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Dropping either condition destroys the implication. The increasing sequence $a_n=n$ has a lower bound one but no upper bound, and no finite limit of a sequence. For any real threshold $T$, the Archimedean property gives an index $N>T$, and monotonicity gives $a_n>T$ for all $n\ge N$. Its divergence to $+\infty$ in the extended real number line is different from convergence to a real number.

Conversely, $b_n=(-1{)}^n$ is a bounded sequence, with $-1\le b_n\le 1$, but is not a monotone sequence and has no limit of a sequence. Its subsequence at even integers is constantly one, and its subsequence at odd integers is constantly minus one. These two subsequential limits cannot both equal a single limit of a sequence. ^[ref]

The ambient real numbers also matter. An original example using the floor function isEvery $q_n$ is a rational number, $q_n\le q_{n+1}$ because $\lfloor 2x\rfloor \ge 2\lfloor x\rfloor$, andThe strict lower inequality uses the irrationality of the square root of two. Hence this bounded sequence converges to $\sqrt{2}$ in the real numbers, but has no limit of a sequence in the rational numbers. Convergence to a rational candidate would also be convergence in the real numbers, contradicting uniqueness of the limit of a sequence. The formulas give $q_1=1$, $q_2=5/4$, $q_3=11/8$, $q_4=11/8$: this counterexample even has a plateau. This illustrates why completeness of the real numbers matters: the rational numbers lack the least-upper-bound property needed to contain the required supremum. ^[ref]

The supremum of an increasing bounded sequence need not be a maximum of its set of terms. For $a_n=1-1/n$, every term is strictly less than one; the supremum and limit of a sequence are one, but there is no maximum. Every candidate term is beaten by the next. The decreasing sequence $1+1/n$ similarly has infimum one without a minimum.

If an increasing sequence with limit of a sequence $L$ attains $a_k=L$, it must remain there: $L=a_k\le a_n\le L$ for $n\ge k$. The decreasing case is identical with inequalities reversed. Thus attainment of the limit of a sequence is equivalent to an eventually constant tail of a sequence for a convergent monotone sequence. The increasing sequence $0,1/2,1,1,\dots$ has both supremum and maximum one. ^[ref]

Every bounded sequence of real numbers contains a subsequence with a finite limit of a sequence. If $|x_n|\le M$ for every positive integer $n$, there are integers $n_1<n_2<\cdots$ and a real number $\ell$ such thatThe increasing indices matter: a subsequence retains the original order and selects terms arbitrarily far along the original sequence. The theorem guarantees the existence of at least one such selection; it does not claim that the entire sequence converges or that the subsequential limit is unique.^[ref]

For example, the bounded sequence $x_n=(-1{)}^n+1/(n+1)$ has two distinct subsequential limits:The even terms form a convergent sequence, and so do the odd terms. The full sequence cannot have a limit of a sequence because its two subsequences have different limits. Its limit superior is $1$ and its limit inferior is $-1$.

This is useful when a construction produces many candidate points but does not immediately produce a convergent sequence. For a continuous function on a nonempty closed bounded feasible region in a finite-dimensional Euclidean space, take a sequence whose objective values approach the infimum. A convergent subsequence supplies a candidate point; a closed subset retains that point, and continuity passes the objective values to the limit of a sequence. This is the mechanism behind attainment in the extreme value theorem. The diagonal subsequence argument extends the real-sequence theorem to finite-dimensional Euclidean space.

The theorem is named for Bernard Bolzano and Karl Weierstrass.^[ref]

For every regular summability matrix $A$, there is a bounded sequence whose transformed row sums do not converge. A gliding hump argument selects successive rows whose mass on already assigned coordinates is small and whose remaining tails are small. On each new finite block, choose signs agreeing or disagreeing with that row alternately. Since row sums tend to one, the transformed values have subsequences uniformly above and below zero. This does not contradict preservation of ordinary convergence, since the constructed sequence is divergent.

Even restricting to increasing sequences in $[0,1]$ with limit of a sequence one gives no common stopping index. For any positive integer $K$, defineThis is an increasing bounded sequence converging to one, but its absolute error is one through index $K$. Since $K$ can exceed any proposed index, the bounded monotone sequence theorem alone cannot say when the absolute error will fall below $1/2$. A zero difference between two successive terms is also insufficient: the initial plateau eventually ends.

There is no universal eventual rate either. Let $(r_n)$ be any proposed positive nonincreasing sequence tending to zero, with $r_n\le 1$. SetThen $(a_n)$ is increasing, lies in $[0,1]$, and converges to one, but for every sufficiently large $n$ its absolute error satisfies $1-a_n=\sqrt{r_n}>r_n$. Indeed, $\sqrt{r_n}/r_n=1/\sqrt{r_n}\to \infty$, so no fixed constant multiple of $r_n$ bounds the absolute error eventually. If a proposed rate is not nonincreasing, replace it, after a finite initial segment if necessary, by the decreasing envelope $s_n=\sup \{r_k:k\ge n\}$; this also tends to zero, and the same construction defeats the original rate. This is an original construction, rather than a quantitative claim attributed to the qualitative theorem.

For a concrete slow example, $a_n=1-1/\log (n+2)$ increases to one, where $\log$ is the natural logarithm. Its absolute error is $1/\log (n+2)$, so making it less than $\epsilon$ requires $n+2>\exp (1/\epsilon )$, using the exponential function. At $\epsilon =0.1$, this first happens at $n=22025$. The reciprocal and geometric examples in increasing and decreasing approximations to a common limit meet the same tolerance at $n=11$ and $n=4$, respectively. All three obey the same bounded monotone sequence theorem.

A monotone sequence of real numbers has a finite limit of a sequence exactly when it is a bounded sequence. In the increasing case, only an upper bound needs to be supplied, because the first term already gives a lower bound; in the decreasing case, only a lower bound needs to be supplied. The limit of a sequence is the exact edge of the set of terms:The upper bound $M$ or lower bound $m$ need not itself be the limit of a sequence. A supremum or infimum identifies the sharp bound. ^[ref]

For example, $a_n=1-1/n$ is increasing and bounded above by one, and $b_n=1+1/n$ is decreasing and bounded below by one. Both have limit of a sequence one, although neither ever equals one. Their absolute errors are exactly $1/n$. They also bracket any sequence $x_n$ satisfying $a_n\le x_n\le b_n$: the squeeze theorem gives $x_n\to 1$, even if $(x_n)$ itself is not a monotone sequence.

The theorem is useful when a recurrence relation gives successive improvements and a bound, before the answer has been identified. It proves existence of a limit of a sequence; further algebra, often using continuity, identifies the answer. A finite plot illustrates the formulas but does not establish the infinite claim. Error bounds for bounded monotone sequences require additional quantitative information.

A real increasing sequence bounded above converges to its supremum; a decreasing sequence bounded below converges to its infimum. For the increasing case, the definition of supremum provides a term above $L-\epsilon$, and every later term remains between that term and $L$. Negation reduces the decreasing case to the increasing case. This elementary result is distinct from the measure-theoretic monotone convergence theorem for integrals.

The existence of the real supremum in this short proof uses the least-upper-bound property, equivalently completeness of the real numbers. The approximation property of a supremum supplies $a_N>L-\epsilon$, and monotonicity keeps every later term within the same tolerance:The corresponding decreasing statement follows from the approximation property of an infimum. The converse uses the fact that every convergent sequence is a bounded sequence. ^[ref]

The sequence is bounded and $|a_{n+1}-a_n|\le \sqrt{n+1}-\sqrt{n}\to 0$ by the mean value theorem. Nevertheless, along the integer parts of $(\pi /2+2\pi j{)}^2$ its terms tend to one, and along the integer parts of $(3\pi /2+2\pi j{)}^2$ they tend to minus one. The square root rounding error is $O(1/j)$. Thus a bounded sequence with vanishing successive increments need not be a convergent sequence.

A Cauchy sequence is a sequence whose sufficiently late terms are all close to one another. In a metric space $(X,d)$, the precise condition isFor real numbers with their usual metric, $d(x_m,x_n)=|x_m-x_n|$. The useful feature is that the definition compares computed terms to other computed terms: it does not require an already known limit of a sequence. The one index $N$ must control every pair of later indices, not merely successive terms. ^[ref].

For an explicit example, take $q_n=1-2^{-n}$ for $n\ge 1$. Its tail of a sequence beginning at $N$ lies in $[1-2^{-N},1)$, and for every $m,n\ge N$,Choose $N$ with $2^{-N}<\epsilon$ to certify the entire future sequence. For $\epsilon =0.1$, $N=4$ works because $2^{-4}=0.0625$. The limit of a sequence is one, but the pairwise bound establishes the Cauchy sequence property without first needing that answer. By contrast, $a_n=(-1{)}^n$ is not a Cauchy sequence: arbitrarily late even integers and odd integers give values separated by two. ^[ref].

Tiny successive changes do not provide this guarantee. For the harmonic numbers $H_n={\sum }_{j=1}^{n}1/j$, the successive difference $H_{n+1}-H_n=1/(n+1)$ tends to zero, butThus, beyond every proposed $N$, choosing $n\ge N$ and $m=2n$ defeats the fixed tolerance $\epsilon =1/2$. Even every fixed positive lag $h$ passes the weaker test: $0<H_{n+h}-H_n\le h/(n+1)\to 0$. The failing lag grows with $n$, as explained more generally by convergence from arbitrary variable-lag increments. ^[ref].

Every convergent sequence in a metric space is a Cauchy sequence: apply the triangle inequality to two late terms and their common limit of a sequence. The reverse implication requires completeness. It holds in the real numbers by completeness of the real numbers, but can fail in the rational numbers: the rational endpoints from the bisection method approaching $\sqrt{2}$ are a Cauchy sequence with no rational limit of a sequence. The existing construction in completeness of the real numbers shows how this happens. A Cauchy sequence is always a bounded sequence; in $\mathbb{R}$, boundedness gives a convergent subsequence by the Bolzano-Weierstrass theorem, and the pairwise condition forces the whole sequence to share its limit. ^[ref].

Suppose the distinct complex numbers ${\beta }_k$ satisfy $|{\beta }_k|\le \delta <1$. The geometric sequences in the displayed formula belong to the absolutely summable sequence space and have dense linear span. Otherwise the Hahn-Banach theorem gives a nonzero annihilating functional, represented by a bounded sequence $b_j$. Its power series $F(z)={\sum }_{j\ge 1}{b}_j{z}^{j-1}$ is holomorphic in the unit disk and vanishes at every ${\beta }_k$. Compactness supplies an accumulation point inside that disk. The identity theorem forces $F=0$ and hence all its coefficients zero, a contradiction. This is a useful form of density by annihilators that converts a Banach-space problem into uniqueness of an analytic function.

For the two sequences in bounded monotone sequence theorem,Their directions and bounds are explicit. One is a succession of lower approximations and the other a succession of upper approximations to the same limit of a sequence. To obtain absolute error strictly below $\epsilon >0$, it suffices to choose an integer $N>1/\epsilon$, and then every $n\ge N$ works. For $\epsilon =0.01$, the first such index is $101$; at $100$ the absolute error equals the tolerance.

A geometric sequence offers a faster example. The recurrence relation $c_1=1/2$, $c_{n+1}=(c_n+1)/2$ gives $c_n=1-2^{-n}$. Thus $(c_n)$ is an increasing bounded sequence with limit of a sequence one and absolute error $2^{-n}$. For $\epsilon =0.01$, $n=7$ already works, since $2^{-7}=1/128<0.01$, whereas $2^{-6}=1/64>0.01$. The bounded monotone sequence theorem guarantees both examples converge; the formulas explain their different speeds.

In the l2 sequence space, let $e_n$ have value one in coordinate $n$ and zero elsewhere. This orthonormal sequence is bounded: $\Vert e_n{\Vert }_2=1$. For distinct indices $n,m$, its terms obey $\Vert e_n-e_m{\Vert }_2=\sqrt{2}$. No subsequence can therefore be a Cauchy sequence, so none converges in the norm. Thus even a complete metric space can have a bounded sequence with no convergent subsequence. Finite dimension is essential to the usual Euclidean space extension, and in a general metric space the relevant replacement is compactness or total boundedness together with completeness.

The limit inferior and limit superior retain the lower and upper long-run behaviour of a real sequence, even when it has no ordinary limit of a sequence. For each tail of a sequence beginning at $N$, defineDiscarding more initial terms raises or preserves the infimum and lowers or preserves the supremum. Hence $(l_N)$ is nondecreasing and $(u_N)$ is nonincreasing. Their limits exist in the extended real number line:For a bounded sequence, both values are finite and $\mathrm{liminf}{a}_n\le \mathrm{limsup}{a}_n$. These bounds use the entire infinite tail of a sequence, rather than the last few observed terms. ^[ref].

Take the original example $a_n=(-1{)}^n(1+1/n)$, beginning $-2,3/2,-4/3,5/4,\dots$. The subsequence at even integers decreases toward one, while the subsequence at odd integers increases toward minus one. If $e_N$ is the first even integer at least $N$ and $o_N$ is the first odd integer at least $N$, thenThus $\mathrm{liminf}{a}_n=-1$ and $\mathrm{limsup}{a}_n=1$. The finite initial extreme $a_1=-2$ disappears from later lower bounds; neither long-run value is attained by any term of this example.

For a bounded sequence, the limit inferior is the smallest subsequential limit and the limit superior is the largest subsequential limit; both are reached by suitable convergent subsequences. Every other subsequential limit lies between them. In the example there are only two subsequential limits, $-1$ and $1$: a convergent subsequence cannot retain infinitely many terms of both parities, because those two branches stay separated. The whole interval between the two bounds need not consist of subsequential limits. ^[ref].

A finite ordinary limit of a sequence exists exactly when the limit inferior and limit superior agree at the same finite value $L$. The bounds $l_N\le a_N\le u_N$ and the squeeze theorem prove the sufficient direction. For a bounded sequence, its tail diameter of a sequence is $D_N=u_N-l_N$, so the limiting gap is zero exactly when the Cauchy sequence condition holds. In the example the gap approaches two, not zero. The same framework allows infinite bounds: $a_n=n$ has both values $+\infty$, whereas $a_n=(-1{)}^{n}n$ has limit inferior $-\infty$ and limit superior $+\infty$. Equality at infinity describes escape rather than convergence to a real value. ^[ref].

For a sequence of real numbers, a common finite limit of a sequence for all monotone subsequences forces the whole sequence to have that limit of a sequence. There is no need to assume a bounded sequence if the hypothesis already includes convergence of every monotone subsequence.

If $a_n$ fails to tend to $L$, select a subsequence with $|a_{n_j}-L|\ge \epsilon$ for some fixed $\epsilon >0$. The monotone subsequence theorem gives a monotone subsequence of this subsequence, which contradicts the common limit of a sequence $L$. Conversely, every subsequence of a convergent sequence has the original limit of a sequence.

For a bounded sequence, the bounded monotone sequence theorem already ensures that all monotone subsequences converge. Only agreement of their limits of sequences remains to be checked.

Begin with a closed interval $I_0=[a,b]$ containing the entire bounded sequence. Split the interval at its midpoint. At least one of the two closed halves contains infinitely many indexed terms; choose such a half and repeat. This gives nested intervals $I_j$ with lengths $(b-a)/2^j$. The nested interval theorem, using completeness of the real numbers, supplies their unique common point $\ell$.

Choose $n_j>n_{j-1}$ with $x_{n_j}\in I_j$. Infinitely many indexed terms in $I_j$ guarantee that the next index can be chosen. Both $x_{n_j}$ and $\ell$ belong to $I_j$, soThis is a short existence proof, not necessarily a finite algorithm for an arbitrary input sequence: deciding which half contains infinitely many future terms may require mathematical knowledge about that sequence.

A real sequence has limit $a\in \mathbb{R}$ if, for every $\epsilon >0$, there is $N$ such that $|a_n-a|<\epsilon$ for all $n\ge N$. It tends to $+\infty$ if, for every real $M$, eventually $a_n>M$; it tends to $-\infty$ if eventually $a_n<M$ for every real $M$.

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a real convergent subsequence. This proves the claim for bounded sequences. If the sequence is unbounded above, every tail remains unbounded above because a finite prefix is bounded. Choose indices $n_j$ increasing with $a_{n_j}>j$. This subsequence tends to $+\infty$. If it is unbounded below, choose $a_{n_j}<-j$ to obtain a subsequence tending to $-\infty$. This proves the extended real subsequence theorem.

For the final example, a bounded sequence with vanishing increments need not converge. TakeThe mean value theorem givesNevertheless, along $n_j=\lfloor (\pi /2+2\pi j{)}^2\rfloor$, the square-root rounding error tends to zero, so $a_{n_j}\to 1$. Along $m_j=\lfloor (3\pi /2+2\pi j{)}^2\rfloor$, it tends to $-1$. These two subsequences have different limits, proving nonconvergence.

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.

The key fact is that convergence of a power series at one nonzero point forces absolute convergence at every smaller modulus. Indeed, if $\sum a_n{w}^n$ converges and $w\ne 0$, its terms form a bounded sequence, say $|a_n{w}^n|\le M$. For $|z|<|w|$,by convergence of a geometric series.

Let $S$ be the set of moduli of points at which the power series converges, and put $R=\sup S$, allowing $R=\infty$. The set contains zero, so its supremum is defined in $[0,\infty ]$. Whenever $|z|<R$, the definition of supremum provides a convergence point $w$ with $|w|>|z|$. The preceding estimate gives absolute convergence at $z$. Whenever $|z|>R$, convergence at $z$ would put $|z|$ in $S$, contradicting its upper bound. This proves the asserted radius of convergence property, including $R=0$ and $R=\infty$. Nothing in the argument decides convergence on $|z|=R$.

For the requested boundary behavior, take the power seriesIts coefficients are $a_{2k}=1/k$ for $k\ge 1$, $a_{2k+1}=0$ and $a_0=0$. At either $z=1$ or $z=-1$ it is the divergent harmonic series. At either $z=i$ or $z=-i$ it is ${\sum }_{k\ge 1}(-1{)}^k/k$, which converges by the alternating series test. In fact its radius of convergence is one, since inside the unit circle the terms are bounded by a geometric series, and outside it they do not tend to zero.

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.

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 every bounded sequence $x$, $x-Sx\in Y$ and $L{|}_Y=0$. Linearity gives $\boxed{L(x)=L(Sx)}$, as required. Thus the same $L$ satisfies all three conditions and is a Banach limit.

The usual positivity condition for a Banach limit follows automatically here. If $x_n\ge 0$ and $a=\Vert x{\Vert }_{\infty }>0$, then $\Vert 1-x/a{\Vert }_{\infty }\le 1$. Since $\Vert L\Vert =L(1)=1$, we have $|1-L(x)/a|\le 1$, whence $L(x)\ge 0$. The case $x=0$ is immediate. This explains why the constructed bounded linear functional is also a positive linear functional extending the ordinary limit while remaining invariant under the left shift on bounded sequences.