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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 generalized limit on real bounded sequences is a positive linear functional of operator norm one that extends the ordinary limit of a sequence. Such a functional can be chosen by the Hahn-Banach theorem: extend the limit on the convergent sequence space while dominating it by the sublinear functional $p(x)={\mathrm{limsup}}_n{x}_n$. Applying the same domination to $-x$ gives the displayed lower bound. A Banach limit additionally has invariance under the left shift on bounded sequences.

A unital C-star algebra is a complex Banach algebra $A$ with a multiplicative identity $1_A$ and a conjugate-linear involution $a\mapsto a^{\ast }$ satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$ and the C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. A C-star isomorphism $\Phi :A\to B$ is a bijective complex-linear map preserving products and the involution. It carries $1_A$ to $1_B$ and is automatically isometric.

For a commutative unital C-star algebra, let $\Omega$ be its maximal ideal space, equivalently the space of Gelfand characters, with the weak-star topology. A Gelfand character has $\chi (1)=1$ and $\Vert \chi \Vert =1$. The equations $\chi (ab)=\chi (a)\chi (b)$ and $\chi (1)=1$ define a closed in the weak-star topology subset of the dual unit ball. The Banach-Alaoglu theorem makes $\Omega$ a compact Hausdorff space. Maximal ideals give characters by the Gelfand-Mazur theorem, and the Banach algebra spectrum satisfies $\sigma (a)=\{\chi (a):\chi \in \Omega \}$.

The Gelfand transform is $\hat{a}(\chi )=\chi (a)$. It preserves products, constants and the involution; the latter uses the fact that characters of a C-star algebra respect the involution. Every element of a commutative C-star algebra is normal, so the C-star identity and the spectral radius formula giveThus its image is a closed self-adjoint subalgebra of $C(\Omega )$, contains the constants, and separates characters. The complex Stone-Weierstrass theorem makes that image dense and therefore all of $C(\Omega )$. This is the Commutative Gelfand--Naimark theorem.

For the convergent sequence space $c$, coordinatewise multiplication satisfies $\Vert \alpha \beta {\Vert }_{\infty }\le \Vert \alpha {\Vert }_{\infty }\Vert \beta {\Vert }_{\infty }$, and coordinatewise complex conjugation is an involution. The constant sequence $1$ is the identity, andIt is complete: a uniform limit of convergent sequences is convergent, so $c$ is a closed linear subspace of the Banach space ${\ell }_{\infty }$. Hence it is a commutative unital C-star algebra.

TakeEvery finite point is isolated, and the neighbourhoods of $\infty$ contain all but finitely many integers. This is a compact Hausdorff space: an open cover has a set containing $\infty$, leaving only finitely many points to cover. DefineContinuity at $\infty$ is exactly convergence of the sequence. Thus $\Phi$ is onto $C(\Omega )$, and pointwise operations show it is a C-star isomorphism with $\Vert \Phi \alpha {\Vert }_{\infty }=\Vert \alpha {\Vert }_{\infty }$. Explicitly, these are all Gelfand characters: each $\chi (e_n)$ is $0$ or $1$ and at most one is $1$. If $\chi (e_n)=1$, multiplication by $e_n$ gives $\chi (\alpha )={\alpha }_n$. If all vanish, density of finitely supported sequences in l-p in $c_0$ gives $\chi {|}_{c_0}=0$, so $\chi (\alpha )=\lim {\alpha }_n$.

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.

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.

Every finitely supported sequence $v$ lies in $Y$: if $v_n=0$ for $n>N$, set $x_n={\sum }_{k=n}^N{v}_k$ for $n\le N$ and $x_n=0$ afterwards. Then $x$ is a bounded sequence and $x-Sx=v$. Consequently $L(v)=0$. The finitely supported sequences are dense in the space of sequences converging to zero $c_0$ for the supremum norm, by truncation, so continuity of the bounded linear functional $L$ gives $L{|}_{c_0}=0$. For $x$ in the real convergent sequence space, with ordinary limit $\ell$, we have $x-\ell 1\in c_0$. Therefore $\boxed{L(x)=\ell L(1)=\underset{n\to \infty }{\lim }{x}_n}$.

For a real or complex matrix, the three displayed conditions are equivalent to being a regular matrix summability method. For necessity, first apply the Uniform boundedness principle to each row's finite partial sums on the convergent sequence space, obtaining absolute summability of that row. Apply the principle again to the full row functionals, obtaining a uniform bound. Coordinate vectors and the constant-one sequence give the remaining conditions. For sufficiency, write $x_j=L+(x_j-L)$ and split the error into a finite head, controlled by columnwise convergence, and a uniformly small tail, controlled by the absolute row bound.

The summability domain is the inverse image of the convergent sequence space under the bounded matrix operator on the l-infinity sequence space. It is therefore a closed linear subspace. The bounded divergence for every regular summability matrix makes it proper, so it has empty interior and is a nowhere dense set. The Baire category theorem then ensures that countably many regular methods miss some common bounded sequence.