Lp Spaces, Inequalities, and Weak Convergence

On a measure space, an $L^p$ norm turns integrability into quantitative control. Hölder’s inequality bounds products and dual pairings, Minkowski’s inequality controls sums, and completeness keeps norm limits inside the same space. Under a sigma-finiteness hypothesis, $L^q$ functions are exactly the continuous linear tests on $L^p$ for $1\leq p<\infty$, where $p$ and $q$ are conjugate exponents. Weak convergence means convergence against every such test; it is enough for linear observables, but it need not preserve norms, point values, or nonlinear expressions.

Every $L^p$ element is an almost-everywhere equivalence class. This is not a technical afterthought: point evaluation is generally undefined, the $L^\infty$ norm is an essential supremum, and changing a representative on a null set changes none of the conclusions below. The QFT example remains a positive Euclidean integral with a fixed finite regulator; it does not construct a continuum functional measure.

Required background. Lebesgue Integration and Convergence Theorems supplies almost-everywhere equivalence, the integral, Fatou’s lemma, and the convergence theorems used in the proofs.

Spaces and norms · Inequalities · Completeness · Duality · Weak convergence · Regulated Gaussian · Exercises

Integrability modulo null sets

Let $(X,\Sigma,\mu)$ be a measure space, and let scalar-valued mean either real- or complex-valued. For $1\leq p<\infty$, first consider measurable functions satisfying

$$\int_X |f|^p\,\mathrm d\mu<\infty.$$

Declare $f\sim g$ when $f=g$ almost everywhere. The space $L^p(X,\mu)$ is the set of equivalence classes $[f]$, with

$$\|[f]\|_p = \left(\int_X |f|^p\,\mathrm d\mu\right)^{1/p}.$$

For $p=\infty$, define

$$\begin{aligned} \|f\|_\infty &= \operatorname*{ess\,sup}_{x\in X}|f(x)| \\ &= \inf\bigl\{M\geq0: |f(x)|\leq M\text{ for almost every }x\bigr\}, \end{aligned}$$

and let $L^\infty(X,\mu)$ consist of the classes with finite essential supremum. These definitions and the passage from measurable functions to almost-everywhere classes are given in Axler 2020, Definitions 7.1, 7.3, and 7.15–7.18, pp. 194–195 and 202–203, PDF.

By convention, the brackets are suppressed and the class $[f]$ is written as $f$. All formulas must nevertheless be independent of the representative. Without the quotient, a function supported on a nonempty null set would be nonzero pointwise but have norm zero, so the norm would fail to distinguish vectors.

Indicator functions and the meaning of the exponent

If $A\in\Sigma$ and $0<\mu(A)<\infty$, then

$$\|\mathbf 1_A\|_p=\mu(A)^{1/p}, \qquad \|\mathbf 1_A\|_\infty=1.$$

If $\mu(A)=0$, the indicator represents the zero element of every $L^p$. If $\mu(A)=\infty$, it is not in $L^p$ for finite $p$, although it remains in $L^\infty$.

The exponent changes what the norm detects:

Exponent	Quantity emphasized	Basic interpretation
$p=1$	$\int \lvert f\rvert\,\mathrm d\mu$	total absolute mass
$p=2$	$\int \lvert f\rvert^2\,\mathrm d\mu$	quadratic size; induced by an inner product
$1<p<\infty$	large values increasingly strongly as $p$ grows	balance between average and peak control
$p=\infty$	the least almost-everywhere upper bound	worst-case size after null sets are ignored

In the complex $L^2$ case, the site convention is

$$\langle f,g\rangle_{L^2} = \int_X f^*g\,\mathrm d\mu,$$

conjugate-linear in the first slot and linear in the second. Its norm is the $L^2$ norm.

Why exponent one is the norm boundary

The formula $\bigl(\int|f|^p\bigr)^{1/p}$ still makes sense for $0<p<1$, but it is not a norm. On a two-point space with counting measure, let $f=(1,0)$ and $g=(0,1)$. Then

$$\|f+g\|_p=2^{1/p}>2=\|f\|_p+\|g\|_p.$$

Thus the triangle inequality fails. The Banach-space, duality, and weak compactness statements on this page all assume $1\leq p\leq\infty$.

Norm inequalities

For $1\leq p,q\leq\infty$, call $p$ and $q$ conjugate exponents when

$$\frac1p+\frac1q=1,$$

with $1/\infty=0$. Thus $1$ and $\infty$ are conjugate, and $2$ is conjugate to itself.

Hölder from scalar Young

Hölder’s inequality. If $f\in L^p(X,\mu)$ and $g\in L^q(X,\mu)$ for conjugate exponents, then $fg\in L^1$ and

$$\|fg\|_1 \leq \|f\|_p\,\|g\|_q.$$

Here is a complete proof. For $1<p,q<\infty$, scalar Young’s inequality says

$$ab\leq \frac{a^p}{p}+\frac{b^q}{q}, \qquad a,b\geq0.$$

If both norms are nonzero, apply it pointwise to

$$a=\frac{|f|}{\|f\|_p}, \qquad b=\frac{|g|}{\|g\|_q}.$$

Integration gives

$$\int_X \frac{|fg|}{\|f\|_p\|g\|_q}\,\mathrm d\mu \leq \frac1p+\frac1q =1.$$

Multiplication by the two norms proves the result. If either norm vanishes, the corresponding function is zero almost everywhere. At the endpoint $(p,q)=(1,\infty)$,

$$\int_X|fg|\,\mathrm d\mu \leq \|g\|_\infty\int_X|f|\,\mathrm d\mu,$$

and the other endpoint is symmetric. Scalar Young and Hölder appear as Axler 2020, Theorems 7.8–7.9, pp. 196–197, PDF.

For complex functions, write the dual pairing as

$$\langle g,f\rangle_{q,p} = \int_X g^*f\,\mathrm d\mu.$$

It is conjugate-linear in $g$, linear in $f$, and satisfies

$$|\langle g,f\rangle_{q,p}| \leq \|g\|_q\|f\|_p.$$

Cauchy–Schwarz is exactly the case $p=q=2$.

Minkowski and the triangle inequality

Minkowski’s inequality. If $1\leq p\leq\infty$ and $f,g\in L^p(X,\mu)$, then

$$\|f+g\|_p \leq \|f\|_p+\|g\|_p.$$

For $p=1$, this follows by integrating $|f+g|\leq|f|+|g|$. For $p=\infty$, the same pointwise inequality holds outside the union of two null exceptional sets, so it gives the essential-supremum bound.

Now suppose $1<p<\infty$ and set $h=f+g$. The elementary estimate

$$|h|^p \leq 2^{p-1}\bigl(|f|^p+|g|^p\bigr)$$

first shows that $h\in L^p$. If $\|h\|_p=0$ there is nothing to prove. Otherwise Hölder, with conjugate exponent $q=p/(p-1)$, gives

$$\begin{aligned} \|h\|_p^p &\leq \int_X |f|\,|h|^{p-1}\,\mathrm d\mu + \int_X |g|\,|h|^{p-1}\,\mathrm d\mu \\ &\leq \bigl(\|f\|_p+\|g\|_p\bigr) \bigl\||h|^{p-1}\bigr\|_q \\ &= \bigl(\|f\|_p+\|g\|_p\bigr)\|h\|_p^{p-1}. \end{aligned}$$

Division by $\|h\|_p^{p-1}$ proves the claim. This is the full argument behind Axler 2020, Theorem 7.14, p. 199, PDF.

Together, the two inequalities authorize three recurring operations:

Operation	Hypothesis	Guaranteed control
Multiply	$f\in L^p$, $g\in L^q$	$fg\in L^1$ and $\lVert fg\rVert_1\leq\lVert f\rVert_p\lVert g\rVert_q$
Add	$f,g\in L^p$	$f+g\in L^p$ and $\lVert f+g\rVert_p\leq\lVert f\rVert_p+\lVert g\rVert_p$
Test an error	$f_n-f\in L^p$, $g\in L^q$	$\lvert\langle g,f_n-f\rangle\rvert\leq\lVert g\rVert_q\lVert f_n-f\rVert_p$

Finite-measure embeddings and their failure

Suppose $\mu(X)<\infty$ and $1\leq r<s\leq\infty$. Then

$$L^s(X,\mu)\subseteq L^r(X,\mu)$$

and

$$\|f\|_r \leq \mu(X)^{1/r-1/s}\|f\|_s.$$

For $s<\infty$, apply Hölder to $|f|^r\mathbf1_X$ with exponents $s/r$ and $s/(s-r)$:

$$\int_X|f|^r\,\mathrm d\mu \leq \left(\int_X|f|^s\,\mathrm d\mu\right)^{r/s} \mu(X)^{1-r/s}.$$

Taking the $r$th root gives the result. This finite-$s$ case is Axler 2020, Theorem 7.10, p. 197, PDF. The case $s=\infty$ follows directly from $|f|\leq\|f\|_\infty$ almost everywhere. In particular, on a probability space the inclusion has norm at most one.

Finite total measure is decisive. First fix $r<s<\infty$ and choose

$$\alpha=\frac12\left(\frac1r+\frac1s\right).$$

On $(0,\infty)$ with Lebesgue measure, define

$$f_{\mathrm{tail}}(x) = x^{-\alpha}\mathbf1_{(1,\infty)}(x).$$

It belongs to $L^s$ but not $L^r$, because $\alpha s>1$ while $\alpha r<1$. On the same space,

$$f_{\mathrm{zero}}(x) = x^{-\alpha}\mathbf1_{(0,1)}(x)$$

belongs to $L^r$ but not $L^s$. Thus on one infinite-measure space, neither inclusion holds universally.

The case $s=\infty$ has the same conclusion with simpler examples. The function $\mathbf1_{(1,\infty)}$ on $(0,\infty)$ lies in $L^\infty$ but not in $L^r$, whereas

$$g(x)=x^{-1/(2r)}\mathbf1_{(0,1)}(x)$$

lies in $L^r$ but is unbounded and therefore does not lie in $L^\infty$.

Completeness and approximation

Riesz–Fischer theorem

Completeness theorem. For every measure space and every $1\leq p\leq\infty$, the normed space $L^p(X,\mu)$ is complete; hence it is a Banach space.

For $1\leq p<\infty$, let $(f_n)$ be Cauchy. Choose a subsequence $(f_{n_k})$ such that

$$\|f_{n_{k+1}}-f_{n_k}\|_p\leq2^{-k}.$$

Define

$$h_m = \sum_{k=1}^m|f_{n_{k+1}}-f_{n_k}|.$$

Minkowski gives $\|h_m\|_p\leq\sum_{k=1}^m2^{-k}\leq1$. Since $h_m\uparrow h$, monotone convergence gives

$$\int_Xh^p\,\mathrm d\mu = \lim_{m\to\infty}\int_Xh_m^p\,\mathrm d\mu \leq1.$$

Thus $h<\infty$ almost everywhere, so the telescoping series

$$f = f_{n_1} + \sum_{k=1}^{\infty}(f_{n_{k+1}}-f_{n_k})$$

converges absolutely almost everywhere. Its tails satisfy

$$\|f-f_{n_m}\|_p \leq \sum_{k=m}^{\infty}2^{-k} \longrightarrow0.$$

The original Cauchy sequence then converges to the same $f$. For $p=\infty$, choose the same rapidly Cauchy subsequence. Outside one countable union of null sets, its successive differences are bounded by $2^{-k}$ pointwise, so the representatives converge uniformly there to an essentially bounded measurable function. The essential-supremum tails obey the same geometric bound.

This proves the theorem. Compare Axler 2020, Theorems 7.20 and 7.24, pp. 204–205, PDF.

Dense simple approximations

For $1\leq p<\infty$, simple functions with finite-measure support are dense in $L^p$. Indeed, truncate the magnitude of $f$, discard the region where $|f|$ is very small, and quantize the remaining bounded range. This produces simple $s_n$ with

$$s_n\longrightarrow f\quad\text{almost everywhere}, \qquad |s_n|\leq|f|.$$

Each nonzero level set of $s_n$ lies inside a set $\{|f|\geq\varepsilon_n\}$ of finite measure, because

$$\varepsilon_n^p \mu\bigl(\{|f|\geq\varepsilon_n\}\bigr) \leq \|f\|_p^p.$$

Dominated convergence applied to $|s_n-f|^p$ gives $\|s_n-f\|_p\to0$. This is the short argument behind Axler 2020, Exercises 7A.17–7A.19, pp. 200–201, PDF.

Measurable simple functions are also dense in $L^\infty$: partition an essentially bounded range into cells of diameter tending to zero. But on an infinite-measure space, simple functions with finite-measure support need not be dense in $L^\infty$. The constant function $1$ stays at $L^\infty$-distance at least $1$ from every such function.

Duality identifies the tests

Let $1\leq p<\infty$, let $q$ be conjugate to $p$, and now assume that $(X,\Sigma,\mu)$ is sigma-finite.

The representation theorem

Duality theorem. Every continuous linear functional $\Lambda:L^p\to\mathbb C$ has a unique $g\in L^q$ such that

$$\Lambda(f) = \int_Xg^*f\,\mathrm d\mu, \qquad \|\Lambda\|=\|g\|_q.$$

The same statement holds over $\mathbb R$ with the conjugation omitted. Hölder proves that every $g\in L^q$ gives a functional of norm at most $\|g\|_q$. For $1<q<\infty$ and $g\neq0$, the function

$$f_g = \frac{|g|^{q-2}g}{\|g\|_q^{q-1}}$$

has $\|f_g\|_p=1$ and $\Lambda_g(f_g)=\|g\|_q$. At the endpoint $p=1$, $q=\infty$, the case $g=0$ is immediate. Otherwise fix $0<\varepsilon<\|g\|_\infty$ and let

$$E_\varepsilon = \{x:|g(x)|>\|g\|_\infty-\varepsilon\}.$$

Sigma-finiteness supplies a measurable $B\subseteq E_\varepsilon$ with $0<\mu(B)<\infty$. The function

$$f_\varepsilon = \frac{g}{|g|} \frac{\mathbf1_B}{\mu(B)}$$

has $L^1$ norm one and $|\Lambda_g(f_\varepsilon)|>\|g\|_\infty-\varepsilon$. Letting $\varepsilon\downarrow0$ proves $\|\Lambda_g\|=\|g\|_q$ in every case.

The dual tests also recover the norm of each $f$. For $1<p<\infty$ and $f\neq0$, define

$$g_f = \frac{|f|^{p-2}f}{\|f\|_p^{p-1}},$$

with value zero where $f=0$. Then $\|g_f\|_q=1$ and

$$\int_Xg_f^*f\,\mathrm d\mu = \|f\|_p.$$

For $p=1$, take $g_f=f/|f|$ where $f\neq0$ and zero elsewhere. Consequently,

$$\|f\|_p = \sup_{\|g\|_q\leq1} \left| \int_Xg^*f\,\mathrm d\mu \right|.$$

Surjectivity is the deeper part. On finite-measure pieces, a functional defines a countably additive measure by $\nu(E)=\Lambda(\mathbf1_E)$. Absolute continuity and the Radon–Nikodym theorem produce a density $g$; the functional bound forces $g\in L^q$. Sigma-finite exhaustion patches the densities uniquely. This is a proof sketch; the full representation argument for $1<p<\infty$ is Axler 2020, Theorem 9.42, pp. 275–277, PDF. The sigma-finite $p=1$ extension is stated there as Axler 2020, Exercise 9B.15, p. 279, PDF, rather than supplied with a full proof.

Endpoint warning

Under the stated hypothesis, $(L^1)^*$ is $L^\infty$. The reverse-looking claim is generally false: $(L^\infty)^*$ can contain continuous functionals that do not arise from integration against an $L^1$ function. Thus testing an $L^\infty$ sequence only against $L^1$ describes a weak-star topology, not the full weak topology. A concrete nonrepresentable functional is constructed in Brezis 2011, §4.3, p. 102, publisher record.

For $1<p<\infty$, applying duality twice shows that $L^p$ is reflexive. The endpoint spaces $L^1$ and $L^\infty$ are not reflexive on the standard infinite-dimensional examples, although finite-dimensional exceptions exist. See Axler 2020, Exercises 9B.16–9B.18, p. 279, PDF and Brezis 2011, §3.5, p. 67, publisher record.

Strong, weak, and weak-star convergence

Continue to assume sigma-finiteness when identifying duals with $L^q$ spaces.

Three modes of convergence

For $1\leq p<\infty$, a sequence converges strongly in $L^p$ when

$$f_n\longrightarrow f \quad\text{strongly in }L^p \quad\Longleftrightarrow\quad \|f_n-f\|_p\longrightarrow0.$$

It converges weakly when every continuous linear test converges. By the duality theorem, this is equivalent to

$$f_n\rightharpoonup f \quad\Longleftrightarrow\quad \int_Xg^*f_n\,\mathrm d\mu \longrightarrow \int_Xg^*f\,\mathrm d\mu \quad \text{for every }g\in L^q.$$

For a sequence in $L^\infty$, the condition

$$\int_Xg^*f_n\,\mathrm d\mu \longrightarrow \int_Xg^*f\,\mathrm d\mu \quad \text{for every }g\in L^1$$

is weak-star convergence, written $f_n\overset{*}{\rightharpoonup}f$. Full weak convergence in $L^\infty$ tests against every element of $(L^\infty)^*$ and is generally stronger.

Mode	What tends to zero	What it directly controls
Strong $L^p$	$\lVert f_n-f\rVert_p$	all $L^q$ pairings, uniformly over the dual unit ball
Weak $L^p$	each pairing $\langle g,f_n-f\rangle$	every fixed continuous linear observable
Weak-star $L^\infty$	each $L^1$ pairing	the chosen predual tests, not every element of $(L^\infty)^*$

The general weak topology and its sequential criterion are developed in Brezis 2011, §3.2 and Proposition 3.5, pp. 57–58, publisher record. Brezis states this chapter over real Banach spaces; the complex version used here replaces real-linear tests by complex-linear tests and uses absolute values in the same norm estimates (Brezis 2011, brief user’s guide, p. viii, publisher record).

What weak convergence preserves

Strong convergence implies weak convergence. Indeed, Hölder gives, for every fixed $g\in L^q$,

$$\left| \int_Xg^*(f_n-f)\,\mathrm d\mu \right| \leq \|g\|_q\|f_n-f\|_p \longrightarrow0.$$

Two further facts require care:

1. A weakly convergent sequence is norm bounded.

2. The norm is weakly lower semicontinuous:

   $$\|f\|_p \leq \liminf_{n\to\infty}\|f_n\|_p.$$

The first is an application of the uniform boundedness principle. For the second, fix $g$ with $\|g\|_q\leq1$. Weak convergence and the dual norm formula give

$$|\langle g,f\rangle| = \lim_{n\to\infty}|\langle g,f_n\rangle| \leq \liminf_{n\to\infty}\|f_n\|_p.$$

Taking the supremum over the dual unit ball proves the claim. Both statements are also part of Brezis 2011, Proposition 3.5, p. 58, publisher record.

A useful extension criterion follows from the same estimate. Suppose $\sup_n\|f_n\|_p<\infty$ and pairings with $f_n$ converge to those with $f$ for every $g$ in a norm-dense subset $D\subset L^q$. For arbitrary $g\in L^q$, choose $d\in D$ close to $g$ and write

$$\begin{aligned} |\langle g,f_n-f\rangle| &\leq |\langle d,f_n-f\rangle| \\ &\quad+ \|g-d\|_q\bigl(\|f_n\|_p+\|f\|_p\bigr). \end{aligned}$$

First send $n\to\infty$, then $d\to g$. The uniform norm bound is what prevents the approximation error from growing with $n$.

Bounded sequences and reflexivity

For $1<p<\infty$, $L^p$ is reflexive. Consequently every norm-bounded sequence in $L^p$ has a weakly convergent subsequence. This conclusion combines weak compactness of the closed unit ball with the nontrivial fact that weak compactness has the required sequential consequence. The abstract machinery is cited rather than proved here; see Brezis 2011, Theorems 3.17–3.19 and Remark 17, pp. 67–70, publisher record.

At $p=1$ and $p=\infty$, boundedness alone gives no general weakly convergent subsequence. Banach–Alaoglu gives weak-star compactness for a dual ball, which is a different statement from weak compactness; compactness and sequential compactness also require separate attention outside metrizable settings. See Brezis 2011, Theorem 3.16, pp. 66–67, publisher record.

Sequences that separate the notions

Weak but not strong. In $\ell^2$, let $e_n$ be the $n$th standard basis vector. For every $g=(g_k)\in\ell^2$,

$$\langle g,e_n\rangle=g_n^*\longrightarrow0,$$

because the coordinates of an $\ell^2$ sequence tend to zero. Hence $e_n\rightharpoonup0$, but $\|e_n\|_2=1$. In particular, weak convergence does not imply convergence of norms or of the nonlinear quantity $\|f_n\|_2^2$.

Weak convergence need not give a pointwise limit either. The functions

$$\psi_n(x)=e^{2\pi i n x}, \qquad x\in(0,1),$$

form an orthonormal sequence in complex $L^2(0,1)$, so Bessel’s inequality implies $\psi_n\rightharpoonup0$. For fixed $x$, however, convergence of $e^{2\pi i n x}$ would force its shifted sequence to have both limits $L$ and $e^{2\pi i x}L$. Since $|L|=1$, this would require $e^{2\pi i x}=1$, which is impossible for $x\in(0,1)$. Thus the sequence has no pointwise limit anywhere on that interval.

An endpoint obstruction. In $\ell^1$, the same coordinatewise limit does not imply weak convergence. Pairing with $g=(1,1,\ldots)\in\ell^\infty$ gives

$$\langle g,e_n\rangle=1.$$

In fact no subsequence $(e_{n_k})$ is weakly convergent: define a bounded test by $g_{n_k}=(-1)^k$ and set its other coordinates to zero. The resulting pairing alternates.

Concentration for interior exponents. On $(0,1)$, let

$$f_n(x)=n^{1/p}\mathbf1_{(0,1/n)}(x), \qquad 1<p<\infty.$$

Then $\|f_n\|_p=1$ and $f_n(x)\to0$ for every $x>0$. If $q$ is conjugate to $p$ and $g\in L^q(0,1)$, then

$$\left|\int_0^1g^*f_n\,\mathrm dx\right| \leq \|g\mathbf1_{(0,1/n)}\|_q \longrightarrow0.$$

The last limit is the absolute continuity of the integral of $|g|^q$. Therefore $f_n\rightharpoonup0$ but not strongly. At $p=1$, the constant test $g=1$ gives $\int f_n=1$, exposing the endpoint failure.

A finite regulated Gaussian bound

Fix $N<\infty$ and let $K$ be a real symmetric positive-definite $N\times N$ matrix. The normalized Euclidean Gaussian measure is

$$\mathrm d\gamma_K(\phi) = \frac{\sqrt{\det K}}{(2\pi)^{N/2}} \exp\!\left(-\frac12\phi^{\mathsf T}K\phi\right) \mathrm d^N\phi.$$

Writing $C=K^{-1}$, the finite-dimensional source integral gives

$$\mathbb E_K[\phi_i\phi_j]=C_{ij}.$$

The normalization, source formula, and covariance identity follow from the finite Gaussian calculation in Zinn-Justin 2021, Chapter 1, §1.1, pp. 1–3, OUP. That source denotes the positive quadratic matrix by $S$ and its inverse by $\Delta$; here they are renamed $K$ and $C$, and the weight is divided by its finite normalization to make $\gamma_K$ a probability measure.

Covariance as an inner-product kernel

For $u,v\in\mathbb R^N$, define the smeared field

$$\Phi_u(\phi)=u^{\mathsf T}\phi.$$

It belongs to $L^2(\gamma_K)$ and

$$\|\Phi_u\|_{L^2(\gamma_K)}^2 = \mathbb E_K[\Phi_u^2] = u^{\mathsf T}Cu.$$

Cauchy–Schwarz applied to $\Phi_u$ and $\Phi_v$ yields the covariance-kernel bound

$$|u^{\mathsf T}Cv| \leq \bigl(u^{\mathsf T}Cu\bigr)^{1/2} \bigl(v^{\mathsf T}Cv\bigr)^{1/2}.$$

This is both a positivity check on the covariance and a quantitative bound on every smeared two-point function.

Convergence of regulated approximations

Let $A_m,A\in L^2(\gamma_K)$. Strong convergence gives the explicit error bound

$$\begin{aligned} &\left| \mathbb E_K[\Phi_u A_m] - \mathbb E_K[\Phi_u A] \right| \\ &\qquad\leq \bigl(u^{\mathsf T}Cu\bigr)^{1/2} \|A_m-A\|_{L^2(\gamma_K)}. \end{aligned}$$

Thus $A_m\to A$ in $L^2$ controls a correlator and supplies a rate whenever the norm error has one. If $A_m\rightharpoonup A$ only weakly, the correlator still converges because $\Phi_u$ is a fixed $L^2$ test, but the definition provides no rate and does not imply $A_m^2\to A^2$.

There is also a direct finite-kernel estimate. Give the index set $\{1,\ldots,N\}$ counting measure, let $p,q$ be conjugate, and suppose $\|C_m-C\|_{\ell^q(\{1,\ldots,N\}^2)}\to0$. Hölder on the finite double sum gives

$$\begin{aligned} |u^{\mathsf T}(C_m-C)v| &\leq \|C_m-C\|_{\ell^q(\{1,\ldots,N\}^2)} \|u\otimes v\|_{\ell^p} \\ &= \|C_m-C\|_{\ell^q} \|u\|_{\ell^p}\|v\|_{\ell^p}. \end{aligned}$$

Hence norm convergence of the regulated kernel controls every fixed smeared matrix element.

The boundaries are essential:

• $N$ is fixed. If $N$ changes, the spaces change and every constant must be controlled uniformly after specifying comparison maps.
• The measure is positive and Euclidean. A Lorentzian weight $e^{iS}$ is not a probability measure to which this argument automatically applies.
• Continuum propagators may be distributions rather than $L^q$ functions. Their pairings require test-function and distribution theory.
• No regulator removal, renormalization, reflection positivity, or continuum symbol $\mathcal D\phi$ follows from a fixed-$N$ estimate.

For the developed physical construction, continue to Regulated Bosonic Field Integrals in Foundations of Quantum Field Theory.

Common pitfalls

Treating an $L^p$ element as a pointwise function. It is an almost-everywhere equivalence class. Point evaluation and values on a named null set require an additional choice of representative.

Using a supremum instead of an essential supremum. A single exceptional point can change the pointwise supremum without changing the $L^\infty$ element or norm.

Calling the expression a norm below exponent one. For $0<p<1$, the triangle inequality fails. Results that use Banach-space completeness or duality cannot be imported unchanged.

Assuming an inclusion without measuring the whole space. The implication $L^s\subseteq L^r$ for $s>r$ needs finite total measure. On an infinite-measure space, either inclusion can fail.

Using nonconjugate exponents in Hölder. The basic two-factor estimate requires $1/p+1/q=1$. Other exponent relations need a different theorem or additional finite-measure input.

Confusing weak with strong convergence. Weak convergence fixes linear pairings, not norms, pointwise behavior, products, or nonlinear observables.

Calling $L^1$ testing of $L^\infty$ weak convergence. Those tests define weak-star convergence. The full dual of $L^\infty$ is generally larger.

Expecting endpoint compactness from boundedness. Bounded sequences in $L^1$ or $L^\infty$ need extra hypotheses for the corresponding weak subsequence statements.

Exporting a regulator-level estimate. Fixed-dimensional norm bounds do not by themselves construct or control a continuum field measure.

Where these controls lead

• Fourier Series, Fourier Transforms, and Plancherel Theory uses $L^1$ and $L^2$ norms, dense approximation, and norm convergence to control transforms.
• Banach and Hilbert Spaces, Completion, and Riesz Representation develops the abstract uniform-boundedness, reflexivity, weak-compactness, and Hilbert-space machinery cited here.
• Weak Solutions, Sobolev Spaces, and Well-Posedness uses weak convergence and lower semicontinuity in variational PDE arguments.
• Test-Function Spaces, Distributions, Support, and Convergence replaces $L^q$ kernels by distributional pairings when singular kernels require them.
• Convolution, Approximate Identities, and Poisson Summation develops Young’s convolution inequality and smoothing; it is distinct from the scalar Young inequality used above.
• Regulated Bosonic Field Integrals supplies the exact finite-regulator physical setting for the Gaussian estimate.

Exercises

Indicator norms and null sets. Let $A\in\Sigma$. Compute $\|\mathbf1_A\|_p$ for finite $p$ and for $p=\infty$. What changes if $A$ is replaced by a set that differs from it by a null set?

Solution

For finite $p$,

$$\|\mathbf1_A\|_p^p = \int_X\mathbf1_A\,\mathrm d\mu = \mu(A).$$

Thus the norm is $\mu(A)^{1/p}$ when the measure is finite. If $\mu(A)=\infty$, the integral expression has extended value $\infty$, so $\mathbf1_A\notin L^p$ for finite $p$. If $\mu(A)>0$, the essential supremum is $1$; if $\mu(A)=0$, the indicator is the zero $L^\infty$ element and its norm is zero. Replacing $A$ by a set equal to it modulo a null set changes no $L^p$ element or norm.

Finite-measure inclusion. Suppose $\mu(X)=M<\infty$ and $1\leq r<s<\infty$. Prove

$$\|f\|_r \leq M^{1/r-1/s}\|f\|_s.$$

What is the bound on a probability space?

Solution

Use Hölder on $|f|^r\mathbf1_X$ with conjugate exponents $a=s/r$ and $a'=s/(s-r)$:

$$\begin{aligned} \|f\|_r^r &\leq \left(\int_X|f|^{ra}\,\mathrm d\mu\right)^{1/a} \left(\int_X1^{a'}\,\mathrm d\mu\right)^{1/a'} \\ &= \|f\|_s^r M^{1-r/s}. \end{aligned}$$

Taking the $r$th root gives the result. If $M=1$, then $\|f\|_r\leq\|f\|_s$.

A norming dual element. Let $1<p<\infty$ and $f\in L^p$ be nonzero. Verify that

$$g_f=\frac{|f|^{p-2}f}{\|f\|_p^{p-1}}$$

has $L^q$ norm one and satisfies $\langle g_f,f\rangle_{q,p}=\|f\|_p$.

Solution

Because $q=p/(p-1)$,

$$(p-1)q=p.$$

Therefore

$$\|g_f\|_q^q = \frac{\int_X|f|^{(p-1)q}\,\mathrm d\mu} {\|f\|_p^{(p-1)q}} = \frac{\|f\|_p^p}{\|f\|_p^p} =1.$$

With the conjugation in the first slot,

$$\langle g_f,f\rangle_{q,p} = \frac{\int_X|f|^p\,\mathrm d\mu}{\|f\|_p^{p-1}} = \|f\|_p.$$

Interior and endpoint basis sequences. Show that $e_n\rightharpoonup0$ in $\ell^2$. Then show that no subsequence of the same standard basis is weakly convergent in $\ell^1$.

Solution

For $g\in\ell^2$, the pairing with $e_n$ is $g_n^*$, which tends to zero because every square-summable sequence has coordinates tending to zero. Hence $e_n\rightharpoonup0$ in $\ell^2$, although its norm stays one.

Given any subsequence $(e_{n_k})$ in $\ell^1$, define $g\in\ell^\infty$ by $g_{n_k}=(-1)^k$ and set all other entries to zero. Then $\langle g,e_{n_k}\rangle=(-1)^k$, so the subsequence is not even weakly Cauchy. Therefore it cannot converge weakly.

A regulated correlator error. For the finite Gaussian measure above, let $A_m\to A$ in $L^2(\gamma_K)$. Prove a bound for

$$\mathbb E_K[\Phi_uA_m]-\mathbb E_K[\Phi_uA]$$

and state what changes if $A_m\rightharpoonup A$ only weakly.

Solution

Cauchy–Schwarz and the covariance identity give

$$\begin{aligned} \left| \mathbb E_K[\Phi_u(A_m-A)] \right| &\leq \|\Phi_u\|_2\|A_m-A\|_2 \\ &= \bigl(u^{\mathsf T}Cu\bigr)^{1/2} \|A_m-A\|_2. \end{aligned}$$

The right side tends to zero and is a quantitative error bound. Weak convergence still makes the left side tend to zero because $\Phi_u$ is one fixed $L^2$ test, but it supplies no norm rate and says nothing by itself about nonlinear expressions such as $A_m^2$.

References

• Sheldon Axler, Measure, Integration & Real Analysis, Graduate Texts in Mathematics 282, Springer (2020), Axler 2020, Chapter 7, pp. 193–210, and §9B, pp. 275–279, author-hosted open-access PDF. $L^p$ definitions, inequalities, completeness, density, dual norm, dual representation, and endpoint qualifications.
• Haïm Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, Universitext, Springer (2011), Brezis 2011, brief user’s guide, p. viii; Chapter 3, §§3.2–3.5, pp. 57–71; and §4.3, p. 102, publisher record. Scalar convention, weak and weak-star topologies, lower semicontinuity, reflexivity, compactness, sequential compactness, and the dual of $L^\infty$.
• Jean Zinn-Justin, Quantum Field Theory and Critical Phenomena, 5th ed., Oxford University Press (2021), Zinn-Justin 2021, Chapter 1, §1.1, pp. 1–3, OUP. Finite-dimensional positive Gaussian integrals, source dependence, and covariance.