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A subset $B$ of a topological vector space is bounded when every neighborhood $U$ of zero absorbs it: $B\subseteq tU$ for all sufficiently large positive $t$. In a locally convex space defined by seminorms $p_{\lambda }$, this is equivalent to ${\sup }_{x\in B}{p}_{\lambda }(x)<\infty$ for every $\lambda$. In particular, a bounded subset of the Schwartz space has a uniform bound for every Schwartz seminorm; it need not have uniformly bounded supports.

In a Hausdorff locally convex space, a separating family of continuous seminorms has $p(x)>0$ for any chosen nonzero $x$ and some $p$. Define a linear functional on its one-dimensional span attaining $p(x)$ and extend it under that seminorm by the real Hahn-Banach theorem. The extension is continuous because it is seminorm-bounded, and distinguishes $x$ from zero. The Hausdorff condition is essential.

A Fréchet space is a Hausdorff locally convex space whose topology is induced by a complete translation-invariant metric. Translation invariance means $d(x+z,y+z)=d(x,y)$; completeness means every Cauchy sequence converges in the space. Use a translation-invariant metric for this condition: an arbitrary compatible metric can change which sequences are Cauchy sequences. ^[ref].

Equivalently, the topology comes from a countable separating family of seminorms $(p_k{)}_{k\ge 0}$ and is complete in their common sense: if $p_k(x_n-x_m)\to 0$ as $n,m\to \infty$ for every fixed $k$, then some $x\in E$ satisfies $p_k(x_n-x)\to 0$ for every $k$. A useful compatible translation-invariant metric isIt is finite because the weights sum to one. The separating condition makes it a metric; without that condition the formula gives only a pseudometric. Its convergence in a metric space and Cauchy sequence conditions are exactly the conditions for all the $p_k$ separately: control finitely many terms, then bound the remaining tail by $2^{-(K+1)}$. Replacing $p_k$ by ${\max }_{0\le j\le k}{p}_j$ gives an increasing family with the same topology.

Every Banach space is a Fréchet space, using its norm as the single measurement. The space of smooth functions on a compact interval is a Fréchet space with derivative seminorms but with no single norm inducing its smooth topology. This is why Fréchet spaces are useful in differential equations: one can require accuracy in every order of derivative, while each particular estimate uses only finitely many orders.

The name honors Maurice Fréchet. His doctoral thesis was published as ^[ref]; an ^[ref] is available. This identifies a historical source without attributing the modern locally convex definition to that paper.

A nonempty compact face of a convex set in a Hausdorff locally convex space contains an extreme point. Chains of compact faces have nonempty compact intersection. A minimal face supplied by the Zorn lemma must be a singleton, since a separating linear functional would otherwise expose a proper maximizer face. This is the existence step in the Krein-Milman theorem.

An extreme point $x$ of a convex set $K$ is one for which $x=(1-t)y+tz$, $y,z\in K$ and $0<t<1$ imply $y=z=x$. It is enough to test midpoint decompositions: a nontrivial interior point of a segment is itself the midpoint of a shorter nontrivial segment contained in that segment.

We prove the Krein-Milman theorem. Assume first that $K$ is nonempty. A face of a convex set $K$ means a convex subset $F$ with the property that whenever an interior point of a segment in $K$ belongs to $F$, both endpoints belong to $F$. The family of nonempty compact faces of $K$, including $K$ itself, has a minimal member by the Zorn lemma: every inclusion-decreasing chain has a nonempty intersection by compactness and the finite intersection property, and that intersection is again a compact face.

Let $M$ be such a minimal face. If it contains two distinct points, the continuous dual of a Hausdorff locally convex space separates them, by the permitted Hahn-Banach theorem. Choose a continuous real linear functional $\ell$ that is not constant on $M$. It attains a maximum on the compact set $M$, andis a nonempty proper compact face of $M$, and hence a face of $K$. Indeed, equality at the maximum in a convex combination forces both endpoint values to be maximal. This contradicts minimality. Thus $M$ is a singleton, and its point is extreme in $K$. The same argument applied within any nonempty compact face proves that every such face contains an extreme point of $K$.

Let $C=\overline{\mathrm{conv}}(\mathrm{ext}K)$. It is contained in $K$, since the latter is closed and convex, and extreme points exist when $K$ is nonempty. If $x_0\in K\setminus C$, the permitted separation theorem provides a continuous real linear functional $\ell$ with $\ell (x_0)>{\sup }_C\ell$. Let $F$ be its maximizer face on $K$. This nonempty compact face contains an extreme point $v$ of $K$, butcontradicting $v\in \mathrm{ext}K\subseteq C$. ThereforeFor empty $K$, the equality holds with the usual empty convex-hull convention. This proves both the extreme-point existence step and the closed-convex-hull step; no compactness of the set of extreme points itself is assumed.

The real absolutely summable sequence space is ${\ell }^1=\{x=(x_j):{\sum }_j|x_j|<\infty \}$ with norm $\Vert x{\Vert }_1={\sum }_j|x_j|$. The real l-infinity sequence space is ${\ell }^{\infty }=\{x=(x_j):{\sup }_j|x_j|<\infty \}$ with norm $\Vert x{\Vert }_{\infty }={\sup }_j|x_j|$. Their completeness is allowed as given; their extreme points still require proof.

For ${\ell }^1$, an interior point of the unit ball is not extreme because small opposite perturbations in one coordinate stay in the ball. At a boundary point with two nonzero coordinates $i,j$, choose $0<\epsilon <\min (|x_i|,|x_j|)$ and putThe two distinct vectors $x+z,x-z$ both have norm one and average to $x$, so $x$ is not extreme. A remaining boundary point must be $\pm e_j$. If $e_j=(y+z)/2$ with $\Vert y{\Vert }_1,\Vert z{\Vert }_1\le 1$, the $j$th coordinate forces $y_j=z_j=1$, exhausting both norm budgets and forcing all other coordinates to zero. The negative case is identical. ThusFor ${\ell }^{\infty }$, if $|x_j|<1$ in even one coordinate, small opposite changes to that coordinate give a nontrivial midpoint decomposition within the ball. If every $x_j=\pm 1$, a midpoint decomposition forces both summands to have that same value in every coordinate, so they equal $x$. HenceThese are precisely the extreme points of real sequence-space unit balls for the two spaces under consideration.

Work over $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$. A Fréchet space is a Hausdorff, complete, metrizable locally convex space. Its topology is generated by a countable separating family of seminorms $(p_n{)}_{n\ge 1}$. A neighbourhood base at zero consists of finite intersectionsReplacing $p_n$ by ${\max }_{j\le n}{p}_j$ makes the family increasing. One compatible translation-invariant metric isCompleteness means that a sequence Cauchy for all these seminorms converges in the space. The separating condition ensures that $d(x,y)=0$ only when $x=y$.

Let $E^{\prime }$ denote the continuous dual space for the original topology $\tau$. The weak topology $\sigma (E,E^{\prime })$ is the coarsest topology making all $f\in E^{\prime }$ continuous. Its basic zero-neighbourhoods areEquivalently, it is generated by the seminorms $x\mapsto |f(x)|$.

To prove the Hausdorff property, fix $z\ne 0$. Some continuous seminorm $p$ has $p(z)>0$. On the one-dimensional span of $z$, define $f_0(az)=ap(z)$. Then $|f_0(az)|=p(az)$. The real or complex Hahn-Banach theorem extends this to $f:E\to \mathbb{F}$ with $|f(x)|\le p(x)$. Thus $f$ is originally continuous and $f(z)\ne 0$. This proves continuous-dual point separation. For distinct $x,y$, inverse images under such an $f$ of disjoint small scalar neighbourhoods of $f(x)$ and $f(y)$ are disjoint weak neighbourhoods. Hence the weak topology is Hausdorff.

If a linear functional is originally continuous, it belongs to $E^{\prime }$ and is weakly continuous by definition. Conversely, $\sigma (E,E^{\prime })$ is coarser than $\tau$, so every weakly continuous scalar function is originally continuous. In particular,This special case of the continuous dual of a weak topology needs no completeness argument.

For the linear map $T:E_1\to E_2$, original continuity implies that $g\circ T\in E_1^{\prime }$ whenever $g\in E_2^{\prime }$. Each such composition is weakly continuous on $E_1$, so the defining property of the target weak topology makes $T$ weak-to-weak continuous.

Conversely, suppose $T$ is weak-to-weak continuous. Every $g\circ T$, $g\in E_2^{\prime }$, is then weakly continuous, and hence continuous for ${\tau }_1$. We show that the graph of a linear operator $\Gamma (T)$ is closed in the original product topology. If $x_n\to x$ in $E_1$ and $T{x}_n\to y$ in $E_2$, thenBy continuous-dual point separation on $E_2$, this forces $y=Tx$. The product is metrizable, so sequential closedness is closedness. The closed graph theorem for Fréchet spaces now gives original continuity of $T$.

The relevant version of that theorem does use completeness of both spaces: a closed graph is a Fréchet space, its first-coordinate projection is a continuous linear bijection onto $E_1$, and the open mapping theorem for Fréchet spaces gives a continuous inverse. Composing with the second-coordinate projection gives $T$. ThusEquality of continuous scalar functionals does not imply equality of the two topologies; the closed graph theorem for Fréchet spaces is the extra step that makes this conclusion valid.

A separated dual pair consists of real vector spaces $E,F$ and a bilinear form $⟨x,y⟩$ that separates both variables: each nonzero $x\in E$ pairs nontrivially with some $y\in F$, and conversely. Thus $F$ embeds in the algebraic dual of $E$, and $E$ embeds in the algebraic dual of $F$.

The weak topology $\sigma (E,F)$ is the coarsest topology making all maps $x\mapsto ⟨x,y⟩$, $y\in F$, continuous. It makes $E$ a Hausdorff locally convex space, with topology generated by the seminorms $p_y(x)=|⟨x,y⟩|$. A neighbourhood base at zero consists of finite intersectionsSeparation ensures that the common zero set of all these seminorms is just zero, giving Hausdorffness.

We determine the continuous dual of a weak topology. Suppose a linear functional $\ell$ is continuous. A basic neighbourhood as above is contained in $\{x:|\ell (x)|<1\}$. If $x$ is in the common kernel of the finitely many evaluations, every scalar multiple of $x$ lies in that neighbourhood, so $\ell (x)=0$. Hence $\ell$ factors through the image ofExtend the resulting linear functional on $T(E)$ to ${\mathbb{R}}^m$. It has the form $z\mapsto {\sum }_j{a}_j{z}_j$, giving $\ell (x)=⟨x,{\sum }_j{a}_j{y}_j⟩$. Conversely every such evaluation is continuous by the definition of the weak topology. Since the pairing separates $F$, this identification is injective, andFor the remaining parts every topological assertion about $E^{\prime }$ or its subsets uses the stipulated weak-star topology $\sigma (E^{\prime },E)$.

Use the Hausdorff convention for a locally convex space: a real vector space $X$ is equipped with a separating family $\mathcal{P}$ of seminorms. Thus each $p$ is nonnegative, subadditive and satisfies $p(tx)=|t|p(x)$, and for every $x\ne 0$ some $p\in \mathcal{P}$ has $p(x)>0$. The topology has a neighborhood basis at $a\in X$ consisting ofEquivalently, it is the coarsest vector-space topology making all these seminorms continuous. The separating condition is precisely what makes this topology Hausdorff. If the Hausdorff requirement were omitted, continuous linear functionals could not distinguish points in the common kernel of the seminorms.

The continuous dual space $X^{\ast }$ consists of all continuous linear functionals $X\to \mathbb{R}$. We first prove the needed Hahn-Banach theorem. Let $p$ be a real-valued sublinear function on $X$, meaning $p(a+b)\le p(a)+p(b)$ and $p(ta)=tp(a)$ for $t\ge 0$, and let $g$ be linear on a vector subspace $M$, with $g(m)\le p(m)$. To extend across $v\notin M$, writeThe necessary bounds on $c$ areThey are compatible becauseTaking one variable equal to zero also shows that $L,U$ are finite. Choose $c\in [L,U]$. The upper bound proves domination when $t>0$, after dividing $m+tv$ by $t$; the lower bound proves it when $t<0$, after dividing by $-t$. Thus $\tilde{g}\le p$ on $M+\mathbb{R}v$. This is the one-dimensional dominated extension of a real linear functional.

Order all dominated extensions of $g$ by extension of their domains. A chain has an upper bound obtained by taking the union of the domains and linear functionals. Zorn's lemma gives a maximal extension, and the one-dimensional construction shows that its domain must be all of $X$. We have therefore proved the real dominated-extension theorem. In particular, when $p$ is a seminorm, domination at both $x$ and $-x$ gives $|f(x)|\le p(x)$.

For $x_0\ne 0$, choose a continuous seminorm $p$ with $p(x_0)>0$, and define $g(t{x}_0)=tp(x_0)$ on its one-dimensional span. Then $|g(t{x}_0)|\le p(t{x}_0)$. Extend by the theorem just proved to $f$ with $|f(x)|\le p(x)$. This bound makes $f$ continuous, and $f(x_0)=p(x_0)>0$. Applying this to the difference of two distinct points proves that $X^{\ast }$ separates the points of $X$, the continuous-dual separation theorem for Hausdorff locally convex spaces.

For separation from a closed linear subspace, choose a basic balanced neighborhood $V$ of zero such that $(x_0+V)\cap Y=\varnothing$. WriteThen $q(x_0-y)\ge 1$ for every $y\in Y$. On $Y+\mathbb{R}{x}_0$, define $g(y+t{x}_0)=t$, which is well-defined since $x_0\notin Y$. For $t\ne 0$,for $t=0$ the inequality is immediate. The proved extension theorem gives a continuous $f$ dominated by $q$, withThis proves separation of a point from a closed linear subspace without using any unproved Hahn-Banach extension or separation result.

All vector spaces in this solution are complex, while convex combinations use real coefficients. An extreme point $x$ of a convex set $C$ is one for which $x=ty+(1-t)z$, $y,z\in C$ and $0<t<1$, forces $y=z=x$. Equivalently, $x$ is not the midpoint of two distinct points of $C$.

The Krein-Milman theorem says that every nonempty compact convex set $K$ in a Hausdorff locally convex space is the closed convex hull of its extreme points. Here is a proof. A face of a convex set $F\subseteq K$ is a convex set such that whenever an interior point of a segment in $K$ belongs to $F$, both endpoints belong to $F$. Consider nonempty compact faces, ordered by reverse inclusion. A chain has nonempty intersection by compactness and the finite intersection property; that intersection is again a compact face. The Zorn lemma therefore gives a minimal nonempty compact face $F$.

If $F$ contains distinct $x,y$, a continuous real linear functional $\ell$ on the underlying real locally convex space distinguishes them. Such a linear functional exists because the space is Hausdorff and locally convex, by the Hahn-Banach theorem. The maximizers of $\ell$ on $F$ form a nonempty proper compact face of $F$. A face of a face is a face of $K$, contradicting minimality. Therefore $F$ is a singleton, yielding an extreme point. The same argument inside any nonempty compact face of $K$ supplies an extreme point of $K$ lying in that face.

Let $D$ be the closed convex hull of the extreme points of $K$. It is a nonempty closed subset of compact $K$, hence compact. If $x_0\in K\setminus D$, the Hahn-Banach separation theorem provides a continuous real linear functional $\ell$ with $\ell (x_0)>{\sup }_D\ell$. Its maximizer set on $K$ is a nonempty compact face, which contains an extreme point $e$ of $K$. But $e\in D$ and $\ell (e)\ge \ell (x_0)>{\sup }_D\ell$, a contradiction. Thus $\boxed{K=\overline{\mathrm{co}}(\mathrm{ext}K)}$, proving Krein-Milman theorem.

For a nonempty compact Hausdorff space $K$, the extreme points of the dual unit ball of C(K) areThis is the permitted description without proof, with $C(K)$ denoting the complex space of continuous functions on a compact space equipped with the supremum norm. If $K$ is empty, $C(K)=\{0\}$ and the closed unit ball of its continuous dual space has the single extreme point $0$ instead.

The complex Banach–Stone theorem states that a surjective complex-linear isometric isomorphism of normed spaces $T:C(K)\to C(L)$ has the formwhere $u\in C(L)$ and $\phi :L\to K$ is a homeomorphism. Conversely, every such map is a surjective complex-linear isometric isomorphism of normed spaces for the supremum norm.

To prove this, suppose first that $K,L$ are nonempty. The Banach-space adjoint $T^{\ast }$ is a bijective isometric isomorphism of normed spaces on the continuous dual spaces, so it bijects their closed unit balls and preserves extreme points. The displayed extreme point description gives uniquelyUniqueness follows by evaluating at the constant function $1$, and then using that continuous functions separate points of a compact Hausdorff space. Evaluation gives the required formula for $Tf$, and $u=T1$ is continuous. Surjectivity of $T^{\ast }$ on extreme points proves surjectivity of $\phi$: the preimage of any ${\delta }_x$ is $\alpha {\delta }_y$ for some $y$. If $\phi (y_1)=\phi (y_2)$, every function in the range of $T$ has equal values at these two points after division by $u$; surjectivity of $T$ and separation of points force $y_1=y_2$. Thus $\phi$ is bijective.

For every $f\in C(K)$, $f\circ \phi =(Tf)/u$ is continuous. The evaluation map $K\to {\mathbb{C}}^{C(K)}$, $x\mapsto (f(x){)}_f$, is a continuous injection of a compact Hausdorff space into a Hausdorff product topology, hence a homeomorphism onto its image. Continuity of every coordinate $f\circ \phi$ proves continuity of $\phi$. A continuous bijection between compact Hausdorff spaces is a homeomorphism. Conversely the weighted-composition formula plainly preserves the supremum norm, and its inverse isIf one compact space is empty, a surjective isometric isomorphism of normed spaces forces the other to be empty, and the empty homeomorphism gives the corresponding trivial case. This completes Banach–Stone theorem.

Neither the space of sequences converging to zero $c_0$ nor $L^1[0,1]$ can be isometrically a continuous dual space of a Banach space. Indeed, every nonzero continuous dual space has a nonempty weak-star compact closed unit ball by Banach-Alaoglu theorem, and Krein-Milman theorem then guarantees an extreme point. A bijective linear isometry preserves extreme points of closed unit balls.

The closed unit ball of $c_0$ has no extreme points. Given $x\in c_0$ with $\Vert x{\Vert }_{\infty }\le 1$, choose $n$ with $|x_n|<1$ and $0<\epsilon <1-|x_n|$. The two distinct elements $x\pm \epsilon e_n$ remain in that closed unit ball and have midpoint $x$.

The closed unit ball of the complex Lp space $L^1[0,1]$ also has no extreme points. An element of norm less than one can be perturbed by a sufficiently small nonzero Lp space element in both directions. If $\Vert f{\Vert }_1=1$, the non-atomic measure $|f(t)|dt$ admits a measurable set $A$ of mass $1/2$; for example the continuous function $s\mapsto {\int }_0^s|f(t)|dt$ attains $1/2$. Put $h=f(1_A-1_{A^c})$. For $0<\epsilon <1$, the distinct functions $f\pm \epsilon h$ both have Lp normwith the two coefficients interchanged for the minus sign. Their midpoint is $f$. Thus neither proposed space is isometrically a Banach dual.

Finally, $[0,1]$ is a connected space, whereas $[0,1]\cup [2,3]$ is disconnected. They cannot be homeomorphic. By Banach–Stone theorem, $C[0,1]$ and $C([0,1]\cup [2,3])$ are not isometrically isomorphic as complex Banach spaces.

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.

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.

Every continuous self-map of a nonempty compact convex subset of a locally convex space has a fixed point. Compactness is in the locally convex topology, which may be weaker than a norm topology. This version permits fixed-point arguments on a weak-star topology compact set of linear functionals. See Hartman's fixed-point chapter, Theorem 0.2.

Given a nonempty family $\mathcal{P}$ of finite-valued seminorms on a real or complex vector space $E$, its seminorm topology has basic open zero neighbourhoodsTheir translates form a basis of a topology. Only finite intersections are required: an infinite intersection might impose a stronger topology. Equivalently, put $q(x)={\max }_i{p}_i(x)/{\epsilon }_i$; then $q$ is a seminorm and $U=\{q<1\}$. See Günther Hörmann, Advanced Functional Analysis, section 4, neighbourhood systems from seminorms for the general construction.

Each defining seminorm is continuous, since $|p(x)-p(y)|\le p(x-y)$. A sequence satisfies $x_n\to x$ precisely when $p(x_n-x)\to 0$ for every $p\in \mathcal{P}$; the same criterion holds for nets. Finitely many measurements can each meet a prescribed tolerance after a common index.

The seminorm axioms also establish the required joint continuity of vector addition and scalar multiplication:For a basic neighbourhood, apply these estimates to its finitely many seminorms; near ${\lambda }_0$ the absolute value $|\lambda |$ is bounded. This gives a locally convex space even when some directions are invisible to every measurement.

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 a locally convex space $E$, the strong dual topology on its continuous dual space $E^{\prime }$ is generated by seminorms $p_B(T)={\sup }_{\phi \in B}|⟨T,\phi ⟩|$, where $B$ runs through the bounded sets in a topological vector space $E$. Convergence therefore means uniform convergence of pairings on every such set. For the Schwartz space, the defining boundedness is boundedness of every Schwartz seminorm, rather than one norm or one common compact support.