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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]

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.

A series over the real numbers or complex numbers converges if and only if its tails satisfy: for every $\epsilon >0$, some $N$ makes $|{\sum }_{k=n+1}^m{a}_k|<\epsilon$ whenever $m>n\ge N$. This says exactly that its partial sums form a Cauchy sequence; sufficiency uses completeness.

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].

In a Banach space, the closed convex hull of a compact set is compact in the norm topology. Given $\epsilon >0$, choose a finite $\epsilon$-net $\{y_i\}$ for $K$. Every convex combination of points of $K$ is within $\epsilon$ in norm of $\mathrm{conv}\{y_i\}$, by replacing its terms by their approximating $y_i$. This finite-dimensional compact convex set has a finite $\epsilon$-net, giving a finite $2\epsilon$-net for $\mathrm{conv}K$. Its closure is therefore totally bounded and complete, hence compact. Completeness of the ambient space is used at this last step.

A real bilinear form $B$ on a normed vector space $X$ is coercive when there is one constant $\alpha >0$ such thatThe lower bound uniformly controls every direction in the chosen norm. Unlike the upper bound defining a bounded bilinear form, it forces positive diagonal values at every nonzero vector. No symmetry is assumed. For complex Hilbert spaces, the usual analogue uses a sesquilinear form and $\mathrm{Re}B(v,v)\ge \alpha \Vert v{\Vert }^2$; it is not the same assertion for a complex bilinear form.

A concrete bounded and coercive example on ${\mathbb{R}}^2$, with the Euclidean norm, isIts quadratic form isThe best lower constant is $\alpha =2$, attained at $(1,0)$. Nevertheless, $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$, so coercivity does not imply a symmetric bilinear form. The symmetric part of a matrix $S=(A+A^T)/2=\mathrm{diag}(2,3)$ determines $q$, while the skew-symmetric matrix part changes mixed pairings without changing $q$. This is the symmetric and alternating parts of a bilinear form decomposition.^[ref]

For the upper constant, direct multiplication gives $A^{T}A=\left(\begin{array}{cc}5& -1\\ -1& 10\end{array}\right)$. Its eigenvalues are $(15\pm \sqrt{29})/2$, so the optimal upper constant, the spectral norm, is $C=\sqrt{(15+\sqrt{29})/2}$. The two constants answer different questions: how large a mixed pairing can be, and how small a self-pairing can be relative to squared norm.

The plotted quadratic forms show the mechanism: $x^2-y^2$ descends in one direction, $x^2$ leaves one direction uncontrolled, and $2x^2+3y^2$ controls both. These are self-pairings; their level sets cannot reveal the skew-symmetric matrix part. Calling $q$ a quadratic energy here refers to this diagnostic quantity, not to a claim that a nonsymmetric equation comes from minimizing an energy functional.

On a real finite-dimensional vector space with any norm, strict positivity $B(v,v)>0$ for every nonzero $v$ does imply coercivity. Indeed the quadratic form is continuous, and its positive minimum on the compact unit sphere supplies $\alpha$. If $B$ is also a symmetric bilinear form, this is positive definiteness. For general $B$, it is the symmetric and alternating parts of a bilinear form decomposition that isolates the positive part. Switching between equivalent norms preserves coercivity, but usually changes its constant.

The infinite-dimensional vector space distinction is explicit on the real l2 sequence space:The Cauchy-Schwarz inequality gives boundedness with $C=1$, and $B(v,v)>0$ for every nonzero vector. But $B(e_n,e_n)=1/n$ with $\Vert e_n{\Vert }_2=1$, so no positive $\alpha$ works. On the span of $e_1,\dots ,e_N$, the best lower constant is $1/N$: each finite restriction is coercive, but the control disappears as $N$ grows. Strict positivity is therefore weaker than a uniform lower bound.

The practical payoff is a stability estimate for a weak formulation. Assume $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is both a bounded bilinear form and a coercive bilinear form, and the load $\ell \in H^{\prime }$ is a bounded linear functional. The Lax-Milgram theorem gives a unique $u\in H$ satisfying $B(u,v)=\ell (v)$ for all $v\in H$, including nonsymmetric $B$.^[ref] The dual norm $\Vert \ell {\Vert }_{H^{\prime }}={\sup }_{v\ne 0}|\ell (v)|/\Vert v\Vert$ measures load size. Testing with $v=u$ gives the short energy estimateFor $u=0$ the bound holds directly; otherwise divide by $\Vert u\Vert$. Subtracting the equations for two loads and using the same argument gives $\Vert u_1-u_2\Vert \le \Vert {\ell }_1-{\ell }_2{\Vert }_{H^{\prime }}/\alpha$. Thus a positive lower constant limits the response to load errors. The upper bound and completeness of the Hilbert space enter the existence theorem; the estimate itself uses the lower bound and the load's dual norm.

For the displayed matrix $A$, take $\ell (v)=f^{T}v$. Then $B(u,v)=\ell (v)$ for every $v$ means $A^{T}u=f$, not $Au=f$, with this ordering of the bilinear form. Since $\det A=7$,For $f=t(1,0{)}^T$, the exact response is $u=t(3,-1{)}^T/7$ and $\Vert u{\Vert }_2=\sqrt{10}|t|/7<|t|/2$. For $f=t(0,1{)}^T$, it is $u=t(1,2{)}^T/7$, with norm $\sqrt{5}|t|/7$. The actual maximal amplification is the operator norm $\Vert (A^T{)}^{-1}{\Vert }_2=\sqrt{2/(15-\sqrt{29})}<1/2$: the coercivity estimate is valid but need not be sharp for a nonsymmetric bilinear form.

A field with a surjective discrete valuation, complete for the associated non-Archimedean metric. Its valuation ring is a discrete valuation ring, and its maximal ideal is generated by a uniformizer. Completeness allows successive congruence solutions to converge to actual roots, which underlies the Hensel lemma. Non-Archimedean local fields are examples; completeness alone does not require the residue field to be finite.

For any nonempty complete metric space $X$, the family of its nonempty closed bounded subsets is a complete metric space under the Hausdorff distance. Boundedness of the entire ambient space is unnecessary in this general form.

For a Cauchy sequence $(F_n)$ of these sets, define $A_n=\overline{{\bigcup }_{m\ge n}{F}_m}$ and ${\delta }_n={\sup }_{p,q\ge n}{d}_H(F_p,F_q)$. A Cauchy sequence in any metric space is bounded, so its tail sets are contained in a common finite-radius neighbourhood of one bounded $F_N$; hence every $A_n$ is bounded. The distance to a set is one-Lipschitz, so estimates pass to the closure, giving $d_H(F_n,A_n)\le {\delta }_n$ and $d_H(A_n,A_m)\le {\delta }_n$ for $m\ge n$. Choose $n_j$ with ${\delta }_{n_j}\le 2^{-j}$. The Summably nested Hausdorff intersection theorem gives $A_{n_j}\to G\ne \varnothing$. The triangle inequality now gives $F_n\to G$, proving completeness.

A contraction mapping $T:X\to X$ on a nonempty complete metric space has exactly one fixed point $x_{\ast }$, and the sequence $x_{n+1}=T(x_n)$ converges to $x_{\ast }$ from every $x_0\in X$. The same constant $0\le L<1$ must satisfy $d(Tx,Ty)\le Ld(x,y)$ for all $x,y\in X$. The statement applies to metric spaces, including closed subsets of a Banach space; neither linearity nor a vector space structure is required.^[ref]

The force of the result is that existence, uniqueness, and a practical approximation procedure arrive together. Repeated application produces corrections bounded by a geometric series, hence a Cauchy sequence; completeness supplies its limit. The Lipschitz continuity of $T$ makes that limit a fixed point. Two different fixed points would have their positive distance multiplied by at most $L<1$ while remaining unchanged, which is impossible.

Let $A_h:X\to Z$, $0<h<1$, be bounded linear operators from a normed space into a Banach space, with $M={\sup }_h\Vert A_h\Vert <\infty$. If $A_{h}d$ converges as $h\to 0^{+}$ for each element of a dense subspace $D$, then $A_{h}x$ converges for every $x\in X$. The limit defines a bounded linear operator $A$ with $\Vert A\Vert \le M$.

For $d\in D$, the triangle inequality givesApproximate $x$ by $d$ and use the Cauchy criterion for a function limit on $D$. The completeness of $Z$ supplies the limit; passing to the limit preserves linearity and the bound. When every limit on $D$ is zero, the inequality $\Vert A_{h}x\Vert \le M\Vert x-d\Vert +\Vert A_{h}d\Vert$ shows $A=0$.

Use radians and take $T(x)=\cos x$ on the closed interval $[0,1]$. This is a complete metric space with distance $|x-y|$, and $T([0,1])=[\cos 1,1]\subset [0,1]$. The mean value theorem and $|T^{\prime }(x)|=\sin x\le \sin 1$ give a contraction mapping with $L=\sin 1\approx 0.841471<1$. Both the interval and its preservation matter; a small derivative by itself would not ensure that all subsequent approximations stay in the chosen space.

Starting at $x_0=0$ gives $x_1=1$, $x_2\approx 0.540302$, $x_3\approx 0.857553$, and $x_4\approx 0.654290$. The cosine is decreasing here, so successive iterates alternate around the unique fixed point $x_{\ast }\approx 0.739085133215161$. The contraction iteration error bound needs no prior knowledge of this numerical value. At step 40, $x_{40}\approx 0.739085078689123$ and the latest-step bound is below $7.20\times 10^{-7}$, certifying error below one millionth for exact iteration. The initial-step bound requires 91 iterations for the same tolerance. The observed step-40 error is only about $5.45\times 10^{-8}$; the guarantees allow less favorable behavior than this example actually exhibits.

This is the beauty of the contraction mapping theorem: a uniform local improvement controls the entire infinite process. Completeness is essential. On the incomplete interval $(0,1)$, the contraction mapping $T(x)=x/2$ keeps every iterate inside but has no fixed point there; its missing limit is zero. Banach's 1922 paper developed abstract methods for different spaces of functions with applications to integral equations, a motivation still visible whenever repeated approximation is used to construct a function.^[ref]

For a normalized discrete valuation, a uniformizer $\pi$, and a chosen set $A$ of representatives of the residue field, every existing nonzero element has a unique displayed expansion, with its first digit not in the maximal ideal. Starting from $y_0={\pi }^{-v(x)}x$, choose $a_j$ representing $y_j$ modulo the maximal ideal and put $y_{j+1}=(y_j-a_j)/\pi$. The error after $r$ digits has valuation at least $v(x)+r$, so the sums converge to $x$ even without completeness. Conversely arbitrary digit sequences need not converge in an incomplete field: in $\mathbb{Q}$ with its three-adic valuation, the digits of $\sqrt{10}\in {\mathbb{Q}}_3$ give a counterexample. Hensel lemma gives this square root, but it is not rational.

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

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

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

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

Functional analysis brings linear algebra and topology together to study vector spaces of functions and the linear operators acting on them. A function space such as $C([0,1])$ has infinitely many independent directions: the monomials $1,x,x^2,\dots$ are linearly independent. Finite linear combinations alone cannot describe all its continuous functions. A useful theory must also say when approximations converge, whether their limits belong to the chosen space, and which operations preserve that norm convergence. Normed vector spaces, Banach spaces, and bounded linear operators provide these three controls.^[ref]

A bounded linear operator $T:X\to Y$ between normed vector spaces satisfies $\Vert Tf{\Vert }_Y\le M\Vert f{\Vert }_X$ for every $f$, with one finite constant $M$. The least possible constant is its operator norm. By linearity, the same estimate controls approximation errors:This is exactly continuity for a linear map between normed vector spaces; neither space needs to be complete. It turns an input accuracy guarantee into an output accuracy guarantee, even when the input has infinitely many degrees of freedom.^[ref]

For a concrete linear functional on $C([0,1])$, take the integral $F(f)={\int }_0^{1}f(x)dx$. In the supremum norm,The constant function $f=1$ attains equality, so $\Vert F\Vert =1$. For $f(x)=x^2$, the output is $F(f)=1/3$; any approximation $g$ with $\Vert g-x^2{\Vert }_{\infty }\le 0.01$ therefore has $|F(g)-1/3|\le 0.01$. The Cauchy-Schwarz inequality also gives $|F(f)|\le \Vert f{\Vert }_2\Vert 1{\Vert }_2=\Vert f{\Vert }_2$ on this interval, again with equality at $1$. Thus integration remains a bounded linear functional under the L2 norm, although point evaluation behaves very differently.

The beauty of this viewpoint is that one estimate can justify an entire approximation procedure. Once a sequence converges in the chosen norm, every bounded linear operator carries its limit along. Completeness ensures that Cauchy approximations actually describe an element of the space. The choice of norm matters: it determines which errors are small and which measurements are stable. The examples below distinguish an average, which ignores sufficiently narrow peaks, from point evaluation, which detects their full height. For applications, this decides whether reducing an overall fitting error guarantees an accurate reported measurement. A small L2 norm of the error controls its integral on this interval, while it gives no such guarantee for a single sampled value.

Multiply two copies of the generating function of the physicists' Hermite polynomials and integrate against $e^{-t^2}$. Completing the square givesComparing coefficients gives the displayed orthogonality and normalization. Equivalently differentiation in $y,z$ at zero justifies coefficient extraction by the Gaussian integrable bound on compact parameter ranges. Hence the unnormalized Hermite functions $e^{-t^2/2}{H}_n(t)$ have norm ${\pi }^{1/4}\sqrt{2^{n}n!}$. This proves their mutual orthogonality; completeness of that family is an additional assertion.

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.

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.

A complex normed vector space has an inducing inner product exactly when its norm satisfies the parallelogram law. Define $B(u,v)=[\Vert u+v{\Vert }^2-\Vert u-v{\Vert }^2]/4$. Subtract two parallelogram identities to obtain $B(u+w,v)+B(u-w,v)=2B(u,v)$. Taking $w=u$ gives doubling, and then replacing $u,w$ by the half-sum and half-difference gives additivity. Integer scaling and division give rational homogeneity; continuity gives real homogeneity. Thus $B$ is a symmetric real-bilinear form. The complex formula $P(u,v)=B(u,v)+iB(u,iv)$ satisfies $P(iu,v)=iP(u,v)$, because $B(iu,v)=-B(u,iv)$ and $B(iu,iv)=B(u,v)$. Symmetry gives conjugate symmetry of $P$, and $P(u,u)=\Vert u{\Vert }^2$ gives positive definiteness. The reverse implication follows by expanding squared norms. Completeness is needed only for the resulting inner product space to be a Hilbert space.

Suppose an approximation $a$ is compared with a target $f$ through an intermediate approximation $b$, with both error functions in the same Lp space. The identity $f-a=(f-b)+(b-a)$ and the Minkowski inequality combine two error budgets without any independence assumption. The L1 norm controls aggregate absolute error, the L2 norm controls the square root of aggregate quadratic loss, and the L-infinity norm controls the largest error almost everywhere. For equally weighted samples with the probability measure $1/n$ on each coordinate, the L2 norm is the root mean square error; the inequality still applies when this normalization is used consistently.

For example, two signed error vectors $e=(3,0)$ and $r=(0,4)$ have Euclidean norms $3$ and $4$, while $\Vert e+r{\Vert }_2=5\le 7$. Reinforcing errors $e=(3,0)$ and $r=(4,0)$ attain the bound $7$; opposing errors $e=(3,0)$ and $r=(-3,0)$ cancel completely. The Minkowski inequality always allows reinforcement. A smaller combined budget requires additional information about the errors, not just their separate norms.

Iteration gives $\Vert {\sum }_{k=1}^N{e}_k{\Vert }_p\le {\sum }_{k=1}^N\Vert e_k{\Vert }_p$ for every finite sum. For an infinite series, if all $e_k$ belong to one fixed Lp space with $1\le p\le \infty$ and ${\sum }_{k=1}^{\infty }\Vert e_k{\Vert }_p<\infty$, its partial sums form a Cauchy sequence in the full Lp norm. Completeness of this Banach space then gives norm convergence to an element of the same Lp space. A finite sum bound alone does not establish convergence, and completeness is an additional ingredient beyond the Minkowski inequality. ^[ref]

A norm on a complex vector space is a map $\Vert \cdot \Vert :X\to [0,\infty )$ that vanishes only at zero, satisfies $\Vert \lambda x\Vert =|\lambda |\Vert x\Vert$, and obeys the triangle inequality. The space is Banach when every Cauchy sequence in the resulting metric converges in $X$.

If $X$ is Banach and ${\sum }_j\Vert x_j\Vert <\infty$, the estimate $\Vert {\sum }_{j=m}^n{x}_j\Vert \le {\sum }_{j=m}^n\Vert x_j\Vert$ shows that its partial sums are Cauchy, hence converge. Conversely let every such series converge and take a Cauchy sequence $(y_n)$. Choose a subsequence with $\Vert y_{n_{k+1}}-y_{n_k}\Vert <2^{-k}$. The series consisting of $y_{n_1}$ followed by these successive differences is absolutely summable in norm, so converges by hypothesis; its partial sums are exactly this subsequence. A Cauchy sequence with a convergent subsequence converges to the same limit, proving the absolute-series criterion for completeness.

On the absolutely summable sequence space ${\ell }^1$, coordinatewise addition and scalar multiplication preserve membership, and summing the scalar triangle inequality verifies its norm axioms. To prove completeness using the criterion, let $x^{(k)}=(x_j^{(k)})\in {\ell }^1$ with ${\sum }_k\Vert x^{(k)}{\Vert }_1<\infty$. For each $j$ the sum $x_j={\sum }_k{x}_j^{(k)}$ converges absolutely, and the permitted interchange of nonnegative double sums givesThus $x\in {\ell }^1$. MoreoverEvery absolutely norm-summable series therefore converges in ${\ell }^1$, proving ${\ell }^1$ is a Banach space.