Banach and Hilbert Spaces, Completion, and Riesz Representation

A norm says when two vectors are close and when an approximation converges. Completeness says that every Cauchy approximation actually has a limit in the space. An inner product adds orthogonality, projections, and coefficients; a complete inner-product space is a Hilbert space. Finally, the Riesz representation theorem says that every continuous linear functional on a complex Hilbert space has the unique form

$$F(\psi)=\langle f_F|\psi\rangle, \qquad \|F\|=\|f_F\|.$$

Here the bra is conjugate-linear and the ket is linear. Consequently the identification $f_F\mapsto F$ is conjugate-linear, not complex-linear. Together, these facts let one begin with a convenient algebraic span of wavepackets or modes, complete it without changing its old distances, and represent every continuous linear amplitude functional on the completed state space by a unique state vector. They do not turn delta-normalized modes, point evaluations, or unbounded operators into Hilbert-space vectors or bounded maps.

Helpful background. Lp Spaces, Inequalities, and Weak Convergence supplies measurable-function equivalence classes, $L^p$ norms, and the distinction between norm and weak convergence.

Normed and Hilbert spaces

Let $\mathbb K\in\{\mathbb R,\mathbb C\}$ be the scalar field; the QFT application uses $\mathbb K=\mathbb C$. A star on a scalar denotes complex conjugation and is the identity when $\mathbb K=\mathbb R$. The site convention is

$$\begin{aligned} \langle \alpha\phi_1+\beta\phi_2|\psi\rangle &= \alpha^*\langle\phi_1|\psi\rangle +\beta^*\langle\phi_2|\psi\rangle,\\ \langle\phi|\alpha\psi_1+\beta\psi_2\rangle &= \alpha\langle\phi|\psi_1\rangle +\beta\langle\phi|\psi_2\rangle. \end{aligned}$$

Thus over $\mathbb C$ the first slot, the bra, is conjugate-linear, while over $\mathbb R$ the inner product is symmetric and bilinear. Some mathematics texts choose the opposite complex slot. The two complex conventions state the same geometry, but the order of the vectors in a formula for a linear functional must be translated.

The continuous dual of a normed space $X$ is denoted

$$X^* = \{F:X\to\mathbb K: F\text{ is linear and continuous}\}.$$

This is not the full algebraic dual. It is also not a space of generalized functions unless a separate topology and dual pairing have been declared. Bounded and compact operators continue on the next operator page; unbounded maps, their domains, and their adjoints continue at Unbounded Operators, Domains, Closure, and Adjoints. The present page supplies the state-space foundation for both routes.

Norms say which approximations are close

A normed vector space is a vector space $X$ with a map $\|\cdot\|:X\to[0,\infty)$ satisfying

$$\begin{aligned} \|x\|=0&\Longleftrightarrow x=0,\\ \|\alpha x\|&=|\alpha|\,\|x\|,\\ \|x+y\|&\leq\|x\|+\|y\|. \end{aligned}$$

The norm supplies the metric $d(x,y)=\|x-y\|$. A sequence $x_n$ converges to $x$ in norm when $\|x_n-x\|\to0$. It is Cauchy when, for every $\varepsilon>0$, there is an $N$ such that

$$\|x_n-x_m\|<\varepsilon \quad\text{whenever}\quad n,m\geq N.$$

Every convergent sequence is Cauchy. The converse is a property of the space, not of the sequence.

A normed space is a Banach space when every Cauchy sequence converges to an element of that space. Completeness is therefore a closure condition on all norm-controlled approximation procedures. It does not assert pointwise convergence, differentiability of the limit, compactness of bounded sets, or convergence in a different norm.

An inner product on a complex vector space is a positive-definite sesquilinear form satisfying

$$\langle\phi|\psi\rangle = \langle\psi|\phi\rangle^*, \qquad \langle\psi|\psi\rangle>0 \quad\text{for }\psi\neq0.$$

The Cauchy–Schwarz inequality,

$$|\langle\phi|\psi\rangle| \leq \|\phi\|\,\|\psi\|,$$

shows that $\|\psi\|=\sqrt{\langle\psi|\psi\rangle}$ is a norm and that the inner product is continuous in both slots. An inner-product space need not be complete; it is often called a pre-Hilbert space. A complete inner-product space is a Hilbert space.

These definitions form two different layers:

Structure	What is supplied	What is not automatic
Normed space	distance, norm convergence, boundedness	limits of all Cauchy sequences
Banach space	a normed space plus completeness	angles and orthogonal projections
Inner-product space	norm plus angles and orthogonality	completeness
Hilbert space	inner-product geometry plus completeness	boundedness of every physically interesting operator

For example, $C([0,1])$ with the supremum norm is Banach. The standard sequence spaces $\ell^p$ are Banach for $1\leq p\leq\infty$, but their standard norm is induced by an inner product only for $p=2$. The diagnostic is the parallelogram law

$$\|x+y\|^2+\|x-y\|^2 = 2\|x\|^2+2\|y\|^2.$$

The Jordan–von Neumann theorem says that a norm comes from an inner product if and only if it obeys this identity. For the standard basis vectors $e_1,e_2$ of $\ell^p$,

$$\|e_1+e_2\|_p^2+\|e_1-e_2\|_p^2 = 2^{1+2/p},$$

whereas the right-hand side of the parallelogram law is $4$. Equality holds only at $p=2$; for $p=\infty$, the left-hand side is $2$. This checks the given norm. It should not be confused with the deeper question of whether a Banach space is linearly isomorphic to some Hilbert space.

Completion adds the missing limits

Let $X_0$ be an incomplete normed space. Its completion is not obtained by declaring that a nonexistent limit was secretly in $X_0$. It is constructed from the approximation data.

Let $\mathcal C(X_0)$ be the vector space of Cauchy sequences in $X_0$ and write

$$(x_n)\sim(y_n) \quad\Longleftrightarrow\quad \|x_n-y_n\|\longrightarrow0.$$

The completion is the quotient

$$\widehat X = \mathcal C(X_0)/{\sim}.$$

Addition and scalar multiplication are defined term by term. The norm is

$$\|[x_n]\|_{\widehat X} = \lim_{n\to\infty}\|x_n\|_{X_0}.$$

The limit exists because the reverse triangle inequality gives

$$\big|\|x_n\|-\|x_m\|\big| \leq \|x_n-x_m\|.$$

The same estimate shows that the norm does not depend on the representative. The map

$$j:X_0\longrightarrow\widehat X, \qquad j(x)=[(x,x,x,\ldots)]$$

is a linear isometry, and $j(X_0)$ is dense in $\widehat X$. A diagonal choice from a Cauchy sequence of equivalence classes proves that $\widehat X$ is complete. More precisely, if $(Y,i)$ is another completion, meaning that $Y$ is Banach and $i:X_0\to Y$ is a linear isometry with dense range, then there is a unique surjective linear isometry $U:\widehat X\to Y$ satisfying $U\circ j=i$. This is the precise sense in which the completion is canonical (Kehle 2025, §1.4, PDF; Teschl 2014, §0.4, PDF).

The completion depends on the norm

Let $c_{00}$ be the vector space of complex sequences with finite support. As an algebraic vector space it is the same in each row below, but its completion changes:

Norm on $c_{00}$	Completion
$\lVert x\rVert_1=\sum_n\lvert x_n\rvert$	$\ell^1$
$\lVert x\rVert_2=(\sum_n\lvert x_n\rvert^2)^{1/2}$	$\ell^2$
$\lVert x\rVert_\infty=\sup_n\lvert x_n\rvert$	$c_0$, the sequences tending to zero

Thus “take the completion” is incomplete unless the norm has already been specified. In infinite dimensions, two reasonable norms can encode different notions of small error and produce different limit spaces.

For a pre-Hilbert space $H_0$, the inner product extends along Cauchy representatives:

$$\langle[x_n]|[y_n]\rangle_{\widehat H} = \lim_{n\to\infty}\langle x_n|y_n\rangle_{H_0}.$$

Cauchy–Schwarz and the boundedness of Cauchy sequences show that this limit exists and is independent of both representatives. Its induced norm is the completed norm, so $\widehat H$ is a Hilbert space.

Closed subspaces and complete subspaces

If $M$ is a linear subspace of a Banach space $X$, then

$$M\text{ is complete in the inherited norm} \quad\Longleftrightarrow\quad M\text{ is closed in }X.$$

For the forward direction, a limit in $X$ of a sequence in $M$ makes that sequence Cauchy in $M$, so completeness puts the limit back in $M$. For the reverse direction, a Cauchy sequence in $M$ converges in $X$, and closedness keeps the limit in $M$.

This statement is frequently the hidden reason for the word “closed” in a Hilbert-space theorem. A proper dense subspace is necessarily incomplete in the inherited norm.

Only bounded linear maps extend automatically

Suppose $D\subseteq X$ is a dense linear subspace, $Y$ is Banach, and $T_0:D\to Y$ is bounded and linear:

$$\|T_0x\|_Y\leq C\|x\|_X.$$

Then there is a unique bounded extension $T:X\to Y$,

$$Tx=\lim_{n\to\infty}T_0x_n \quad \text{for any }x_n\in D\text{ with }x_n\to x,$$

and $\|T\|=\|T_0\|$. Boundedness makes $(T_0x_n)$ Cauchy and makes the answer independent of the approximating sequence (Kehle 2025, §2.1, PDF; Teschl 2014, §0.5, PDF).

Without a bound there is no such conclusion. Differentiation on $C^1([0,1])\subset C([0,1])$, with both domain and ambient space measured by the supremum norm, is unbounded: for $f_n(x)=\sin(n\pi x)$,

$$\|f_n\|_\infty=1, \qquad \|f_n'\|_\infty=n\pi.$$

Completion therefore extends continuous structure. It does not erase the domain questions of an unbounded differential operator.

Hilbert geometry gives projections and expansions

For a subset $M\subseteq H$, define its orthogonal complement by

$$M^\perp = \{\psi\in H: \langle m|\psi\rangle=0 \text{ for every }m\in M\}.$$

It is always a closed linear subspace. If $M$ itself is a closed linear subspace of a Hilbert space, every $\psi\in H$ has a unique decomposition

$$\psi=P_M\psi+(1-P_M)\psi, \qquad P_M\psi\in M, \quad (1-P_M)\psi\in M^\perp.$$

Equivalently, $P_M\psi$ is the unique vector in $M$ nearest to $\psi$. This is the projection theorem.

The role of completeness is visible in its proof. Put

$$d=\inf_{m\in M}\|\psi-m\|$$

and choose $m_n\in M$ with $\|\psi-m_n\|\to d$. The parallelogram identity gives

$$\begin{aligned} \|m_n-m_k\|^2 &= 2\|\psi-m_n\|^2 +2\|\psi-m_k\|^2\\ &\quad -4\left\| \psi-\frac{m_n+m_k}{2} \right\|^2\\ &\leq 2\|\psi-m_n\|^2 +2\|\psi-m_k\|^2 -4d^2. \end{aligned}$$

Hence $(m_n)$ is Cauchy. Hilbert completeness gives a limit, and closedness of $M$ keeps it in $M$. Varying the minimizer along $m+t v$, for $v\in M$, with real $t$ proves orthogonality in the real case. In the complex case, also taking $t=is$ with $s\in\mathbb R$ makes both real and imaginary parts vanish. Thus $\psi-P_M\psi\in M^\perp$. The Pythagorean identity proves uniqueness.

Closedness cannot be omitted. In $\ell^2$, the subspace $c_{00}$ is dense but not closed. For

$$\psi=(1,\tfrac12,\tfrac13,\ldots)\in\ell^2\setminus c_{00},$$

the distance from $\psi$ to $c_{00}$ is zero, yet there is no nearest vector in $c_{00}$. The truncations approach $\psi$, but their limit is precisely the vector missing from the subspace.

For an arbitrary linear subspace $M$,

$$M^{\perp\perp}=\overline M, \qquad H=\overline M\oplus M^\perp.$$

Thus the direct sum is $H=M\oplus M^\perp$ exactly when $M$ is closed.

An orthonormal family $(e_j)_{j\in J}$ satisfies

$$\langle e_j|e_k\rangle=\delta_{jk}.$$

For every finite subset $K\subset J$, Bessel’s inequality reads

$$\sum_{j\in K} |\langle e_j|\psi\rangle|^2 \leq \|\psi\|^2.$$

If the closed linear span of the family is all of $H$, it is an orthonormal basis, and

$$\begin{aligned} \psi &= \sum_{j\in J} \langle e_j|\psi\rangle e_j,\\ \|\psi\|^2 &= \sum_{j\in J} |\langle e_j|\psi\rangle|^2. \end{aligned}$$

The first sum converges in Hilbert norm; the second equality is Parseval’s identity. Even when $J$ is uncountable, a fixed vector has at most countably many nonzero coefficients. A Hilbert space is separable when it has a countable dense subset, equivalently a countable orthonormal basis. Every separable infinite-dimensional complex Hilbert space is unitarily isomorphic to $\ell^2(\mathbb C)$; in the real case it is orthogonally isomorphic to $\ell^2(\mathbb R)$. Either isomorphism depends on a choice of orthonormal basis.

An orthonormal basis is not a Hamel basis. In an infinite-dimensional Hilbert space, a typical vector is an infinite norm-convergent sum, not a finite algebraic combination. Completion is exactly what licenses those limits (Teschl 2014, §§1.1–1.3, PDF).

Riesz turns bounded linear questions into vectors

For a normed space $X$, a linear functional $F:X\to\mathbb K$ is continuous if and only if it is bounded. Its norm is

$$\|F\| = \sup_{\|\psi\|\leq1}|F(\psi)| = \sup_{\psi\neq0} \frac{|F(\psi)|}{\|\psi\|}.$$

The continuous dual $X^*$ is always Banach, even when $X$ is incomplete (Kehle 2025, §2.1, PDF). For a general Banach space, however, its dual need not look like the original space. For example,

$$(\ell^1)^*\cong\ell^\infty, \qquad F_a(x)=\sum_{n=1}^{\infty}a_nx_n.$$

This identification is isometric (Kehle 2025, §4.3, PDF). Hilbert geometry gives a much stronger result.

Riesz representation theorem for Hilbert spaces. Let $H$ be a real or complex Hilbert space and let $F\in H^*$. There is a unique $f_F\in H$ such that

$$F(\psi)=\langle f_F|\psi\rangle \quad \text{for every }\psi\in H,$$

and

$$\|F\|=\|f_F\|.$$

Proof. If $F=0$, take $f_F=0$. Otherwise $M=\ker F$ is a closed proper subspace because $F$ is continuous. Choose $z$ with $F(z)\neq0$ and form

$$u = \frac{z-P_Mz}{\|z-P_Mz\|}.$$

Then $u\in M^\perp$, $\|u\|=1$, and $F(u)\neq0$. Every $\psi\in H$ decomposes as

$$\psi = \left( \psi-\frac{F(\psi)}{F(u)}u \right) + \frac{F(\psi)}{F(u)}u,$$

with the expression in parentheses belonging to $M$. Set

$$f_F=F(u)^*u.$$

Because the bra slot is conjugate-linear,

$$\langle f_F|u\rangle = F(u),$$

and the decomposition gives $\langle f_F|\psi\rangle=F(\psi)$. If two vectors represent $F$, their difference is orthogonal to every vector, including itself, so they are equal. Finally, Cauchy–Schwarz gives $\|F\|\leq\|f_F\|$, while evaluating on $f_F/\|f_F\|$ gives the reverse inequality. This proves the theorem.

Completeness is essential. On $c_{00}$ with the $\ell^2$ norm,

$$F(x)=\sum_{n=1}^{\infty}\frac{x_n}{n}$$

is bounded by Cauchy–Schwarz, but its representing vector $(1,\frac12,\frac13,\ldots)$ belongs to the completion $\ell^2$, not to $c_{00}$.

The resulting Riesz map

$$\begin{aligned} J:H&\longrightarrow H^*,\\ f&\longmapsto Jf, \qquad (Jf)(\psi)=\langle f|\psi\rangle \end{aligned}$$

is an isometric bijection, but over $\mathbb C$ it obeys

$$J(\alpha f+\beta g) = \alpha^*Jf+\beta^*Jg.$$

It is therefore conjugate-linear. Writing $H=H^*$ without this qualification hides a real convention-dependent step. Over $\mathbb R$, the map is linear.

The Riesz map should also not be confused with the canonical evaluation map

$$\iota:H\longrightarrow H^{**}, \qquad (\iota\psi)(F)=F(\psi),$$

which is linear over both scalar fields. Hilbert spaces are reflexive, so $\iota$ is surjective, but that statement and $J:H\to H^*$ have different types.

The theorem concerns continuous linear functionals on a Hilbert space. It is not the Riesz–Markov–Kakutani theorem representing functionals on spaces of continuous functions by measures, and it does not represent arbitrary discontinuous algebraic functionals or distributional evaluations by vectors of $H$ (Kehle 2025, §5.2, PDF; Teschl 2014, §1.3, PDF).

Completing one-particle mode spans

The page’s controlled QFT application is a massive scalar one-particle space. It illustrates the mathematical construction; the Poincaré classification, spin and helicity labels, and physical normalization choices belong to One-Particle States: Mass, Spin, and Relativistic Normalization.

Discrete modes first

Suppose a finite-volume problem supplies a countable orthonormal family of one-particle modes $\{|n\rangle\}_{n\in\mathbb N}$. Begin with the algebraic span

$$\mathcal D_0 = \left\{ \sum_{n=1}^{N}c_n|n\rangle: N<\infty \right\}.$$

For two finite sums,

$$\left\langle \sum_n c_n|n\rangle \middle| \sum_n d_n|n\rangle \right\rangle = \sum_n c_n^*d_n.$$

The partial sums

$$|\psi_N\rangle = \sum_{n=1}^{N}c_n|n\rangle$$

are Cauchy exactly when

$$\sum_{n=1}^{\infty}|c_n|^2<\infty,$$

because

$$\|\psi_N-\psi_M\|^2 = \sum_{M<n\leq N}|c_n|^2.$$

Completing $\mathcal D_0$ therefore produces $\ell^2$ coefficients and includes norm limits which are not finite mode sums. No new inner products among the old finite sums are introduced.

The infinite-volume mass shell

Now work in $d$-dimensional Lorentzian spacetime with the site’s $(+,-,\ldots,-)$ metric. For $m>0$, put

$$E_{\mathbf p} = \sqrt{|\mathbf p|^2+m^2}$$

and use the positive-energy mass-shell measure

$$\mathrm d\mu_m(\mathbf p) = \frac{\mathrm d^{d-1}\mathbf p} {(2\pi)^{d-1}2E_{\mathbf p}}.$$

On smooth compactly supported wavepackets define

$$\langle f|g\rangle_1 = \int \mathrm d\mu_m(\mathbf p)\, f(\mathbf p)^*g(\mathbf p).$$

The one-particle Hilbert space for this scalar example is the completion

$$\mathcal H_1 = \overline{C_c^\infty(\mathbb R^{d-1})}^{\,\|\cdot\|_1}.$$

Equivalently, after the usual almost-everywhere identification, this is the corresponding $L^2$ space. Indeed, $\mathrm d\mu_m$ is a locally finite regular Borel measure, continuous compactly supported functions are dense in its $L^2$ space, and uniform approximation on compact sets by smooth compactly supported functions proves the displayed density (Teschl 2014, §0.6, pp. 36–37, PDF). A state is represented by a square-integrable wavefunction, not by its values at individual momenta. Norm convergence means

$$\|f_n-f\|_1^2 = \int\mathrm d\mu_m(\mathbf p)\, |f_n(\mathbf p)-f(\mathbf p)|^2 \longrightarrow0.$$

It does not require pointwise convergence.

The common momentum-ket notation is a distributional shorthand. With covariant normalization,

$$\langle\mathbf p|\mathbf q\rangle = (2\pi)^{d-1}2E_{\mathbf p}\, \delta^{(d-1)}(\mathbf p-\mathbf q),$$

one writes formally

$$\begin{aligned} |f\rangle &= \int\mathrm d\mu_m(\mathbf p)\, f(\mathbf p)|\mathbf p\rangle,\\ \mathbf 1_{\mathcal H_1} &= \int\mathrm d\mu_m(\mathbf p)\, |\mathbf p\rangle\langle\mathbf p|. \end{aligned}$$

These formulas reproduce $\langle f|g\rangle_1=\int\mathrm d\mu_m\,f^*g$ (Schwartz 2014, §2.3.1). They do not imply $|\mathbf p\rangle\in\mathcal H_1$: its displayed “norm” contains $\delta^{(d-1)}(0)$.

Riesz representation makes the distinction exact. For every $g\in\mathcal H_1$,

$$F_g(f)=\langle g|f\rangle_1$$

is a bounded linear functional with $\|F_g\|=\|g\|_1$, and every bounded linear functional on $\mathcal H_1$ is of this form. By contrast,

$$f\longmapsto f(\mathbf p_0)$$

is not even well defined on $L^2$ equivalence classes and is not bounded in the $L^2$ norm. The formal momentum bra $\langle\mathbf p_0|$ is therefore not in the continuous Hilbert dual identified by Riesz. Generalized kets and distributional state spaces need the finer topology developed later in Locally Convex, Nuclear, and Rigged Hilbert Spaces.

What the structure establishes

Norm, completeness, and inner product answer different parts of the principal question:

• the norm fixes the approximation and error notion;
• Banach completeness keeps every norm-Cauchy approximation inside the state or function space;
• an inner product adds orthogonality, nearest-point projections, and coefficient expansions;
• Hilbert completion turns a pre-Hilbert mode span into a state space containing all square-summable norm limits;
• Riesz representation identifies continuous linear amplitudes with bras arising from unique Hilbert vectors, with a conjugate-linear identification over $\mathbb C$; and
• bounded linear maps extend across dense completions, while unbounded operations retain separate domain questions.

The most consequential misuse is to infer that every formal mode or linear expression is a Hilbert vector or continuous functional. The norm and the declared topology decide that question.

This page is hard preparation for both Bounded, Compact, and Integral Operators and Unbounded Operators, Domains, Closure, and Adjoints. Its exact physical continuation is the one-particle-state treatment.

Common pitfalls

Complete is not closed without an ambient space. Completeness is a property of a metric space. Closedness is a property of a subset of another topological space. A subspace of a Banach space is complete in the inherited norm exactly when it is closed.

Completion is not independent of the norm. The same algebraic space $c_{00}$ completes to $\ell^1$, $\ell^2$, or $c_0$ under three different norms. Always state the norm before naming a completion.

A Banach space is not automatically Hilbert. Completeness supplies limits, not an inner product. The parallelogram law tests whether the given norm comes from Hilbert geometry.

The complex Riesz map is not linear. With bras conjugate-linear, $f\mapsto\langle f|\cdot\rangle$ is conjugate-linear. The representing formula for a linear functional is $F(\psi)=\langle f_F|\psi\rangle$, not $\langle\psi|f_F\rangle$.

Closedness is part of the projection theorem. A vector may have distance zero from a dense proper subspace without having a nearest vector in that subspace.

Delta-normalized modes are not normalizable states. They are useful generalized objects. Physical one-particle vectors are square-integrable wavepackets, and point evaluation is not a bounded functional on $L^2$.

Exercises

Classify four spaces

Classify each space as normed, Banach, pre-Hilbert, or Hilbert, using every applicable label:

1. $c_{00}$ with the $\ell^2$ norm;
2. $\ell^2$ with its standard inner product;
3. $C([0,1])$ with the supremum norm; and
4. $C([0,1])$ with the $L^2$ inner product.

Solution

1. $c_{00}$ is a normed pre-Hilbert space, but it is not complete, so it is neither Banach nor Hilbert. Its Hilbert completion is $\ell^2$.
2. $\ell^2$ is complete in its inner-product norm, so it is both Banach and Hilbert.
3. $C([0,1])$ with the supremum norm is Banach. That norm fails the parallelogram law, so this is not a Hilbert space with the stated norm.
4. $C([0,1])$ with the $L^2$ inner product is a pre-Hilbert space but is not complete. Its completion is $L^2([0,1])$, whose elements are almost-everywhere equivalence classes rather than necessarily continuous functions.

Find the missing projection hypothesis

Let $M=c_{00}\subset\ell^2$ and $\psi=(1,\tfrac12,\tfrac13,\ldots)$. Show that $\inf_{m\in M}\|\psi-m\|_2=0$, but that no minimizing $m\in M$ exists. Which hypothesis of the projection theorem fails?

Solution

Let

$$m_N=(1,\tfrac12,\ldots,\tfrac1N,0,0,\ldots).$$

Then

$$\|\psi-m_N\|_2^2 = \sum_{n>N}\frac1{n^2} \longrightarrow0,$$

so the infimum is zero. A minimizer would have norm distance zero from $\psi$, hence would equal $\psi$, but $\psi\notin c_{00}$. The missing hypothesis is that $M$ be closed.

Check the Riesz scaling

Let $F(\psi)=\langle f|\psi\rangle$ on a complex Hilbert space. Compute the representing vector for $\alpha F+\beta G$ in terms of the representing vectors $f$ and $g$. Explain why the Riesz map is conjugate-linear.

Solution

We need $h$ such that

$$\langle h|\psi\rangle = \alpha\langle f|\psi\rangle +\beta\langle g|\psi\rangle.$$

Because the bra slot is conjugate-linear,

$$h=\alpha^*f+\beta^*g$$

gives

$$\langle\alpha^*f+\beta^*g|\psi\rangle = \alpha\langle f|\psi\rangle +\beta\langle g|\psi\rangle.$$

Thus $J^{-1}(\alpha F+\beta G)=\alpha^*f+\beta^*g$, equivalently $J(\alpha f+\beta g)=\alpha^*Jf+\beta^*Jg$.

Complete a one-particle mode span

Let $\{|n\rangle\}$ be orthonormal and let $c_n=1/n$. Do the finite sums

$$|\psi_N\rangle = \sum_{n=1}^{N}\frac1n|n\rangle$$

converge in the completed one-particle space? Is their limit in the algebraic mode span? What would change for $c_n=1/\sqrt n$?

Solution

For $M<N$,

$$\|\psi_N-\psi_M\|^2 = \sum_{M<n\leq N}\frac1{n^2}.$$

Since $\sum_n n^{-2}$ converges, the partial sums are Cauchy and have a limit in the Hilbert completion. The limit is not a finite mode sum, so it does not belong to the algebraic span.

For $c_n=1/\sqrt n$, the squared norm of the partial sum is $\sum_{n\leq N}1/n$, which diverges. The sequence is not Cauchy and does not define a vector in the Hilbert completion.

References

• Christoph Kehle, Introduction to Functional Analysis, PDF, lecture notes for MIT 18.102, Spring 2025, especially §§1.4, 2.1, 4.3, and 5.1–5.3. These sections develop metric completion, bounded extension, Hilbert completion, projection, Riesz representation, Bessel’s inequality, Parseval’s identity, and separability. It uses the same linear-in-the-second-slot convention as this page.
• Matthew D. Schwartz, Quantum Field Theory and the Standard Model, Cambridge University Press, 2014, especially §2.3.1, supports the covariant normalization and resolution of the identity for scalar one-particle momentum states. The page uses those formulas only to illustrate Hilbert completion and hands their developed physical interpretation to Foundations.
• Gerald Teschl, Mathematical Methods in Quantum Mechanics: With Applications to Schrödinger Operators, PDF, second edition, Graduate Studies in Mathematics 157, American Mathematical Society, 2014, especially §§0.3–0.6 and §§1.1–1.3. These sections establish inner-product geometry, density, Jordan–von Neumann criterion, completion, bounded extension, orthonormal expansions, projection, and Hilbert-space Riesz representation. Its conjugate-linear-first-slot convention agrees with the site convention.