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A Banach limit is a positive linear functional on the real l-infinity sequence space that extends the ordinary limit on the convergent sequence space, has operator norm one, and is invariant under the left shift on bounded sequences. The Hahn-Banach theorem constructs one by separating the constant sequence $1$ from the subspace $(I-S){\ell }^{\infty }$, whose distance from $1$ is one. The functional is generally not unique.

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 an arbitrary set $\Gamma$ and scalar field $\mathbb{F}$, this is the space of bounded functions $u:\Gamma \to \mathbb{F}$ with the supremum norm $\Vert u\Vert ={\sup }_{\gamma }|u(\gamma )|$. Uniform limits of Cauchy sequences prove it is a Banach space. For $\Gamma =\mathbb{N}$ it is the l-infinity sequence space; for the empty index set it is the zero space, with norm zero.

The displayed sublinear functional on the real l-infinity sequence space agrees with the ordinary limit on convergent sequences. A dominated extension by the Hahn-Banach theorem therefore defines a generalized limit bounded between the lower and upper limits of the Cesàro means. For the left shift on bounded sequences $S$, the averages of $Sx-x$ equal $(x_{N+1}-x_1)/N$, which tend to zero. Domination applied to both signs gives $L(Sx-x)=0$, hence shift invariance. Positivity, agreement on constant sequences and the operator norm one follow from the bounds, so $L$ is a Banach limit.

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.

The left shift is a bounded linear operator of operator norm one on the l-infinity sequence space. For $y=x-Sx$, telescoping gives $N^{-1}{\sum }_{n=1}^N{y}_n=(x_1-x_{N+1})/N\to 0$. This identity is useful in constructing a Banach limit.

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.

Use the dominated form of the real Hahn-Banach theorem. Let $p:V\to \mathbb{R}$ be a sublinear functional on a real vector space, meaning $p(x+y)\le p(x)+p(y)$ and $p(tx)=tp(x)$ for $t\ge 0$. If a linear functional $f$ on a vector subspace $M$ satisfies $f\le p$ there, then it has a linear extension $F$ to $V$ with $F\le p$ everywhere.

Here is the one-dimensional extension step. For $z\notin M$, the value $a=F(z)$ must satisfyThis interval is nonempty: for any $m,n\in M$,so every candidate lower bound is at most every candidate upper bound. The candidates with $m=n=0$ also bound the endpoints by finite numbers, so a real $a$ can be chosen. Define $F(m+tz)=f(m)+ta$. For $t>0$, the upper bound, applied to $m/t$, gives $F(m+tz)\le p(m+tz)$ after multiplication by $t$. For $t<0$, the lower bound applied to $m/(-t)$ gives the same conclusion after multiplication by $-t$; $t=0$ is the original hypothesis. Thus the extension remains dominated.

Order all dominated extensions by extension of their domains and values. Every chain has the dominated linear union as an upper bound. The Zorn lemma supplies a maximal extension. If its domain were not all of $V$, the step just proved would enlarge it, a contradiction. This proves the real Hahn-Banach extension theorem. In particular, taking $p(x)=\Vert f\Vert \Vert x\Vert$ on a normed vector space and applying the domination to both $x$ and $-x$ yields the usual norm-preserving extension of a bounded linear functional.

For generalized limits, take $V={\ell }^{\infty }(\mathbb{N};\mathbb{R})$, the l-infinity sequence space, and $M=c$, the convergent sequence space. Let $f(x)={\lim }_n{x}_n$ on $c$ and $p(x)={\mathrm{limsup}}_n{x}_n$ on $V$. This $p$ is finite and sublinear, and agrees with $f$ on $c$. The dominated extension $L$ therefore satisfiesThus a generalized limit exists, withIt is positive, agrees with the ordinary limit on every convergent sequence, and satisfies $|L(x)|\le \Vert x{\Vert }_{\infty }$. Since $L(1)=1$, its operator norm is exactly one.

If generalized limits are additionally required to be shift invariant, the same theorem gives the stronger Banach limit. Replace the dominating functional byIt is again sublinear and agrees with $f$ on $c$, by the Cesàro theorem for convergent sequences. The resulting $L_C$ lies between the lower and upper limits of the Cesàro means. These in turn lie between the ordinary lower and upper limits of a bounded sequence, so $L_C$ is a positive, norm-one extension of the ordinary limit. For the left shift on bounded sequences $S$, the means of $Sx-x$ equal $(x_{N+1}-x_1)/N\to 0$. Applying domination to both signs forces $L_C(Sx-x)=0$. HenceThis Cesàro construction of a Banach limit proves existence even under the stronger convention, rather than leaving shift invariance implicit.

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 convex set $K$ is an extreme point if $x=(1-t)a+tb$ with $a,b\in K$ and $0<t<1$ forces $a=b=x$. It is enough to test midpoints: a nontrivial interior point of a segment is the midpoint of a smaller nontrivial segment.

Here is a proof of the Krein-Milman theorem. The empty-set case is immediate, so take $K\ne \varnothing$. A face of a convex set is a convex subset $F\subseteq K$ such that an interior convex combination lying in $F$ has both endpoints in $F$. Consider nonempty compact faces. They include $K$. Every inclusion chain has nonempty intersection, because its closed subsets of the compact set $K$ have the finite intersection property. The intersection remains a compact face. The Zorn lemma, ordered by reverse inclusion, therefore supplies a minimal nonempty compact face $F$.

If $F$ contains two distinct points, the Hausdorff locally convex hypotheses and the allowed separation theorem supply a continuous real linear functional $\ell$ taking different values there. Its maximizer set in $F$ is nonempty and compact. It is convex, and if a convex combination attains the maximum, both endpoint values attain it too; hence it is a face of $F$. A face of a face is a face of $K$, by applying the defining endpoint property twice. This maximizer face is proper because $\ell$ is nonconstant, contradicting minimality. Thus $F$ is a singleton, whose point is extreme in $K$. The same minimal compact face argument inside any nonempty compact face shows that each such face contains an extreme point of $K$.

Let $H=\overline{\mathrm{conv}}(\mathrm{ext}K)$. Since $K$ is compact, hence closed in the Hausdorff space, and convex, $H\subseteq K$. It is nonempty, closed and convex. If $p\in K\setminus H$, the allowed strict separation theorem supplies a continuous linear functional withThe maximizer face of $\ell$ on $K$ contains an extreme point $e$ by the preceding argument. Then $e\in H$ but $\ell (e)={\max }_K\ell \ge \ell (p)>{\sup }_H\ell$, a contradiction. ThereforeAll closures refer to the given locally convex topology, and compactness is what justified both maxima and chain intersections.

For the real l-infinity sequence space, if $|x_j|<1$ for any coordinate, choose $0<ϵ\le 1-|x_j|$. The two distinct sequences $x\pm ϵe_j$ belong to the closed unit ball and have midpoint $x$, so $x$ is not extreme. Conversely, if every $x_j$ is one or minus one, writing $x=(1-t)a+tb$ for unit-ball sequences forces $a_j=b_j=x_j$ at every coordinate, because an endpoint of $[-1,1]$ cannot be a nontrivial convex average of its points. ThusFor the space of sequences converging to zero, every unit-ball sequence has some coordinate with $|x_j|<1$, since its coordinates tend to zero. The same opposite single-coordinate perturbations still belong to $c_0$, givingThese are extreme points of real sequence-space unit balls. There is no conflict with Krein-Milman: the norm-closed unit ball of $c_0$ is not compact in the norm topology, as the coordinate vectors have mutual distance one.

The Baire category theorem states that, in a nonempty complete metric space, a countable intersection of open dense sets is dense. Equivalently, a nonempty open set cannot be covered by countably many nowhere dense sets.

For the Uniform boundedness principle, let $X$ be a Banach space, $Y$ a normed vector space, and $(T_{\alpha })$ a family of bounded linear operators for which ${\sup }_{\alpha }\Vert T_{\alpha }x\Vert <\infty$ for every $x\in X$. PutEach $F_n$ is closed, being an intersection of inverse images of closed balls, and $X={\bigcup }_n{F}_n$. By the Baire category theorem, some $F_n$ has interior. Choose a closed ball $\overline{B}(x_0,r)\subseteq F_n$ with $r>0$. For $\Vert h\Vert \le r$, both $x_0$ and $x_0+h$ belong to $F_n$, and linearity gives $\Vert T_{\alpha }h\Vert \le 2n$. Scaling yieldsCompleteness is needed for the domain, not the codomain.

We next prove the Silverman-Toeplitz theorem. Write $c$ for the real convergent sequence space with the supremum norm. It is a Banach space: the l-infinity sequence space is complete by coordinatewise passage to limits of norm-Cauchy sequences, and $c$ is closed in it. Indeed, if $x^{(m)}\to x$ uniformly and each $x^{(m)}$ converges, approximate $x$ within $\epsilon /3$ by one $x^{(m)}$ and make that sequence's tail oscillation smaller than $\epsilon /3$. Then the tail of $x$ is Cauchy and therefore converges in $\mathbb{R}$.

Suppose $A$ is a regular summability matrix. Fix a row $i$ and consider the finite bounded linear functionalsFor every $x\in c$ their values converge, by the definition of regularity, and are therefore bounded. Their operator norms are exactly ${\sum }_{j\le N}|a_{ij}|$: the upper bound is the triangle inequality, and the finite sign sequence $x_j=\mathrm{sgn}(a_{ij})$ for $j\le N$, zero afterwards, attains it. Applying the Uniform boundedness principle to this row shows ${\sum }_j|a_{ij}|<\infty$.

Consequently the full row functional $A_i:c\to \mathbb{R}$ is continuous and has norm ${\sum }_j|a_{ij}|$, using the same finite sign tests and then letting $N\to \infty$. Regularity makes $(A_i(x){)}_i$ convergent, hence bounded, for every $x\in c$. A second application of the Uniform boundedness principle gives a common row bound $M$. Testing regularity on a coordinate vector $e_j$ and on the constant-one sequence gives, respectively, convergence to zero down each column and convergence of row sums to one. Thus the necessary conditions are

Conversely assume these conditions and let $x_j\to L$. The sequence is bounded, so every row sum defining $A_i(x)$ converges absolutely. WriteGiven $\epsilon >0$, choose $N$ such that $|x_j-L|<\epsilon /(M+1)$ for $j>N$. The tail of the first sum has modulus at most $M\epsilon /(M+1)<\epsilon$, uniformly in $i$. Its finite head tends to zero by columnwise convergence, and $L(s_i-1)\to 0$. Since $\epsilon$ is arbitrary, $A_i(x)\to L$. The three conditions are necessary and sufficient for regularity.

For the averaging matrix $B$, every absolute row sum and every row sum equals $1$. A fixed column eventually has entry $1/r\to 0$. Thus $B$ is a regular summability matrix, and its transformed values are the Cesàro means. To exhibit a bounded sequence without a Cesaro limit, put $N_0=0$, $N_m=2^{2^m}$ for $m\ge 1$, and setAt the end of the $m$th block,The averages at even block endpoints approach $1$, while those at odd endpoints approach $-1$. Hence $w$ is bounded and has no $B$-limit.

For an arbitrary regular summability matrix $A$, use a gliding hump argument; this also handles negative entries. Put $N_0=0$ and $i_0=0$. Inductively choose $i_m>i_{m-1}$ so large thatThis is possible by the finite-column convergence and convergence of the row sums to one. Since the selected row is absolutely summable, choose $N_m>N_{m-1}$ withDefine a single bounded sequence on the successive disjoint blocks byusing $\mathrm{sgn}(0)=0$. The row's absolute mass on its own block is greater than $3/4-1/8=5/8$, while all coordinates outside that block contribute a term of modulus less than $1/8$, irrespective of the signs subsequently assigned. ThereforeEven and odd selected rows cannot tend to the same limit. Every row series does converge absolutely because $\Vert y{\Vert }_{\infty }\le 1$. This proves bounded divergence for every regular summability matrix.

The row bound makes $T_A:{\ell }^{\infty }\to {\ell }^{\infty }$, $T_{A}x=(A_i(x){)}_i$, a bounded linear operator with $\Vert T_A\Vert \le M$. The summability domain of a regular matrix isSince $c$ is closed, this is a closed linear subspace. It is proper by the constructed $y$. A proper vector subspace of a normed vector space has empty interior: a ball in the subspace can be translated to a ball about zero, and scaling that ball would put every vector in the subspace. ThusFor a countable family of regular matrices, the Baire category theorem on the Banach space ${\ell }^{\infty }$ givesAny vector in this complement has none of the indicated matrix limits. The matrices in this clause must be regular, as in the preceding discussion; without that implicit qualification even a single zero matrix sums every bounded sequence to zero.

Finally take any bounded sequence $x$. By the Bolzano-Weierstrass theorem it has a subsequence $x_{n_i}\to L$ with $n_i$ strictly increasing. Define the selector matrixIts absolute row sums and row sums are $1$, and each fixed column is eventually zero. The Silverman-Toeplitz theorem makes it regular, while $A_i(x)=x_{n_i}\to L$. Thus every bounded sequence is summable by some regular matrix depending on that sequence. This is subsequence selection as regular matrix summability; it reverses the quantifier order of the impossibility result for a fixed matrix.

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.

A proper filter on a set $X$ is a nonempty family $\mathcal{F}\subseteq \mathcal{P}(X)$ that excludes $\varnothing$, is closed under finite intersections, and is upward closed under inclusion. In particular $X\in \mathcal{F}$. In a topological space, convergence of a filter $\mathcal{F}\to x$ means that every neighbourhood of $x$ belongs to $\mathcal{F}$. An ultrafilter is a proper filter maximal under inclusion; equivalently, it contains exactly one of $A$ and $X\setminus A$ for every subset $A\subseteq X$.

To justify the equivalence, if $A\notin \mathcal{U}$ and every member of $\mathcal{U}$ met $A$, adjoining $A$ would generate a larger proper filter. Maximality therefore supplies $F\in \mathcal{U}$ with $F\cap A=\varnothing$, and upward closure gives $X\setminus A\in \mathcal{U}$. Both complementary sets cannot belong to a proper filter. Conversely, a filter with this dichotomy cannot acquire any new set without also containing its complement, which would give an empty intersection.

The ultrafilter characterization of compact Hausdorff spaces, stated without proof, is: a topological space is compact and Hausdorff exactly when every ultrafilter on it converges to a unique point.

For a finite set partition $X=A_1\cup \cdots \cup A_n$, if no cell belonged to $\mathcal{U}$, the ultrafilter dichotomy would put all $X\setminus A_j$ in $\mathcal{U}$. Their finite intersection is empty, a contradiction. Thus some $A_j\in \mathcal{U}$, and at most one cell can belong because distinct cells are disjoint. The converse is: a proper filter on a set meeting every finite partition by containing one cell is an ultrafilter. Apply the partition hypothesis to $X=A\cup (X\setminus A)$ to obtain the dichotomy, then the maximality argument above. Empty cells can be omitted, and the cases $A=\varnothing$ or $A=X$ are automatic.

For a function $f:X\to Y$, the image of a filter isSince preimages preserve inclusions and finite intersections, this is a proper filter on a set $Y$. Equivalently, it is the filter generated by $f(F)$ for $F\in \mathcal{F}$: if $f^{-1}(B)\in \mathcal{F}$, then its image is contained in $B$, while $f(F)\subseteq B$ implies $F\subseteq f^{-1}(B)$. For an ultrafilter $\mathcal{U}$, the identity $f^{-1}(Y\setminus B)=X\setminus f^{-1}(B)$ shows that $f_{\ast }\mathcal{U}$ contains exactly one of $B,Y\setminus B$. Therefore the pushforward filter of an ultrafilter is an ultrafilter.

Now let $x\in {\ell }^{\infty }$, the l-infinity sequence space, and put $a_x(n)=n^{-1}{\sum }_{j=1}^n{x}_j$. All these Cesàro means lie in the closed disk $K_x=\{z:|z|\le \Vert x{\Vert }_{\infty }\}$, a compact Hausdorff space. The pushforward filter $a_{x\ast }\mathcal{U}$ contains $K_x$; restricting it to that disk and applying the ultrafilter theorem gives a unique limitFor real sequences the disk can be replaced by a closed interval. This is the ultrafilter construction of a Banach limit.

The identities $a_{\alpha x+\beta y}(n)=\alpha a_x(n)+\beta a_y(n)$ give linearity of $l$. Explicitly, the two sets on which $|a_x(n)-l(x)|<\delta$ and $|a_y(n)-l(y)|<\delta$ both belong to $\mathcal{U}$, hence so does their intersection. On that intersection the difference from $\alpha l(x)+\beta l(y)$ is at most $(|\alpha |+|\beta |)\delta$, proving $l(\alpha x+\beta y)=\alpha l(x)+\beta l(y)$. If $x_n\ge 0$, its means lie in $[0,\Vert x{\Vert }_{\infty }]$, so $l(x)\ge 0$. Thus $l$ is a positive linear functional. The bound $|l(x)|\le \Vert x{\Vert }_{\infty }$ gives $\Vert l\Vert \le 1$, while the constant-one sequence satisfies $l(1)=1$. Hence $\boxed{\Vert l\Vert =1}$.

Because $\mathcal{U}$ contains no finite sets, its dichotomy puts every cofinite set in $\mathcal{U}$. If $x_n\to L$ ordinarily, the Cesàro theorem for convergent sequences follows directly by splitting the sum: for a fixed $N$ with $|x_j-L|<\epsilon$ for $j>N$,The finite head tends to zero, so $a_x(n)\to L$ ordinarily. Every sufficiently long tail belongs to $\mathcal{U}$, making its ultralimit also $L$. Thus $\boxed{x_n\to L⟹l(x)=L}$.

Finally, for the left shift on bounded sequences $S$,Taking ultralimits along the same cofinite-containing ultrafilter gives $\boxed{l(Sx)=l(x)}$. This establishes every defining property of a Banach limit.

For $y=x-Sx\in Y$, the telescoping arithmetic means satisfybecause $x$ is a bounded sequence. Hence $\Vert 1-y{\Vert }_{\infty }\ge |1-N^{-1}{\sum }_{n=1}^N{y}_n|$ for every $N$, and passage to the limit gives $\Vert 1-y{\Vert }_{\infty }\ge 1$. Taking $y=0$ shows $d(1,Y)=1$. The proved Hahn-Banach distance formula therefore yields a bounded linear functional $L$ on the real l-infinity sequence space withWe verify that this single linear functional also has the two remaining properties.

We prove the required Hahn-Banach theorem for a seminorm directly. First suppose the vector space is real. At an intermediate extension stage, let $g:E\to \mathbb{R}$ be a linear functional with $|g(y)|\le p(y)$, and let $z\notin E$. Extending to $E+\mathbb{R}z$ amounts to choosing $c=f(z)$. The upper domination $f\le p$ requiresThe interval is nonempty: for all $y,w\in E$,Its endpoints are finite since inserting zero gives bounds $-p(z)$ and $p(z)$, while every lower candidate is below every upper candidate. Choose $c$ in the interval and put $f(y+tz)=g(y)+tc$. For $t>0$, the upper bound with $w=y/t$ proves $f(y+tz)\le p(y+tz)$. For $t<0$, writing $s=-t>0$ and applying the lower bound with $y/s$ proves the same inequality. The case $t=0$ is already known. Applying this domination to $-x$ and using $p(-x)=p(x)$ gives $|f(x)|\le p(x)$.

Partially order all dominated linear functional extensions of the original $g$ by extension of their domains and values. They form a nonempty set. The union along any chain is a well-defined dominated linear functional on a vector subspace, so every chain has an upper bound. The Zorn lemma gives a maximal extension. If its domain were not $X$, the preceding one-dimensional construction would enlarge it, a contradiction. Thus the required extension exists on $X$.

For a complex vector space, apply the proved real result to $\mathrm{Re}g$ on the underlying real vector subspace. Let $U:X\to \mathbb{R}$ be the resulting real linear functional, with $|U(x)|\le p(x)$, and setIt is additive and real-linear, and $f(ix)=if(x)$, so it is complex-linear. On $Y$, the identity $\mathrm{Re}g(iy)=-\mathrm{Im}g(y)$ proves $f(y)=g(y)$. Choose $\alpha$ with $|\alpha |=1$ and $\alpha f(x)=|f(x)|$ when $f(x)\ne 0$. ThenThe bound is immediate if $f(x)=0$. This proves the real and complex seminorm forms of Hahn-Banach theorem without invoking any version of that theorem.

For the distance to a set assertion, let $d=d(z,Y)>0$. Define on $Y+\mathbb{K}z$ the linear functional $g(y+\alpha z)=\alpha d$. The decomposition is unique because $z\notin Y$, and for $\alpha \ne 0$,For $\alpha =0$ the bound is immediate. Apply the just-proved Hahn-Banach theorem with $p(x)=\Vert x\Vert$ to get $f\in X^{\ast }$, $f{|}_Y=0$, $f(z)=d$, and $\Vert f\Vert \le 1$. Choose $y_n\in Y$ with $\Vert z-y_n\Vert \to d$. Sinceand $d>0$, passage to the limit gives $\Vert f\Vert \ge 1$. Thus $\boxed{f{|}_Y=0,f(z)=d(z,Y),\Vert f\Vert =1}$. This version of the Hahn-Banach distance formula needs neither a closed $Y$ nor an attained distance to a set.

For the Banach limit construction below, take the real l-infinity sequence space, let $S(x_1,x_2,\dots )=(x_2,x_3,\dots )$ be its left shift on bounded sequences, put $Y=(I-S){\ell }^{\infty }$, and let $1=(1,1,\dots )$. This is a vector subspace and its distance to a set from $1$ determines the construction.

The Hahn-Banach theorem for bounded linear functionals says that a bounded linear functional on any linear subspace of a normed vector space extends to the whole space with its norm unchanged. No closedness or completeness of the subspace is required. Question 1 uses real-valued ${\ell }_{\infty }(\Gamma )$, so its operator assertions are read over the real scalar field. The analogous complex statements use complex-valued indexed functions.

For $x\ne 0$, define $g(tx)=t\Vert x\Vert$ on its one-dimensional span. This is a norm-one functional, so Hahn-Banach theorem supplies an extension $f\in X^{\ast }$ withThe canonical embedding into the bidual is $Jx(f)=f(x)$. It is linear and $\Vert Jx\Vert \le \Vert x\Vert$. The bounded linear functional just constructed gives the reverse inequality for nonzero $x$, and the zero case is immediate. Thus $\boxed{\Vert Jx\Vert =\Vert x\Vert }$ and $J$ is injective.

For the coordinate functional representation of an operator into bounded indexed functions, let $e_{\gamma }$ evaluate a coordinate and put $f_{\gamma }=e_{\gamma }T$. If $T$ is bounded and linear, then $f_{\gamma }\in X^{\ast }$ and $\Vert f_{\gamma }\Vert \le \Vert T\Vert$. Conversely, if $M={\sup }_{\gamma }\Vert f_{\gamma }\Vert <\infty$, the formula $(Tx)(\gamma )=f_{\gamma }(x)$ defines a bounded scalar function for each $x$, is linear, and satisfies $\Vert Tx{\Vert }_{\infty }\le M\Vert x\Vert$. Combining the two estimates givesFor an empty index set both spaces/families have norm bound zero; the supremum of the empty nonnegative family is taken as zero.

Choose $\Gamma =B_{X^{\ast }}$ and $Tx=(f(x){)}_{f\in B_{X^{\ast }}}$. The Hahn-Banach theorem makes $\Vert Tx{\Vert }_{\infty }=\Vert x\Vert$, so this is a linear isometric embedding into bounded scalar functions on an index set.

If $X$ is nonzero and separable, choose a dense sequence $u_n$ in its unit sphere and supporting functionals $f_n$ with $\Vert f_n\Vert =f_n(u_n)=1$. For a unit vector $u$ and $ϵ>0$, some $u_n$ satisfies $\Vert u-u_n\Vert <ϵ$, whence $|f_n(u)|\ge 1-ϵ$. Thus $(f_n)$ is a countable norming family and $x\mapsto (f_n(x))$ is an isometry into the l-infinity sequence space. Completeness of $X$ is not needed.

If instead $X=E^{\ast }$ for a separable normed vector space $E$, choose a dense sequence $v_n$ in $B_E$. The evaluations $f_n(x)=x(v_n)$ lie in $B_{X^{\ast }}$ and continuity of $x\in E^{\ast }$ gives ${\sup }_n|x(v_n)|=\Vert x\Vert$. This again gives $\Gamma =\mathbb{N}$, even when $E^{\ast }$ is not separable. If $X=\{0\}$, use zero coordinates throughout.

To prove 1-injectivity, take $T:Y\to {\ell }_{\infty }(\Gamma )$ on a subspace of $Z$. Extend every coordinate $f_{\gamma }\in Y^{\ast }$ to ${\tilde{f}}_{\gamma }\in Z^{\ast }$ by Hahn-Banach theorem, keeping its norm. Their uniform bound defines $\tilde{T}z=({\tilde{f}}_{\gamma }(z))$, and the coordinate norm identity yieldsSelection of these extensions uses the usual choice convention for arbitrary index sets. This verifies the lambda-injective normed space definition with $\lambda =1$.

For the final retraction characterization of lambda-injectivity, suppose first that $X$ is lambda-injective and $i:X\to Z$ is a linear isometry. The inverse $i(X)\to X$ has norm one when $X\ne 0$; extend it to $P:Z\to X$ with $\Vert P\Vert \le \lambda$. Then $Pi=I_X$, and $iP$ is a bounded projection onto $i(X)$. For the zero space take $P=0$.

Conversely, suppose every linear isometry out of $X$ has such a left inverse. Fix an isometry $j:X\to {\ell }_{\infty }(\Gamma )$ as above and a left inverse $P$ with $\Vert P\Vert \le \lambda$. For any $T:Y\to X$, extend $jT$ to $S:Z\to {\ell }_{\infty }(\Gamma )$ by 1-injectivity. Then $\tilde{T}=PS$ extends $T$, and $\Vert \tilde{T}\Vert \le \lambda \Vert S\Vert =\lambda \Vert T\Vert$. Therefore $X$ is lambda-injective exactly when every isometric embedding $i:X\to Z$ admits a bounded map $P:Z\to X$ satisfying

The coordinate evaluations $e_n:x\mapsto x_n$ belong to the dual unit ball of the l-infinity sequence space and separate its points. Therefore the countable separating family metrizes a weakly compact set result makes any nonempty weakly compact $K\subset {\ell }_{\infty }$ a compact metric space in its weak topology. Choose finite $1/n$-nets and take their countable union $D$; it is weakly dense in $K$.

It follows that $K\subset {\overline{D}}^w\subset {\overline{\mathrm{conv}D}}^w$. By Mazur theorem, the last set is ${\overline{\mathrm{conv}D}}^{\Vert \cdot \Vert }$. Finite convex combinations of members of $D$ with nonnegative rational coefficients summing to one form a countable norm-dense subset of $\mathrm{conv}D$: for each fixed finite list, approximate its coefficients in the simplex by rational coefficients, and use the norm triangle inequality. Thus its norm closure is a separable metric space containing $K$.

Every subset of a separable metric space is separable: balls centred on a countable dense set with positive rational radii form a countable base; intersect this base with the subset and choose one point from each nonempty intersection. Consequently the weakly compact subsets of l-infinity are norm separable conclusion isThe empty case is immediate. The convexification matters: the chosen weakly dense set $D$ itself need not be norm dense in a nonconvex $K$.

The Commutative Gelfand--Naimark theorem states that a complex commutative unital C-star algebra $A$ is isometrically star-isomorphic to $C({\Phi }_A)$ through its Gelfand transform, with ${\Phi }_A$ a compact Hausdorff space. The l-infinity sequence space is such an algebra under coordinatewise multiplication and conjugation, with the supremum norm. Thereforegives the required isometric isomorphism.

Uniqueness as a Banach space representation uses the Banach–Stone theorem, rather than just uniqueness as an algebra representation. To justify the relevant theorem, the Riesz-Markov-Kakutani representation theorem givesthe extreme points of the dual unit ball of C(K). A norm-one measure whose variation measure is not concentrated at one point splits into two normalized restrictions to disjoint sets of positive variation, and so is not extreme. Conversely, equality in the variation bound shows that any decomposition of $\alpha {\delta }_t$ into the average of two dual-unit-ball elements forces both to be $\alpha {\delta }_t$: after removing the phase, the two measures must be positive and their average is concentrated at $t$.

If $U:C(K)\to C(L)$ is a surjective linear isometry, its dual map preserves these extreme points and their scalar orbits. Hence $U^{\ast }{\delta }_s=u(s){\delta }_{h(s)}$ for a bijection $h:L\to K$ and $|u(s)|=1$. Evaluating at $1$ gives the continuous function $u=U1$, whileshows that $h$ is continuous, since continuous functions determine the topology of a compact Hausdorff space. It is consequently a homeomorphism. Thus any other compact space representing ${\ell }^{\infty }$ is homeomorphic to this $K$. The Banach–Stone theorem is also stated in Leonard Tomczak's notes on András Zsák's functional analysis lectures.

Embed $\mathbb{N}$ by the evaluation characters $\iota (n)(a)=a_n$. They are distinct because the coordinate indicator functions $e_n$ distinguish them. Moreover, $e_n^2=e_n$, so every character has $\chi (e_n)\in \{0,1\}$. If $\chi (e_n)=1$, thenand hence $\chi =\iota (n)$. It follows that $\{\iota (n)\}=\{\chi :|\chi (e_n)-1|<1/2\}$ is open in $K$. Thus $\iota$ is a homeomorphism from the discrete natural numbers onto its image.

To prove density, suppose $D=\overline{\iota (\mathbb{N})}\ne K$. The Urysohn lemma provides a nonzero $v\in C(K)$ vanishing on $D$. Write $v=\Gamma (a)$. Then $a_n=v(\iota (n))=0$ for all $n$, so $a=0$ and $v=0$, a contradiction. Every bounded function $b:\mathbb{N}\to \mathbb{C}$ is an element of ${\ell }^{\infty }$, and $\Gamma (b)$ is its continuous extension to $K$. Density makes this extension unique. ThereforeThis identifies $K$ with the Stone-Čech compactification of the natural numbers.

For $x=(x_n)\in {\ell }^p$ with $1\le p<\infty$, letThenbecause the defining nonnegative series converges. Thus $F=c_{00}$ is dense in the l-p sequence space for every finite $p$; this is the density of finitely supported sequences in l-p.

It is not dense in the l-infinity sequence space. For the constant sequence $1=(1,1,\dots )$ and any finitely supported $y$, some tail of $y$ is zero, soHence $1$ does not belong to the closure of $F$.

A sequence space is a vector space of scalar sequences, closed under coordinatewise addition and scalar multiplication. Examples include the l-p sequence space, the l-infinity sequence space and spaces of convergent sequences.