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Use a nonzero complex unital Banach algebra $B$ with submultiplicative norm, and write its unit as $e$. The complex scalar field is the usual convention in this spectral statement; for a real algebra one uses its complexification. We derive the needed analytic facts from norm-convergent series and scalar complex analysis.

If $\Vert a\Vert <1$, completeness makes the Neumann series ${\sum }_{j=0}^{\infty }{a}^j$ converge. Multiplying its finite partial sums by $e-a$ gives $e-a^{m+1}$, which tends to $e$. Hence its sum is the inverse of $e-a$. If $b$ is invertible, then $b+h=b(e+b^{-1}h)$ is invertible whenever $\Vert b^{-1}h\Vert <1$. This proves that the invertible elements form an open set and that inversion is locally norm-continuous.

Define the Banach algebra spectrum and the resolvent of an element byFor $|\lambda |>\Vert x\Vert$, the Neumann series givesThus $\sigma (x)$ is bounded by $\Vert x\Vert$ and is closed, because invertibility is open. Also $R(\lambda )\to 0$ in norm at infinity; for example its norm is at most $\Vert e\Vert /(|\lambda |-\Vert x\Vert )$, since $\Vert e\Vert \ge 1$ in a nonzero submultiplicative unital algebra. The standard normalization $\Vert e\Vert =1$ is not needed.

For points where the Banach algebra resolvent exists $\lambda ,\mu$, multiply out the inverses to get the resolvent identityFor ${\lambda }_0$ outside the Banach algebra spectrum, put $R_0=R({\lambda }_0)$. Locally,This series and its differentiated series converge uniformly on every smaller disk. It therefore proves $R$ is a Banach-space-valued holomorphic function, establishing complex differentiability in the Banach-space norm and gives $R^{\prime }=-R^2$, without assuming a Banach-algebra-valued analytic theorem. In particular every continuous complex linear functional $\varphi$ makes $\varphi (R(\lambda ))$ a scalar holomorphic function.

The Banach algebra spectrum is nonempty. Otherwise $R$ would be defined and analytic everywhere. Its local norm continuity and decay at infinity make it globally bounded. For every continuous linear functional $\varphi$, the scalar entire function $\varphi (R(\lambda ))$ is bounded and tends to zero at infinity, so it is identically zero: the scalar Cauchy estimate for its derivative on circles of radius $r$ is $M/r$, which tends to zero, and the limiting value then fixes its constant as zero. Such functionals separate points by the Hahn-Banach theorem. Explicitly a bounded real functional can be extended from the real span of any nonzero element using its norm, and complexifying it by $\varphi (v)=\ell (v)-i\ell (iv)$ gives a bounded complex linear functional nonzero on that element. Therefore $R$ itself would be zero, contradicting $(\lambda e-x)R=e$. We have proved the Banach algebra spectrum is a nonempty compact subset of the complex plane.

We next establish the spectral radius formula. The power norms are submultiplicative. If some $x^m=0$, their root norms are eventually zero. Otherwise, for any fixed $m$, write $n=qm+r$ with $0\le r<m$ to obtainTaking roots gives ${\mathrm{limsup}}_n\Vert x^n{\Vert }^{1/n}\le \Vert x^m{\Vert }^{1/m}$. Taking the infimum in $m$, and using that every root norm is at least that infimum, proves existence ofFor $|\lambda |>{\rho }_0$, the root test makes $\sum x^j/{\lambda }^{j+1}$ converge, even when $|\lambda |\le \Vert x\Vert$. The same telescoping multiplication proves it is a two-sided inverse. HenceFor the reverse bound fix $R>r_{\sigma }$. If $R_1>\max (R,\Vert x\Vert )$, integrate the uniformly convergent exterior series term by term on $|\lambda |=R_1$, obtainingThe same integral equals the one on $|\lambda |=R$: applying an arbitrary continuous linear functional reduces contour deformation through the Banach algebra spectrum-free annulus to the scalar Cauchy theorem, and separation of points restores equality in $B$. This proves the resolvent Cauchy coefficient formula rather than assuming vector-valued analytic calculus. Using the norm bound for the allowed vector integral givesTaking $n$th roots yields ${\rho }_0\le R$, and letting $R\downarrow r_{\sigma }$ finishes the proof:

For the first example take the commutative algebra of dual numbers $B=\mathbb{C}[\epsilon ]/({\epsilon }^2)$ with norm $\Vert a+b\epsilon \Vert =|a|+|b|$. Completeness follows from finite dimension, and the product estimate makes the norm submultiplicative. For $x=\epsilon$, $x\ne 0$ but $x^2=0$, so $\rho (x)=0$. Directly, $\lambda e-\epsilon$ has inverse ${\lambda }^{-1}e+{\lambda }^{-2}\epsilon$ for $\lambda \ne 0$, and its Banach algebra spectrum is $\{0\}$. For the second example take $B=\mathbb{C}$ with its absolute-value norm and $x=1$. Its Banach algebra spectrum is $\{1\}$, giving $\rho (x)=\Vert x\Vert =1$.

We construct the finite evaluation representation explicitly by Lagrange interpolation. This supplies all the polynomial facts needed for the bound.

Define the Chebyshev polynomial by $T_0=1,T_1=x$ andInduction shows that $T_m$ has degree $m$; the cosine addition identity also proves $T_m(\cos \theta )=\cos (m\theta )$. Therefore $\Vert T_n{\Vert }_{[-1,1]}=1$, and $T_n(-x)=(-1{)}^n{T}_n(x)$ follows from the recurrence. The case $n=0$ is immediate, since both $P$ and $T_0$ are constant.

For $n\ge 1$, choose the descending extremal nodesso $T_n(x_j)=(-1{)}^j$. Define the fundamental interpolation polynomialsThey satisfy ${\ell }_j(x_k)={\delta }_{jk}$. Consequentlyfor every polynomial of degree at most $n$: the difference has degree at most $n$ and vanishes at the $n+1$ distinct nodes, so it is zero. The root count used here follows by successively factoring $(t-x_j)$ from a polynomial having such a root; a nonzero degree-$n$ polynomial cannot have more than $n$ distinct roots. Thus the interpolation identity has been proved.

If $u>1$, every numerator factor in ${\ell }_j(u)$ is positive, while exactly $j$ denominator factors are negative. Hence $\mathrm{sgn}{\ell }_j(u)=(-1{)}^j$. Applying interpolation to $T_n$ itself givesPutting $M={\sup }_{[-1,1]}|P|$ in the interpolation formula yieldsFor $u<-1$, apply the proved positive-side inequality to $Q(t)=P(-t)$ at $-u>1$, and use the parity identity. The result isThis is sharp, with equality for $P=T_n$ or its scalar multiples.

In functional terms, the evaluation map $E_u:P\mapsto P(u)$ on the normed polynomial space has $\Vert E_u\Vert =|T_n(u)|$: the inequality gives the upper bound and $T_n$ attains it. For $u>1$, its normalized version has the exact representationThus Chebyshev interpolation represents external evaluation using $n+1$ nodes. The necessary finite representation and its norm are derived directly, so no convex-hull theorem is being invoked without proof.

Linearizing the ideal magnetohydrodynamic induction equation about a static magnetic field and using $u^{\prime }={\partial }_t\xi$ gives the fluid Lagrangian displacement-generated magnetic field perturbation $b=\nabla \times (\xi \times B)$. The two solenoidal vector field conditions reduce the curl identity toIts divergence vanishes because it is a curl. Directly, the two products of first derivatives in ${\partial }_i{b}_i$ cancel after exchanging their indices, while the remaining terms differentiate $\nabla \cdot \xi$ or $\nabla \cdot B$, both zero. Similarly the integrated linear continuity equation gives ${\rho }^{\prime }=-\nabla \cdot (\rho \xi )$, so $\boxed{{\rho }^{\prime }=-\xi \cdot \nabla \rho }$.

Let $\Pi =p+B^2/(2{\mu }_0)$ be the magnetohydrodynamic total pressure. Magnetostatic equilibrium and the momentum equation in the linearized ideal magnetohydrodynamic equations areThere is no perturbed gravitational potential here because the field is fixed. Multiply by ${\xi }_i^{\ast }$ and integrate. The pressure term vanishes by the assumed boundary cancellation and $\nabla \cdot {\xi }^{\ast }=0$. Thuswhere $\mathcal{M}=-\int {\xi }_i^{\ast }(B_j{\partial }_j{b}_i+b_j{\partial }_j{B}_i)dV$.

Integrate the first magnetic term along $B$, using $\nabla \cdot B=0$, then insert the expression for $b$. This gives the positive term $\int |(B\cdot \nabla )\xi {|}^{2}dV$ and two mixed derivative terms. For clarity, the troublesome term transforms asThe first term on the right cancels the other mixed term. Combining the remaining derivative with the term containing ${\partial }_j{B}_k{\partial }_k{B}_i$ givesDifferentiating the equilibrium relation now combines the gradient of mass density and magnetostatic equilibrium terms:We obtain the incompressible magnetic displacement energy identityThe two Hessian matrices are real symmetric, making their Hermitian form evaluations real. The last term is a squared norm, and $I>0$ for a nonzero displacement in fluid with positive mass density. Thus $\boxed{{\sigma }^2\text{ is real}}$. It may be negative, in which case one temporal branch has exponential growth; real squared frequency is not itself a proof of stability.

For the atmosphere with a horizontal magnetic field, $\Phi =gz$ has zero Hessian matrix and the equilibrium equation isConsequently the interchange stability of an incompressible magnetized atmosphere is governed by the numeratorFor ${\xi }_z=\sin ky$, the magnetic tension term is zero: these fluid Lagrangian displacements interchange neighboring magnetic field lines without varying along them. A mass density increasing with height gives negative energy and instability; decreasing mass density gives positive energy for these vertical interchanges. If ${\rho }_z\le 0$ everywhere, the displayed form is nonnegative for every admitted incompressible displacement. If the gradient changes sign, suitably localized divergence-free interchange trials detect an unstable inverted layer; For example, take ${\xi }_z=f(z)e^{iky}$ and ${\xi }_y=i{f}^{\prime }(z)e^{iky}/k$, with $f$ supported in that layer. Their divergence is zero, and neither component varies along the magnetic field.

For ${\xi }_z=\sin kx$, magnetic field lines are bent. Averaging over a full horizontal wavelength gives the trial Rayleigh quotientunder the stipulated admissible boundary/integrability conditions. It is a trial quotient, not a claim that this displacement is an exact eigenfunction for every profile. mass density decreasing upward is stable here too. For inverted mass density, magnetic tension can stabilize sufficiently short waves along the magnetic field, but does not remove the transverse interchange instability. Locally the two contributions have the familiar scale $-g{\rho }_z/\rho +k^2{v}_A^2$, with the Alfvén speed $v_A=B/\sqrt{{\mu }_0\rho }$.

In a unital Banach algebra, the Neumann series gives $(1-u{)}^{-1}={\sum }_{k=0}^{\infty }{u}^k$ whenever $\Vert u\Vert <1$: the series converges in norm by submultiplicativity and completeness, and multiplication of its partial sums on either side gives $1-u^{N+1}\to 1$. Consequently, if $g$ is invertible and $\Vert g^{-1}\Vert \Vert h-g\Vert <1$, writeThe second factor has a Neumann series inverse, so $h$ is invertible. This exhibits a ball around each $g$ in the group of invertible elements of a Banach algebra, proving that group is an open set.

The same series proves continuity of inversion in a Banach algebra, with the useful quantitative estimateThe right side tends to zero as $h\to g$. Since applying inversion twice gives the original element, it is a bijection whose inverse is the same continuous map. Thus inversion is a homeomorphism of the open group of invertibles onto itself.

For the final unheaded request, defineIt is open by the preceding argument and contains $0$. If ${\lambda }_0$ were a boundary point, a sequence ${\lambda }_n\in E$ would converge to ${\lambda }_0$, while openness gives ${\lambda }_0\notin E$. Then $1-{\lambda }_{n}a\to 1-{\lambda }_{0}a$ is a noninvertible limit of invertibles. The one-sided-inverse argument in part (ii) shows that this limit has neither a left inverse of an algebra element nor a right inverse of an algebra element, contradicting the hypothesis. Therefore $E$ has empty boundary; it is both open and closed in the connected complex plane, and $E=\mathbb{C}$.

For every nonzero $\mu$, put $\lambda ={\mu }^{-1}$. The identity $\mu 1-a=\mu (1-\lambda a)$ proves invertibility, so the spectrum of an element $a$ is contained in $\{0\}$. It is nonempty: if the resolvent of an element existed for all $\mu \in \mathbb{C}$, it would be an entire algebra-valued function tending to zero at infinity by the Neumann series. Applying any bounded linear functional and then the Liouville theorem would make it zero; the Hahn-Banach theorem separates points and would force the resolvent itself to be zero, contradicting its inverse equation. Hence

First, a maximal left ideal $L$ is closed. Its norm closure is again a left ideal, since multiplication is continuous. If that closure were all of $A$, some $\ell \in L$ would satisfy $\Vert 1-\ell \Vert <1$ and would be invertible by the Neumann series. But a proper left ideal cannot contain an invertible element: ${\ell }^{-1}\ell =1$ would put $1$ in $L$. Maximality therefore forces $\overline{L}=L$.

Consequently $X=A/L$ is a nonzero quotient Banach space. It is a simple module for left multiplication by $A$, because the inverse image of any submodule under $A\to A/L$ is a left ideal containing $L$, and hence is either $L$ or $A$. Since $La\subseteq L$, right multiplication defines a well-defined bounded module endomorphismIndeed, changing $u$ by an element of $L$ changes $ua$ by an element of $L$, and taking the infimum over representatives proves the quotient norm bound. Right multiplication commutes with the left action by associativity.

Choose $\lambda$ in the nonempty spectrum of an element of $T$ in the Banach algebra $\mathcal{B}(X)$. Both the kernel and range of $T-\lambda I$ are submodules. If this module endomorphism were nonzero, simplicity would force its kernel to be zero and its range to be all of $X$. It would be a bounded bijection of Banach spaces, with bounded inverse by the bounded inverse theorem, contradicting the choice of $\lambda$. Thus $T=\lambda I$. Applying this identity to $1+L$ yields $a-\lambda 1\in L$. If also $a-\mu 1\in L$, then $(\lambda -\mu )1\in L$; since $1\notin L$, $\lambda =\mu$. ThereforeThis is the scalar right action on a maximal-left-ideal quotient; it does not assume that $L$ is a two-sided ideal or that $A/L$ is an algebra.

Now consider the center of an associative algebra $Z$. For each $u\in A$, the map $z\mapsto zu-uz$ is a continuous linear map, so its kernel is closed. Their intersection $Z$ is therefore closed. The identity belongs to $Z$, as do sums and scalar multiples of its elements. If $z,w\in Z$, then for every $u\in A$,Thus $zw\in Z$. Also $zw=wz$, since $z$ commutes with every element of $A$, including $w$. Hence the center is a closed commutative unital subalgebra, and is a Banach algebra in the inherited norm.

For every $z\in Z$, the inclusion $Lz=zL\subseteq L$ allows the first argument to apply. Let $\chi (z)$ be the unique scalar for which $z-\chi (z)1\in L$. Equivalently, the induced right multiplication on $A/L$ is $T_z=\chi (z)I$. Sincethe map $\chi :Z\to \mathbb{C}$ is a complex-linear unital character of an algebra. It is surjective because $\chi (\alpha 1)=\alpha$, and its kernel is precisely $L\cap Z$. ConsequentlyIndeed any ideal properly containing the kernel would have a nonzero image in the field $\mathbb{C}$ and thus would be all of $Z$. The algebra character is also continuous: $|\chi (z)|=\Vert \chi (z)I\Vert =\Vert T_z\Vert \le \Vert z\Vert$, although continuity is not needed for maximality.

An algebra involution is a map $x\mapsto x^{\ast }$ satisfyingThus it is a conjugate-linear map, reverses multiplication, and has square equal to the identity map. These are algebraic axioms; norm continuity, when included in a convention for a Banach star algebra, is an additional requirement and is not used here. A Hermitian element of a star algebra satisfies $h^{\ast }=h$.

Conjugate linearity gives $0^{\ast }=0$. The map $\ast$ is surjective, so every $u$ is $v^{\ast }$ for some $v$. The identities $(v1{)}^{\ast }=1^{\ast }{v}^{\ast }=v^{\ast }$ and $(1v{)}^{\ast }=v^{\ast }{1}^{\ast }=v^{\ast }$ show that $1^{\ast }$ is a two-sided identity; uniqueness of the identity gives $1^{\ast }=1$. Therefore both zero and the identity are Hermitian algebra elements.

For the final unheaded request, first prove equality of the two spectra for a Hermitian element of a star algebra $h\in B$. Write $D=\mathbb{C}\setminus {\sigma }_A(h)$. The spectrum is a compact subset of the real axis by the assumption that $A$ is a Hermitian Banach algebra, so $D$ is connected: points above and below the real axis can be joined by paths passing beyond the endpoints of an interval containing the spectrum; points of the real axis outside the spectrum can first move a short distance vertically. DefineFor large $|\lambda |$, the Neumann series ${\lambda }^{-1}{\sum }_{n\ge 0}(h/\lambda {)}^n$ converges in the closed unital star-subalgebra $B$, so $E$ is nonempty. It is relatively open in $D$ by the same series applied to a perturbation of an already invertible element of $B$. It is relatively closed in $D$, since the resolvent of an element is continuous in $A$ and $B$ is closed. Connectedness gives $E=D$. Conversely an inverse in $B$ is an inverse in $A$, soThis proves the needed spectrum in a closed unital subalgebra equality for Hermitian algebra elements without presupposing that their spectra in $B$ are real.

Now suppose $b\in B$ is invertible in $A$. Both $b{b}^{\ast }$ and $b^{\ast }b$ are Hermitian algebra elements, since reversing products and applying $\ast$ twice fixes them. They are invertible in $A$, by part (ii), so the preceding equality of spectra of Hermitian algebra elements makes them invertible in $B$. Part (ii), applied within $B$, then makes $b$ invertible in $B$. Applying this to $b-\lambda 1$ for every scalar $\lambda$, and using the automatic converse for an inverse already in $B$, proves spectral permanence for Hermitian Banach algebras:Only closedness, a common identity, closure under the algebra involution, and real ambient spectra of Hermitian algebra elements were used; the C-star identity was not assumed.

A linear operator between Banach spaces is a compact operator when it maps the unit ball into a relatively compact set, equivalently when every bounded sequence of inputs has an image subsequence converging in norm. Such an operator is bounded. For a bounded linear operator on a complex Banach space, the resolvent set consists of the complex $\lambda$ for which $T-\lambda I$ is bijective with bounded inverse; its complement is the spectrum. The point spectrum consists of those $\lambda$ for which $T-\lambda I$ has a nonzero kernel, namely the eigenvalues.

On a Hilbert space, $T$ is self-adjoint if $T=T^{\ast }$, or equivalently $⟨Tx,y⟩=⟨x,Ty⟩$ for all vectors. The given prescription extends uniquely toIt is bounded with norm one. The real diagonal entries prove self-adjointness by the inner-product identity. Define the finite-rank truncation $T_{m}x={\sum }_{j\le m}{x}_j{e}_j/j$. Then $\Vert T-T_m\Vert =1/(m+1)\to 0$. To see compactness directly, extract successively convergent subsequences of the finitely many first coordinates from a bounded input sequence; the uniform tail bound just displayed makes the resulting image subsequence Cauchy. Hence $T$ is compact and self-adjoint.

Each $1/j$ is an eigenvalue with eigenspace $\mathbb{C}{e}_j$. If $Tx=\lambda x$, each coordinate satisfies $(1/j-\lambda )x_j=0$, so there are no other eigenvalues; in particular zero is not an eigenvalue. But $\Vert T{e}_j\Vert =1/j\to 0$, which contradicts a bounded inverse at zero. Thus zero is in the spectrum.

For $\lambda \notin \{0,1,1/2,1/3,\dots \}$, its distance from this closed set is positive. The diagonal formulathen defines a bounded inverse. This proves the complete description

For a normalized trial state in the quadratic-form domain of a self-adjoint Hamiltonian operator bounded below, its expectation of $H$ is at least the ground energy $E_0$. Minimizing this Rayleigh quotient over a tractable trial family therefore supplies an upper bound.

For the two states, set $a_{\pm }=(a_1\pm a_2)/\sqrt{2}$. Their quotient isThe strict assumption $ϵ<sE$ excludes $|s|=1$, since equality in the overlap Cauchy-Schwarz bound would then force the two states to be proportional and $ϵ=sE$. Thus $|s|<1$. The two generalized eigenvalues are $(E\pm ϵ)/(1\pm s)$, and their difference has the sign of $ϵ-sE$. The symmetric trial is the lower one, so

For the single attractive delta well, the supplied wavefunction has norm one because $\int \lambda e^{-2\lambda |x|}dx=1$. Its kinetic energy in quadratic-form notation is ${\hslash }^2\int |{\psi }_1^{\prime }{|}^2/(2m)={\hslash }^2{\lambda }^2/(2m)$, and its delta potential energy is $-K|{\psi }_1(0){|}^2=-K\lambda$. With $\lambda =mK/{\hslash }^2$ this givesIts derivative jump is ${\psi }_1^{\prime }(0+)-{\psi }_1^{\prime }(0-)=-2\lambda {\psi }_1(0)$, exactly the delta boundary condition, so it is an eigenstate. A decaying bound state on the two half-lines must have a common decay rate fixed by this same jump; there is only this one negative-energy state, confirming it is the ground state.

For $R>0$, integrate the overlap on the three regions $x<0$, $0<x<R$, $x>R$. Put $a=\lambda R$. The result isThe second equality is the single-well energy plus the extra delta expectation; the third follows by applying either single-well eigenvalue equation in the cross matrix element. Moreover $ϵ-sE=K\lambda e^{-a}[(1+a)e^{-2a}-1]<0$, so the preceding bound applies. Substitution givesFor widely separated wells, each localized orbital is close to an isolated-well ground state and the low-energy subspace is accurately their span; the symmetric combination captures the leading tunneling energy lowering. In the limit both exact and trial energies approach $E_B$. This illustrates electronic molecular binding: sharing an orbital between attractive centers can lower its energy below the separated-center value. A full equilibrium bond length would also require the repulsion of the positive centers.

Put $\zeta =e^{2\pi i/7}$ and induce the characters ${\chi }_r(a)={\zeta }^r$ of $H=⟨a⟩$. Since $H$ is normal, the induced value outside $H$ is zero, while at $a^m$ it is ${\zeta }^{rm}+{\zeta }^{2rm}+{\zeta }^{4rm}$. The nonzero $r$ split into the two doubling orbits, so the two distinct induced characters have class valueswhere $A=\zeta +{\zeta }^2+{\zeta }^4$ and $B={\zeta }^3+{\zeta }^5+{\zeta }^6=\overline{A}$. Summing seventh roots gives $A+B=-1$, and multiplication gives $AB=2$, hence $|A{|}^2=|B{|}^2=2$. Their character norms are $(9+3\cdot 2+3\cdot 2)/21=1$, proving irreducibility. Together with the three degree-one characters, the sum of squared degrees is $3+2\cdot 9=21$. Thus these are all the remaining irreducible characters.

The continuous dual space $X^{\ast }$ consists of bounded complex-linear functionals $\varphi :X\to \mathbb{C}$, with $\Vert \varphi \Vert ={\sup }_{\Vert x\Vert \le 1}|\varphi (x)|$. For $1\le s<\infty$, ${\ell }^s$ consists of sequences with ${\sum }_j|x_j{|}^s<\infty$, norm $({\sum }_j|x_j{|}^s{)}^{1/s}$; ${\ell }^{\infty }$ consists of bounded sequences with supremum norm.

For $a\in {\ell }^p$, define ${\varphi }_a(x)={\sum }_j{a}_j{x}_j$ on ${\ell }^q$. Hölder's inequality gives absolute convergence and $\Vert {\varphi }_a\Vert \le \Vert a{\Vert }_p$. For a finite truncation, choose $x_j=\overline{a_j}|a_j{|}^{p-2}/({\sum }_{j\le N}|a_j{|}^p{)}^{1/q}$ for $j\le N$, zero elsewhere, interpreting the numerator as zero when $a_j=0$. Its $q$-norm is one and its pairing equals $({\sum }_{j\le N}|a_j{|}^p{)}^{1/p}$. Letting $N$ grow proves equality of the norms.

Conversely, for a bounded $\varphi$ let $a_j=\varphi (e_j)$. The same finite test vector gives ${\sum }_{j\le N}|a_j{|}^p\le \Vert \varphi {\Vert }^p$, so $a\in {\ell }^p$. Finite truncations converge in ${\ell }^q$, and continuity gives $\varphi (x)={\sum }_j{a}_j{x}_j$. Thus $\boxed{({\ell }^q{)}^{\ast }\cong {\ell }^p\text{ isometrically}.}$

For ${\ell }^1$, boundedness gives $|a_j|\le \Vert \varphi \Vert$, while any bounded sequence defines an absolutely convergent pairing bounded by $\Vert a{\Vert }_{\infty }\Vert x{\Vert }_1$. Density of finite sequences again proves surjectivity, and single-coordinate tests give norm equality. Hence $\boxed{({\ell }^1{)}^{\ast }\cong {\ell }^{\infty }\text{ isometrically}.}$ The reversed argument fails for ${\ell }^{\infty }$: finite sequences are not dense in its supremum norm; their closure is $c_0$. For example the limit functional on convergent sequences extends by the Hahn-Banach theorem to ${\ell }^{\infty }$, vanishes on every $e_j$, yet takes value one on the constant-one sequence. It cannot be represented by pairing with an ${\ell }^1$ coefficient sequence.

Working with measurable functions modulo equality almost everywhere, the sequence is Cauchy in Lp space if for every $ϵ>0$ there exists $N$ such that $\Vert f_n-f_m{\Vert }_p<ϵ$ for all $m,n\ge N$. This is a statement about integral norm distance, not pointwise convergence. Without quotienting null functions the displayed quantity is a seminorm, so the conventional $L^p$ normed space uses these equivalence classes.