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For a nonempty radially open convex set $C$ disjoint from a vector subspace $U$, there is a nonzero linear functional $L$ whose kernel of a linear map contains $U$ and avoids $C$. Choose $a\in C$ and the Minkowski functional $p$ of $C-a$. On $U+\mathbb{R}a$, set $g(u+ta)=-t$. Disjointness implies $p(u-a)\ge 1$, which gives $g\le p$ for negative $t$; nonnegativity of $p$ gives it for nonnegative $t$. The algebraic Hahn-Banach theorem extends $g$ to $L\le p$. For $x\in C$, $L(x)+1=L(x-a)<1$, so $L(x)<0$. Also $L(a)=-1$ and $L{|}_U=0$.

A locally convex space is a topological vector space with a neighbourhood basis at zero consisting of convex sets. Equivalently, its topology is a seminorm topology generated by some family $\mathcal{P}$ of seminorms. We allow non-Hausdorff spaces here; a separating family of seminorms gives precisely the Hausdorff case. Every normed vector space is locally convex, but smooth convergence generally requires a whole family of measurements.

The geometric and measurement descriptions fit together: seminorm sublevel sets are convex, while the Minkowski functional of an open, convex zero neighbourhood invariant under all scalar multiplications with $|\lambda |\le 1$ is a continuous seminorm. Such neighbourhoods can be chosen as a neighbourhood basis. ^[ref].

Local convexity makes Hahn-Banach theorem and separation of a point and an open convex set available. It turns geometric distinctions into continuous linear functionals, connecting convex sets to the continuous dual space. For a seminorm topology generated by $\mathcal{P}$, a linear functional $\ell$ is continuous exactly when there are $p_1,\dots ,p_m\in \mathcal{P}$ and $C>0$ withThus one continuous linear functional needs only finitely many of the available measurements. For example, point evaluation of a derivative on the space of smooth functions on a compact interval is controlled by a suitable derivative seminorm.

Suppose $B$ contains zero, is convex, and meets every line $\mathbb{R}v$, $v\ne 0$, in the finite symmetric open segment $\{tv:|t|<\lambda (v)\}$ with $\lambda (v)>0$. Then its Minkowski functional is a norm, with $p_B(v)=1/\lambda (v)$ and $B=\{x:p_B(x)<1\}$. Absolute homogeneity follows by rescaling the segment. If $s>p_B(x)$ and $t>p_B(y)$, convexity puts $(x+y)/(s+t)$ in $B$, giving $p_B(x+y)<s+t$; passing to the infima proves the triangle inequality. Conversely the open unit ball uniquely determines any norm by the displayed formula. No prior topology on the vector space is needed for these radial hypotheses.

For a bounded convex open set $U$ containing zero and invariant under $x\mapsto -x$, its Minkowski functional is a norm. Inner and outer Euclidean balls give positive finite bounds; symmetry and scaling give absolute homogeneity; convexity gives the triangle inequality. Openness makes $U=\{p_U<1\}$, while boundedness prevents nonzero vectors from having zero gauge.

A norm $p$ is nonnegative with $p(x)=0$ exactly when $x=0$, satisfies $p(ax)=|a|p(x)$, and obeys $p(x+y)\le p(x)+p(y)$. The Euclidean norm has positivity and absolute homogeneity immediately. For the triangle inequality, positivity of $\Vert x-ty{\Vert }_2^2$ for all real $t$ gives the Cauchy-Schwarz inequality $|x\cdot y|\le \Vert x{\Vert }_2\Vert y{\Vert }_2$ (with $y=0$ immediate). Hencewhich proves the triangle inequality after taking square roots.

For the prescribed set, use the Minkowski functionalSince $U$ is open and contains zero, some Euclidean ball of radius $r>0$ lies inside it. Since it is bounded, it lies inside a ball of radius $R$. These inclusions imply $\Vert x{\Vert }_2/R\le p_U(x)\le \Vert x{\Vert }_2/r$, so the functional is finite and positive for nonzero $x$; it is zero at zero. Symmetry gives $p_U(-x)=p_U(x)$, and substitution in the infimum gives $p_U(ax)=a{p}_U(x)$ for $a>0$, hence absolute homogeneity for all real scalars.

Convexity and $0\in U$ imply $sU\subseteq U$ for $0\le s\le 1$. Consequently, for $a>p_U(x)$ and $b>p_U(y)$, one has $x/a,y/b\in U$. Convexity then givesso $p_U(x+y)\le a+b$. Letting $a\downarrow p_U(x)$ and $b\downarrow p_U(y)$ proves the triangle inequality.

Finally, if $p_U(x)<1$, there is a $t<1$ with $x\in tU\subseteq U$. Conversely, if $x\in U$ and $x\ne 0$, openness permits $(1+\delta )x\in U$ for some $\delta >0$, so $p_U(x)\le 1/(1+\delta )<1$; zero is immediate. Thus $\boxed{U=\{x:p_U(x)<1\}}$, completing the norm from a bounded symmetric convex neighbourhood construction.

Let $p$ be the Minkowski functional of $E$:The contained ball makes $E$ absorbing, so $p$ is finite, nonnegative and satisfies $p(v)\le \Vert v\Vert /ϵ$. It is positively homogeneous. To check subadditivity, if $v\in sE$ and $w\in tE$, the defining property of a convex set gives $(v+w)/(s+t)\in E$, whence $p(v+w)\le s+t$; let $s,t$ decrease to the respective infima. Thus $p$ is a sublinear functional. We have $p(e)\le 1$ for every $e\in E$. Also $p(x)\ge 1$: otherwise $x\in tE$ for some $0<t<1$, which would put $x$ in $E$ because $E$ is a convex set and $0\in E$.

We prove the dominated form of the Hahn-Banach theorem needed here. Suppose $g$ is linear on a subspace $M$ and $g\le p$ there. To extend it to $M+\mathbb{R}v$ with $v\notin M$, choose the value $c$ at $v$ betweenEvery lower candidate is at most every upper candidate, becauseTaking $m=n=0$ among the candidates shows the two endpoints are finite: the lower supremum is at least $-p(-v)$ and at most $p(v)$, and the upper infimum lies between them and $p(v)$. Define $\tilde{g}(m+tv)=g(m)+tc$. For $t>0$, the upper inequality applied to $m/t$ gives domination by $p(m+tv)$. For $t<0$, apply the lower inequality to $m/(-t)$ and multiply by $-t$. At $t=0$ domination already holds. Hence this is a dominated one-dimensional extension.

Order all dominated extensions by inclusion of their domains. Chain unions preserve linearity and domination, so Zorn's lemma gives a maximal extension. The one-dimensional argument shows its domain is the whole vector space. This proves the requisite dominated Hahn-Banach theorem, rather than invoking an unproved separation form.

On the line $\mathbb{R}x$, start with $g(tx)=tp(x)$. For $t\ge 0$ this equals $p(tx)$; for $t<0$ it is nonpositive and therefore at most $p(tx)$. Extend it to $T\le p$ on $V$. ThenApplying domination to $v$ and $-v$ also givesThus $T$ is a continuous linear functional, andThis proves separation from an absorbing convex set with no closedness assumption on $E$. The proof also gives the norm-preserving real extension theorem used earlier: take $p(v)=C\Vert v\Vert$ when the original functional has norm at most $C$.

We first prove the algebraic extension step needed for Hahn-Banach separation theorem. Suppose $p$ is a finite sublinear functional on a real vector space and $g$ is a linear functional on a vector subspace $W$, with $g\le p$ there. For $v\notin W$, choose a real number $c$ betweenThis interval is nonempty: for any $w,z\in W$,Taking $w=0$ or $z=0$ also bounds the endpoints by finite numbers. Define $\tilde{g}(w+tv)=g(w)+tc$. For $t>0$, the upper bound for $c$, applied to $w/t$, proves $\tilde{g}(w+tv)\le p(w+tv)$. For $t<0$, writing $t=-s$ and using the lower bound at $w/s$ proves the same inequality. The case $t=0$ is the original domination. Thus a dominated linear functional extends over one further direction. Order all dominated extensions by extension of their domains. A chain has its union as an upper bound, so Zorn's lemma supplies a maximal extension. The one-direction argument shows its domain is the whole vector space. This proves the required algebraic Hahn-Banach theorem, rather than assuming the separation result.

Choose $a\in A$, and put $C=A-a$. This is a radially open convex set containing zero. It is absorbing: along each line through zero, sufficiently small multiples of any vector lie in $C$. Its Minkowski functionalis finite and nonnegative. Positive homogeneity follows by rescaling $t$. If $x\in sC$ and $y\in tC$, convexity gives $x+y\in (s+t)C$, proving subadditivity by taking infima. Moreover,Indeed, $p(x)<1$ gives $x\in sC$ for some $s<1$, hence $x\in C$ by convexity and $0\in C$. Conversely, radial openness at $x\in C$ permits $(1+ϵ)x\in C$ for some $ϵ>0$, giving $p(x)<1$.

Since $A\cap U=\varnothing$, $a\notin U$ and $u-a\notin C$ for every $u\in U$. Define a linear functional on $U+\mathbb{R}a$ byIt is well defined because $a\notin U$. For $t<0$, write $s=-t$; then $p(u+ta)=sp(u/s-a)\ge s=g(u+ta)$. For $t\ge 0$, domination follows from $g(u+ta)\le 0\le p(u+ta)$. The extension step therefore gives $L:V\to \mathbb{R}$ with $L\le p$, $L{|}_U=0$ and $L(a)=-1$. If $x\in A$, thenso $L(x)<0$. ConsequentlyBecause $L(a)\ne 0$, this kernel of a linear map has codimension one and is a hyperplane. This establishes the geometric Hahn-Banach theorem for radially open sets without assuming a norm or a topology on $V$.

For the convex function, use its strict epigraph in $V\times \mathbb{R}$. It is nonempty and convex, and does not contain $(0,0)$. It is also radially open: on every affine line, a finite convex function is continuous. To justify this last fact, convexity orders secant slopes; on a smaller interval the slopes are bounded above and below by slopes to two fixed exterior endpoints, giving a local Lipschitz bound. Thus $t\mapsto f(x+ty)-\beta -ts$ is continuous, and a strict negative value persists near $t=0$.

Apply the separation theorem with the zero subspace. A nonzero separating linear functional has form $L(x,\beta )=g(x)+c\beta$. Its values on the convex epigraph cannot have both signs, since a line segment would then meet its kernel. Choose the sign so that $L>0$ on the epigraph. At $x=0$ and $\beta >0$ this gives $c>0$. Letting $\beta$ decrease to $f(x)$ yields $g(x)+cf(x)\ge 0$. ThereforeThe resulting linear functional need not be nonzero: for example, the zero functional is the only linear minorant of $f(x)=x^2$ on $\mathbb{R}$.

Now suppose the norm and continuity hypotheses hold. Domination at $x$ and $-x$ givesIf $x_n\in \ker l$ and $x_n\to x$ in norm, apply this inequality to $x-x_n$. Since $l(x-x_n)=l(x)$ and both bounding values tend to $f(0)=0$, we obtain $l(x)=0$. The kernel of a linear map is therefore closed.

Finally, a linear functional with closed kernel on a normed vector space is continuous. If $l=0$, this is immediate. Otherwise choose $v$ with $l(v)=1$ and put $M=\ker l$. Closedness gives $d=\mathrm{dist}(v,M)>0$. For $l(x)\ne 0$, $x/l(x)-v\in M$, so $d\le \Vert x\Vert /|l(x)|$. The same resulting bound holds when $l(x)=0$:This proves continuity and explains exactly how closedness of the kernel is used.

The interior of the diamond is $B=\{(x_1,x_2):|x_1|+|x_2|<1\}$. Its Minkowski functional, or the uniqueness argument already proved, gives the L1 norm:The final unheaded set $C$ cannot be an open unit ball of a norm, because every norm ball is convex by homogeneity and the triangle inequality. Both $p=(9/10,0)$ and $q=(0,9/10)$ lie in $C$: their squared-distance expression is $101/100>1$. Their midpoint has that expressionso $(p+q)/2\notin C$. Thus $\boxed{\text{no norm has }C\text{ as its open unit ball}}$. The strict inequalities in the definition are respected by this explicit counterexample.

The unheaded preliminary requests are addressed before the first example. A norm is a function $p:V\to [0,\infty )$ satisfying $p(x)=0$ exactly for $x=0$, $p(ax)=|a|p(x)$, and $p(x+y)\le p(x)+p(y)$. Its induced metric is $d(x,y)=p(x-y)$. Translation $T_v$ preserves this distance, and scalar multiplication satisfies $d(ax,ay)=|a|d(x,y)$, so both are continuous; multiplication by zero is constant.

The open unit ball determines the norm uniquely through its Minkowski functional:because $x/t\in B$ is equivalent to $p(x)<t$. This also works for $x=0$.

Under the given radial and convexity hypotheses, set $p_B(0)=0$ and $p_B(v)=1/\lambda (v)$ for $v\ne 0$. The finite symmetric segment along each line gives positivity, absolute homogeneity and $v\in B⟺p_B(v)<1$. If $s>p_B(x)$ and $t>p_B(y)$, then $x/s,y/t\in B$, and the stated convexity condition yields $(x+y)/(s+t)\in B$. Hence $p_B(x+y)<s+t$; taking $s\downarrow p_B(x)$ and $t\downarrow p_B(y)$ proves the triangle inequality. Thus this is a norm determined by a convex radial unit ball. The case $V=\{0\}$ is immediate.

For the coordinate box, the Minkowski functional is the supremum norm:Indeed $x/t\in B$ exactly when $t>{\max }_i|x_i|$.

The Hahn-Banach separation theorem for two convex sets states that disjoint nonempty convex sets $A,B$ in a real locally convex space, with $A$ open, admit a nonzero continuous linear functional $\ell$ and $\alpha \in \mathbb{R}$ such thatThere need not be a positive uniform gap between the two sets.

For a proof, set $C=A-B$. It is open and convex, and $0\notin C$. Choose $c_0\in C$, let $V=C-c_0$, and let $p$ be the Minkowski functional of $V$. Because $V$ is open, convex and contains zero, $p$ is a finite sublinear functional, and $V=\{x:p(x)<1\}$. Put $z=-c_0$. Since $z\notin V$, $p(z)\ge 1$. On $\mathbb{R}z$ define ${\ell }_0(tz)=tp(z)$. For $t\ge 0$ this equals $p(tz)$; for $t<0$ it is nonpositive and hence at most $p(tz)$. The Hahn-Banach theorem extends it to a linear $\ell$ with $\ell \le p$. Both $p(x)$ and $p(-x)$ are bounded near zero, so this domination makes $\ell$ continuous; it is nonzero because $\ell (z)=p(z)\ge 1$. For $c\in C$,Thus $\ell (a)<\ell (b)$ for every $a\in A,b\in B$. Define $\alpha ={\sup }_{a\in A}\ell (a)$, which is finite because any fixed $b\in B$ bounds it above. Openness of $A$ and nonzeroness of $\ell$ ensure $\ell (a)<\alpha$ for every $a\in A$, and $\alpha \le \ell (b)$ for every $b\in B$. This proves the stated separation.

For the sequence criterion, suppose $f\in S_{X^{\ast }}$ and $f(x_i)\ge \varrho$. If each $t_i\ge 0$, thenConversely, the assumed inequality implies that every convex combination of the $x_i$ has norm at least $\varrho$. Hence $D=\{x:\Vert x\Vert <\varrho \}$ is disjoint from $C=\mathrm{conv}\{x_i:i\ge 1\}$. Apply the separation theorem with the open set $D$ to obtain $\ell \ne 0$ andThen $f=\ell /\Vert \ell \Vert$ has norm one and satisfies $f(x_i)\ge \varrho$ for every $i$. ThereforeThe PDF confirms the “if and only if” that is garbled in the TeX transcription.

Let $p_C(x)=\inf \{t>0:x\in tC\}$ be the Minkowski functional of $C$. Openness, convexity, and $0\in C$ make $p_C$ sublinear, with $p_C(x)<1$ for $x\in C$ and $p_C(x_0)\ge 1$. Define a linear functional on $\mathbb{R}{x}_0$ by $f(t{x}_0)=t{p}_C(x_0)$. The real Hahn-Banach theorem extends it to $X$ with $f\le p_C$. Hence

Apply this separation to the open ball of radius $\Vert x_0\Vert$ and rescale to obtain $f\in S_{X^{\ast }}$ with $f(x_0)=\Vert x_0\Vert$. For a closed subspace $Y$, apply it toThe separator must vanish on $Y$ because $C$ contains every translate along $Y$, and normalization gives

Let $F\subseteq X^{\ast }$ be finite-dimensional and $\Phi \in B_{X^{\ast \ast }}$. ConsiderIf $\Phi {|}_F$ did not belong to $S((1+\epsilon )B_X)$, finite-dimensional strict separation would produce some $f\in F$ withcontradicting $\Vert \Phi \Vert \le 1$. Thus there is $x\in X$ with $\Vert x\Vert <1+\epsilon$ and $f(x)=\Phi (f)$ for every $f\in F$.

Given a basic weak-star neighbourhood of $\Phi$ in $B_{X^{\ast \ast }}$, apply this result to the finite-dimensional span of its defining functionals and then replace $x$ by $x/(1+\epsilon )$. As $\epsilon \downarrow 0$, the resulting points of $J(B_X)$ enter that neighbourhood. Hence $J(B_X)$ is weak-star dense in $B_{X^{\ast \ast }}$, proving Goldstine theorem.

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:

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.

A subset $C$ of a real vector space is radially open when, for each $x\in C$ and each vector $v$, the points $x+tv$ remain in $C$ for all sufficiently small real $t$. No ambient topology is required. If $C$ is also convex and contains zero, it is absorbing and its Minkowski functional $p_C(x)=\inf \{s>0:x\in sC\}$ is finite and sublinear, with $C=\{p_C<1\}$. Convexity gives subadditivity; radial openness permits a slight dilation of any point of $C$, giving the strict inequality.

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 $C$.