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The Banach space ${\ell }^1$ consists of absolutely summable scalar sequences, with norm $\Vert x{\Vert }_1={\sum }_n|x_n|$. Its continuous dual is ${\ell }^{\infty }$, under the pairing ${\sum }_n{x}_n{y}_n$. Finitely supported sequences are norm dense, which identifies a linear functional from its values on coordinate vectors. The space of sequences converging to zero inside this dual is a norming subspace of infinite codimension.

For example, $x=(2^{-1},2^{-2},\dots )$ belongs to the absolutely summable sequence space and has norm one, by the geometric series formula. Merely tending to zero is insufficient: $(1/n)$ tends to zero but its series of absolute values diverges. Thus this sequence space records total absolute size, rather than just the eventual size of individual coordinates. Its completeness ensures that a Cauchy sequence in this total-error norm has an absolutely summable limit.^[ref]

The analytic Wiener algebra consists of the power series $f(z)={\sum }_{n\ge 0}{a}_n{z}^n$ with ${\sum }_{n\ge 0}|a_n|<\infty$, identified with their values on the unit circle. The norm is $\Vert f\Vert =\sum |a_n|$. Discrete convolution gives multiplication and a submultiplicative norm; completeness follows from the absolutely summable sequence space. Each series converges uniformly on the closed unit disk and is holomorphic in its interior.

A Banach space is a normed vector space over the real numbers or complex numbers that is complete for the metric $d(x,y)=\Vert x-y\Vert$. Explicitly, whenever a sequence $(x_n)$ satisfiesthere is an element $x$ of the same space with $\Vert x_n-x\Vert \to 0$. The hypothesis says that the approximations eventually agree to any requested accuracy; completeness says that they describe an actual object in the space. The Cauchy sequence condition can be checked before anyone knows what that object is.

This makes a Banach space a natural setting for constructing solutions by successive improvements. If corrections $v_k$ have ${\sum }_k\Vert v_k\Vert <\infty$, their partial sums form a Cauchy sequence by the triangle inequality, and completeness supplies their sum. In fact, convergence of every such absolutely summable vector series characterizes a Banach space. The contraction mapping theorem uses precisely this mechanism: each correction is smaller than the preceding one by a fixed factor.^[ref]

Every normed vector space of finite dimension is a Banach space. Infinite-dimensional examples include the absolutely summable sequence space and the l-infinity sequence space; the choice of norm remains part of the specification. A collection of increasingly accurate finite descriptions can require an infinite description for its limit, as the finitely supported sequence example below demonstrates. Completion of a normed space adds these missing limits while preserving all original distances.

For the real l-infinity sequence space, every coordinate of an extreme point must be an endpoint of $[-1,1]$; any slack coordinate permits opposite perturbations. For the real absolutely summable sequence space, a unit vector with two nonzero coordinates permits a transfer of mass between them, so only signed coordinate vectors are extreme. For the real l2 sequence space, the parallelogram law makes every unit vector extreme. In all three cases interior points of the unit ball are nonextreme.

Equip the vector space $c_{00}$ of finitely supported sequences with the norm inherited from the absolutely summable sequence space. Consider the truncationsFor $M>N$, the geometric series givesTherefore $(x^{(N)})$ is a Cauchy sequence. Any possible norm limit must have coordinate $k$ equal to $2^{-k}$, since $|x_k-y_k|\le \Vert x-y{\Vert }_1$. That forces infinitely many nonzero coordinates, so no limit lies in $c_{00}$. There is nothing wrong with the approximations: the space is missing their limit.

Inside the absolutely summable sequence space, the limit is $x=(2^{-k}{)}_{k\ge 1}$ and its exact truncation error is $2^{-N}$. Four retained coordinates leave error $1/16$; ten leave $1/1024$. More generally, truncating any element of the absolutely summable sequence space leaves a tail whose norm tends to zero. Together with completeness, this identifies that Banach space as the completion of a normed space $c_{00}$ with this norm. Finite storage and an infinite mathematical object fit together through a controlled error.

A matrix summability method takes the ordinary limit of a sequence of the row sums $A_i(x)$, whenever every row series converges and the resulting sequence has a finite limit. For rows in the absolutely summable sequence space with uniformly bounded norms, the matrix defines a bounded linear operator from the l-infinity sequence space to itself.

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.

A point $x$ of a set $S$ is an extreme point if $x=(1-t)y+tz$, with $y,z\in S$ and $0<t<1$, forces $y=z=x$. For a convex set it is enough to test midpoint decompositions: a nontrivial segment containing $x$ in its interior contains a smaller segment with midpoint $x$. The nonempty convex set ${\mathbb{R}}^2$ has no extreme points, since every $x$ is the midpoint of $x+e_1$ and $x-e_1$.

For the real l-infinity sequence space, a vector in the closed unit ball is extreme precisely when every coordinate is $1$ or $-1$. If $|x_j|<1$ for some $j$, choose $0<\delta \le 1-|x_j|$. The distinct vectors $x\pm \delta e_j$ remain in the unit ball and have midpoint $x$, so $x$ is not extreme. Conversely suppose all coordinates have modulus one and $x=(y+z)/2$, with $\Vert y{\Vert }_{\infty },\Vert z{\Vert }_{\infty }\le 1$. In each real interval $[-1,1]$, an endpoint can be a midpoint only if both entries equal that endpoint. Thus $y_j=z_j=x_j$ for every $j$ and $y=z=x$. Therefore

For the real absolutely summable sequence space, any point with $\Vert x{\Vert }_1<1$ is nonextreme: choose a nonzero coordinate perturbation of norm less than $1-\Vert x{\Vert }_1$. If $\Vert x{\Vert }_1=1$ and two coordinates $x_j,x_k$ are nonzero, take $0<\delta <\min (|x_j|,|x_k|)$ and putBoth perturbations preserve the signs of those two coordinates. One adds $\delta$ to one absolute value and subtracts it from the other, so $\Vert x+h{\Vert }_1=\Vert x-h{\Vert }_1=1$. Hence such an $x$ is not extreme. The only remaining candidates are $\pm e_j$. If $e_j=(y+z)/2$ with $\Vert y{\Vert }_1,\Vert z{\Vert }_1\le 1$, its $j$th coordinate forces $y_j=z_j=1$. That consumes the entire norm allowance, forcing all other coordinates to vanish. The argument for $-e_j$ is identical. Thus

For the real l2 sequence space, interior points of the unit ball again permit sufficiently small opposite perturbations and are not extreme. If $\Vert x{\Vert }_2=1$ and $x=(y+z)/2$, write $y=x+h$, $z=x-h$. The parallelogram law givesThe left side is at most one, so $h=0$ and $y=z=x$. ConsequentlyThese are the extreme points of real sequence-space unit balls: signs in ${\ell }^{\infty }$, coordinate spikes in ${\ell }^1$, and the whole sphere in ${\ell }^2$. The restriction to real scalars matters for the first description; the complex supremum ball has arbitrary unit-modulus coordinates instead.

Suppose the weak-star closure of $Z$ were a proper linear vector subspace of $X^{\ast }$. The Hahn-Banach separation theorem would give a nonzero weak-star continuous linear functional vanishing on it. A continuous dual of a weak-star topology consists precisely of evaluations at points of $X$: continuity bounds the linear functional by finitely many evaluations, so it factors through their finite-dimensional coordinate map and is a linear combination of them. Thus some $0\ne x\in X$ would satisfy $z(x)=0$ for all $z\in Z$. The norming inequality would imply $c\Vert x\Vert \le 0$, a contradiction. Every norming subspace of $X^{\ast }$ is weak-star dense.

For the infinite-codimension example, takeHere $X$ is the absolutely summable sequence space and $c_0$ is the space of sequences converging to zero, a norm-closed vector subspace of ${\ell }^{\infty }$. The dual identification is $y(x)={\sum }_n{x}_n{y}_n$: a bounded sequence defines a linear functional of norm $\Vert y{\Vert }_{\infty }$, and every linear functional on ${\ell }^1$ has this form by evaluating on the coordinate vectors and using density of finitely supported sequences.

For $x\in {\ell }^1$, use the finitely supported sequence $y^{(N)}$ whose first $N$ entries are $\mathrm{sgn}(x_n)$ and whose remaining entries vanish. It belongs to $c_0$, has norm at most one, andThe reverse inequality follows from $|y(x)|\le \Vert y{\Vert }_{\infty }\Vert x{\Vert }_1$. Hence $c_0$ is 1-norming for ${\ell }^1$, an instance of vanishing sequences norm the summable sequence space.

To prove infinite codimension, take disjoint infinite setsTheir indicator sequences have linearly independent classes in ${\ell }^{\infty }/c_0$. Indeed, a finite combination has the constant value $a_j$ on $A_j$; if it tends to zero, every $a_j$ must vanish. Thus the quotient is infinite-dimensional.

Finite truncations of the sign or phase sequence of $x\in {\ell }^1$ belong to $c_0$ and attain increasing partial sums of ${\sum }_n|x_n|$. Thus the space of sequences converging to zero is 1-norming for the absolutely summable sequence space. Its codimension in ${\ell }^{\infty }$ is infinite: indicators of pairwise disjoint infinite subsets have linearly independent classes modulo $c_0$. Norming does not require finite codimension.