }_j\in Z^{\prime }$ without increasing its norm, and then

For an arbitrary finite-dimensional subspace $M$, choose a basis $z_1,\dots ,z_n$ and these extended coordinate functionals. Thenis closed. Every $z\in Z$ has the decompositionApplying every ${\ell }_j$ shows $M\cap N=\{0\}$, so

Suppose first that $X$ is separable, and choose a norm-dense sequence $(x_n)$ in its unit ball. On the dual unit ball defineUniform boundedness on the unit ball and density of the $x_n$ show that this metric induces the weak-star topology. Conversely, if $B_{X^{\ast }}$ is weak-star metrizable, the Banach-Alaoglu theorem makes it a compact metric space. Hence $C(B_{X^{\ast }})$ is separable. The evaluation mapis an isometry by the Hahn-Banach theorem. A subspace of a separable metric space is separable, so $X$ is separable. This proves the weak-star metrizability criterion for a dual ball.

If $X$ has a countable weakly dense subset $D$, the rational linear span of $D$ is weakly dense. Its norm closure is a convex set, so Mazur theorem says that its weak and norm closures agree. Thus $X$ is norm separable. The weak-star compact metric ball $B_{X^{\ast }}$ consequently has a countable weak-star dense subset, and the union of its integer dilates is weak-star dense in $X^{\ast }$. Therefore $X^{\ast }$ is weak-star separable.

It need not be weakly separable. Take $X={\ell }^1$, whose dual is ${\ell }^{\infty }$. A weakly separable normed space is norm separable by the preceding convex-closure argument, whereas ${\ell }^{\infty }$ is not norm separable.

If the Banach space $X$ is reflexive, its closed unit ball identifies with the weak-star compact ball of $X^{\ast \ast }$, hence is weakly compact. Conversely, if $B_X$ is weakly compact, its canonical image $J(B_X)$ is weak-star compact and therefore weak-star closed in $X^{\ast \ast }$. Goldstine theorem says it is weak-star dense in $B_{X^{\ast \ast }}$, soScaling proves that $J$ is surjective and $X$ is reflexive. This is the weak compactness characterization of reflexivity.

The Krein-Milman theorem says that a nonempty compact convex subset of a locally convex space is the closed convex hull of its extreme points. For reflexive $X$, the ball $B_X$ is weakly compact, sowhere weak and norm closure agree for the convex hull by Mazur's theorem.

For the final claim, let $\mathcal{H}$ be the set of functions ${\mathbb{Z}}^2\to [0,1]$ with the mean-value property. It is a compact convex subset of the product $[0,1{]}^{{\mathbb{Z}}^2}$. If $f$ is extreme, its four unit translates also lie in $\mathcal{H}$, and the mean-value identity writes $f$ as their average. Extremality forces every translate to equal $f$, so $f$ is constant. Every extreme point is therefore constant. Krein--Milman now makes every member of $\mathcal{H}$ a limit of convex combinations of constant functions, and hence constant. This is the bounded harmonic function theorem on the integer lattice.

Regard the space of continuous functions vanishing at infinity $C_0({\mathbb{R}}^n)$ as a subspace of $L^{\infty }({\mathbb{R}}^n)$. Point evaluation at the origin,is a bounded linear functional of norm one on this subspace because a continuous function's supremum and essential supremum agree. By the Hahn-Banach theorem, ${\ell }_0$ extends to a bounded linear functional

Suppose that some $g\in L^1({\mathbb{R}}^n)$ represented this extension:Choose $\phi \in C_c({\mathbb{R}}^n)$ with $0\le \phi \le 1$, $\phi (0)=1$, and support in the unit ball, and set ${\phi }_k(x)=\phi (kx)$. Then $\ell ({\phi }_k)=1$ for every $k$. On the other hand, ${\phi }_k(x)\to 0$ for almost every $x$ and $|{\phi }_{k}g|\le |g|$. The dominated convergence theorem givesa contradiction. Thus $\ell$ is a singular functional on L infinity and has no $L^1$ density.

First suppose that $X$ is separable. Choose a norm-dense sequence $(x_n)$ in $B_X$. On every norm-bounded subset of $X$, a countable norm-dense subset $(f_n)$ of $B_{X^{\ast }}$ in the weak-star topology separates points, andmetrizes the weak topology. Indeed, convergence for all $f_n$, boundedness, and weak-star density imply convergence for every $f\in X^{\ast }$. Thus a weakly compact set $K$ is a compact metric space and hence is sequentially compact.

For general $X$, take a sequence $(x_n)$ in $K$ and letThe space $Y$ is separable and norm closed. The Hahn-Banach theorem shows both that $Y$ is weakly closed in $X$ and that its own weak topology is the subspace topology inherited from $X$. Hence $K\cap Y$ is weakly compact and, by the separable case, contains a weakly convergent subsequence of $(x_n)$. Its limit lies in $K\cap Y$. Therefore every weakly compact subset of a Banach space is weakly sequentially compact, which is one direction of the Eberlein-Šmulian theorem.

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.

In a unital C-star algebra $A$, an element is Hermitian when $x=x^{\ast }$, unitary when $x^{\ast }x=x{x}^{\ast }=1$, and normal when $x^{\ast }x=x{x}^{\ast }$. It is a Positive element of a C-star algebra when $x=y^{\ast }y$ for some $y$, equivalently when it is Hermitian and its spectrum lies in $[0,\infty )$.

Suppose first that $\Vert \tau \Vert =\tau (1)$, and normalize so that both equal one. If $x=x^{\ast }$, the C-star identity givesWriting $\tau (x)=a+ib$, we obtainfor every real $t$, which forces $b=0$. Decomposing an arbitrary $x$ into its Hermitian real and imaginary parts now gives

Let $x=x^{\ast }$. Every character of an algebra on the commutative unital Banach algebra generated by $x$ has norm and value at one equal to one, so the preceding argument makes its value on $x$ real. The character description of the spectrum of an element therefore gives $\sigma (x)\subset \mathbb{R}$. Iterating the C-star identity,so $\Vert x\Vert =r(x)$. If $c=\Vert x\Vert$, then $x+c1$ has nonnegative spectrum and is positive, as is $c1$; henceis a difference of positive elements.

If $0\le x$ and $\Vert x\Vert \le 1$, then spectral translation givesThus $1-x$ is positive and, by the norm--spectral-radius equality for Hermitian elements, $\Vert 1-x\Vert \le 1$.

Return to a norm-one $\tau$ with $\tau (1)=1$. For $0\le x\le 1$, the preceding paragraph and reality on Hermitian elements giveScaling proves that $\tau$ is a positive functional on a C-star algebra. Every character has norm and value at one equal to one, so every character is positive. On $C[0,1]$, the functionalis positive but is not multiplicative, and hence is not a character.

Conversely, let $\tau$ be positive. Writing a Hermitian element as a difference of positive elements shows that $\tau$ is real on Hermitian elements. Positivity offor every $\lambda \in \mathbb{C}$ says that the associated quadratic polynomial is nonnegative. Minimizing it in $\lambda$ gives the Cauchy--Schwarz inequalityTaking $y=1$ yields $|\tau (x){|}^2\le \tau (x^{\ast }x)\tau (1)$. Since $0\le x^{\ast }x\le \Vert x{\Vert }^{2}1$, positivity givesTogether with $\tau (1)\le \Vert \tau \Vert$, this proves $\Vert \tau \Vert =\tau (1)$.

Convex combinations preserve positivity and value one at the identity, so the state space $S(A)$ is convex. If $x$ is normal, the unital C-star subalgebra $C^{\ast }(1,x)$ is commutative and its Gelfand transform identifies it with $C(\sigma (x))$. Choose $\lambda \in \sigma (x)$ with $|\lambda |=\Vert x\Vert$. Evaluation at $\lambda$ is a state taking $x$ to $\lambda$. Its norm-preserving Hahn--Banach extension to $A$ still takes $1$ to one, so the norm criterion makes the extension a state $\tau$ satisfying $|\tau (x)|=\Vert x\Vert$.

The state space is nonempty, convex, and weak-star compact by the Banach-Alaoglu theorem. The Krein-Milman theorem gives an extreme point, so a pure state on a C-star algebra exists. For positive $x$, the preceding norm-attainment result makesa nonempty weak-star compact convex set. It is a face: if a convex combination has the maximal possible value on $x$, each summand does. An extreme point of $F$ exists by Krein--Milman and, because $F$ is a face, is extreme in $S(A)$. It is the required pure state.

The Hahn-Banach separation theorem for two convex sets states that if $A$ and $B$ are disjoint nonempty convex subsets of a real locally convex space $X$ and $A$ is open, then there are a continuous linear functional $f\in X^{\ast }$ and a real number $c$ such thatTo prove it, formThis is open and convex and does not contain zero. The separation of a point and an open convex set, proved from the Minkowski functional and the Hahn-Banach theorem, gives a nonzero continuous $f$ with $f(c)<0$ for every $c\in C$. Hence $f(a)<f(b)$ for all $a,b$. Taking $c$ between ${\sup }_{A}f$ and ${\inf }_{B}f$ gives the stated form.

For a dual pair $(E,F)$, the topology $\sigma (E,F)$ has a neighbourhood base at zero consisting ofEvery $f\in F$ is continuous by definition. Conversely, if a linear functional $g$ is continuous, some such neighbourhood lies in $\{|g|<1\}$. Therefore ${\bigcap }_j\ker f_j\subseteq \ker g$. Elementary linear algebra then gives $g\in \mathrm{span}\{f_1,\dots ,f_n\}\subseteq F$. Thus

For a normed space $X$, the weak topology is $\sigma (X,X^{\ast })$; on $X^{\ast }$ the weak-star topology is $\sigma (X^{\ast },X)$, using the canonical image of $X$ in $X^{\ast \ast }$. If $X$ is reflexive, $X^{\ast \ast }=J_X(X)$, so the weak and weak-star topologies on $X^{\ast }$ coincide. Conversely, if they coincide, every $x^{\ast \ast }\in X^{\ast \ast }$ is weak-star continuous. The dual-pair result says that every such functional is evaluation at some $x\in X$, so $J_X$ is onto and $X$ is reflexive.

For $A\subseteq E$, every $a\in A$ and zero satisfy every inequality defining $A^{\circ \circ }$, so $A\cup \{0\}\subseteq A^{\circ \circ }$. The latter is an intersection of weakly closed convex half-spaces, hence containsIf $x\notin C$, choose an open convex neighbourhood $V$ of zero with $(x+V)\cap C=\varnothing$. Applying the separation theorem to $x+V$ and $C$ gives $f\in F$ withBecause $0\in C$, the supremum is nonnegative. After multiplying $f$ by a positive scalar, $f\le 1$ on $C$ while $f(x)>1$. Thus $f\in A^{\circ }$ but $x\notin A^{\circ \circ }$. We conclude with the Bipolar theorem for a dual pair:

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-Šmulian 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-Šmulian 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.

For a complex unital Banach algebra and a circle enclosing its element's Banach algebra spectrum, the displayed identity holds for $n\ge 0$. On a sufficiently large circle, expand the Neumann series and integrate it term by term. At any resolvent point the local series $R({\lambda }_0+h)={\sum }_{j\ge 0}(-h{)}^{j}R({\lambda }_0{)}^{j+1}$ proves norm analyticity. Apply continuous complex linear functionals and the scalar Cauchy theorem to deform the large circle to any radius greater than the maximum spectral modulus; the Hahn-Banach theorem separates points and restores the vector integral identity. Its norm bound $\Vert x^n\Vert \le R^{n+1}{\max }_{|\lambda |=R}\Vert R(\lambda )\Vert$ gives the reverse inequality in the spectral radius formula.

If a convex set $E$ in a real normed vector space contains $B(0,ϵ)$, every $x\notin E$ admits a continuous linear functional $T$ with $Tx\ge 1\ge Te$ for all $e\in E$, even if $E$ is not closed. Its Minkowski functional $p(v)=\inf \{t>0:v\in tE\}$ is sublinear functional, satisfies $p(v)\le \Vert v\Vert /ϵ$, $p(e)\le 1$, and $p(x)\ge 1$. Extend $tx\mapsto tp(x)$ by the dominated Hahn-Banach theorem. The resulting domination $T\le p$ applied to $v$ and $-v$ makes $T$ continuous and supplies the separation inequalities.

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.

Suppose $q$ is a sublinear functional on the bounded real sequences, dominated by the limit superior and nonpositive on every shift difference $x-Sx$. The Hahn-Banach theorem then yields a Banach limit. Indeed $0\le q(x)+q(-x)$ implies $q(x)\ge -q(-x)\ge \mathrm{liminf}{x}_n$, so $q$ agrees with the ordinary limit on convergent sequences. Extend this linear limit functional by Hahn-Banach to a linear $\psi \le q$. Domination on $x-Sx$ and on its negative forces $\psi (x)=\psi (Sx)$. The two-sided limit bounds give positivity, norm one and agreement with ordinary limits. In finite-shift averaging constructions, the needed bound on shift differences follows by telescoping consecutive averages to $(x_{n+1}-x_{n+k+1})/k$.