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✓ 21 results · all verified · 1 also independently AI-judged

Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs. The 20 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

The $L^{p}$ Spaces Holder Minkowski and Riesz Fischer

1 · Prerequisites

2 · Summary

This first $L^{p}$ page fixes the quotient-first convention used everywhere later: elements of $L^{p} (μ)$ are almost-everywhere classes, not pointwise functions. It proves the three structural landmarks the later pages actually consume: Holder, Minkowski, and Riesz-Fischer; records the finite-counting-space agreement seams against the published finite-dimensional inequalities; and keeps the comparison block together with the deliberately small $0 < p < 1$ appendix.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Conjugate exponents, including the endpoint conventions

Definition

Numbers $p, q \in [1, \infty]$ are conjugate exponents when

$\frac{1}{p} + \frac{1}{q} = 1,$

with the convention $1 / \infty : = 0$ . Thus the endpoint pairs $(1, \infty)$ and $(\infty, 1)$ are conjugate, and if $1 < p < \infty$ then the conjugate of $p$ is

$q = \frac{p}{p - 1} .$

No other endpoint pair is conjugate: $(1, 1)$ gives $2$ , while $(\infty, \infty)$ gives $0$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

The essential supremum of a measurable function with respect to a measure

Definition

Let $(X, A, μ)$ be a measure space, and let $f : X \to R$ be measurable. The essential supremum of $∣ f ∣$ with respect to $μ$ is

$∥ f ∥_{\infty, μ} : = \inf {M \in [0, \infty] : ∣ f ∣ \leq M μ -almost everywhere} .$

When the measure is fixed from context, this is written $∥ f ∥_{\infty}$ . The phrase essentially bounded means $∥ f ∥_{\infty} < \infty$ .

The later proposition The essential supremum is attained as the least essential bound proves that when $∥ f ∥_{\infty} < \infty$ , the inequality $∣ f ∣ \leq ∥ f ∥_{\infty}$ itself holds almost everywhere and that this bound is the least essential bound.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-31Open item page →

The function space $L^{p} (μ)$ for $0 < p < \infty$

Definition

Let $(X, A, μ)$ be a measure space and let $p$ be a real number with $0 < p < \infty$ . Write $M (X, μ)$ for the measurable functions $f : X \to R$ . For $f \in M (X, μ)$ define the extended-valued $p$ -functional

$∥ f ∥_{p} : = {\begin{cases} {(\int ∣ f ∣^{p} d μ)}^{1 / p}, & \int ∣ f ∣^{p} d μ < \infty, \\ + \infty, & \int ∣ f ∣^{p} d μ = + \infty . \end{cases}$

The finite power uses Real powers for positive bases, with the zero-base positive-exponent convention, while the second clause avoids applying real powers to the extended value $+ \infty$ . The integral is the nonnegative Lebesgue integral of The nonnegative Lebesgue integral.

The class

$L^{p} (μ) : = {f \in M (X, μ) : ∥ f ∥_{p} < \infty}$

is the measurable-function space used on this page before passing to almost-everywhere equivalence classes.

For $1 \leq p < \infty$ , the pointwise operations make $L^{p} (μ)$ a real vector space by $L^{p}$ and $L^{\infty}$ are vector spaces for $p \geq 1$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

The space $L^{\infty} (μ)$ of essentially bounded measurable functions

Definition

Let $(X, A, μ)$ be a measure space. The class

$L^{\infty} (μ) : = {f : X \to R : f measurable and ∥ f ∥_{\infty} < \infty}$

of essentially bounded measurable functions is the page's $p = \infty$ space. Its size is measured by the essential-supremum functional of The essential supremum of a measurable function with respect to a measure.

The later theorem $L^{p}$ and $L^{\infty}$ are vector spaces for $p \geq 1$ proves that $L^{\infty} (μ)$ is a real vector space under pointwise addition and scalar multiplication.

DefinitionDefinition: Literature-sourcedProof: Not applicableaudited 2026-08-31Open item page →

The null subspace of measurable functions that vanish almost everywhere

Definition

Let $(X, A, μ)$ be a measure space. Define

$N (μ) : = {f : X \to R : f measurable and f = 0 μ -almost everywhere} .$

This is the family of null representatives that will be quotiented out of each $L^{p} (μ)$ and of $L^{\infty} (μ)$ . The linear-subspace claim implicit in that phrasing is proved in Null functions form a linear subspace and are exactly the zero-seminorm class.

DefinitionDefinition: Literature-sourcedProof: Not applicableverified 2026-09-24 (gpt-6-sol)Open item page →

The space $L^{p} (μ)$ as the quotient by null functions

Definition

Let $(X, A, μ)$ be a measure space.

For $0 < p < \infty$ , define an equivalence relation on $L^{p} (μ)$ by $f \sim g ⟺ f - g \in N (μ),$ and write $L^{p} (μ)$ for the set of classes $[f]$ .
For $p = \infty$ , define the same relation on $L^{\infty} (μ)$ and again write $L^{\infty} (μ)$ for the set of classes.

Thus an element of $L^{p} (μ)$ is an almost-everywhere equivalence class of measurable representatives.

When $1 \leq p \leq \infty$ , the displayed set quotient agrees with the usual quotient-vector-space construction of The quotient vector space $V / W$ and its canonical projection.

For $0 < p < 1$ , the same class notation is used; under its stated choice hypothesis, the later item The $L^{p}$ distance for $0 < p < 1$ is a complete translation-invariant metric supplies the metric structure rather than a normed-space structure.

RemarkRemark: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Elements of $L^{p}$ are equivalence classes, so pointwise statements require a representative

Elements of The space $L^{p} (μ)$ as the quotient by null functions are classes $[f]$ , not chosen functions. So formulas such as " $f (x) = 0$ " or " $f (x) \to g (x)$ " are not meaningful until a representative is named. Quantities invariant under changing $f$ on a null set, such as $∥ [f] ∥_{p}$ , are meaningful on the quotient; pointwise assertions are not.

RemarkRemark: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

$ℓ^{p}$ is the $L^{p}$ space of counting measure

Remark

On $(N, P (N), #)$ with counting measure, every function $f : N \to R$ is measurable. Writing $a_{k} : = f (k)$ , one has

$\int ∣ f ∣^{p} d # = \sum_{k = 0}^{\infty} ∣ a_{k} ∣^{p} (0 < p < \infty),$

by the counting-measure integral dictionary, so $L^{p} (#)$ is exactly the usual sequence class $ℓ^{p}$ . Also

$∥ f ∥_{\infty} = \sup_{k \in N} ∣ a_{k} ∣,$

because a subset of $N$ has counting measure zero only when it is empty. Hence the quotient by almost-everywhere equality does nothing: for counting measure on $N$ , equality almost everywhere means equality everywhere.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-23 (gpt-6-sol)Open item page →

Holder's inequality for integrals, including the endpoint cases

Statement

Let $(X, A, μ)$ be a measure space, let $p, q \in [1, \infty]$ be conjugate exponents, and let $f, g$ be measurable real-valued functions.

If $1 \leq p < \infty$ and $q < \infty$ with $f \in L^{p} (μ)$ and $g \in L^{q} (μ)$ , then $\int ∣ f g ∣ d μ \leq ∥ f ∥_{p} ∥ g ∥_{q} .$
If $p = 1$ and $q = \infty$ with $f \in L^{1} (μ)$ and $g \in L^{\infty} (μ)$ , then $\int ∣ f g ∣ d μ \leq ∥ f ∥_{1} ∥ g ∥_{\infty} .$
If $p = \infty$ and $q = 1$ with $f \in L^{\infty} (μ)$ and $g \in L^{1} (μ)$ , then $\int ∣ f g ∣ d μ \leq ∥ f ∥_{\infty} ∥ g ∥_{1} .$

In every case the right-hand side is finite, so $f g$ is integrable.

Facts & Assumptions

Given: A measure space $(X, A, μ)$ , conjugate exponents $p, q \in [1, \infty]$ , and measurable real-valued functions $f, g$ in the spaces named in the relevant clause of the Statement.

[L1]

Conjugate exponents are defined in Conjugate exponents, including the endpoint conventions.

[L2]

For $0 < r < \infty$ , membership in $L^{r} (μ)$ means $\int ∣ h ∣^{r} d μ < \infty$ , while $L^{\infty} (μ)$ means finite essential supremum (The function space $L^{p} (μ)$ for $0 < p < \infty$ , The space $L^{\infty} (μ)$ of essentially bounded measurable functions).

[L3]

A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

[L4]

If $h \in L^{\infty} (μ)$ , then $∣ h ∣ \leq ∥ h ∥_{\infty}$ almost everywhere (The essential supremum is attained as the least essential bound).

[L5]

Young's inequality says $u v \leq u^{p} / p + v^{q} / q$ for $u, v \geq 0$ when $1 < p, q < \infty$ are conjugate (Young's inequality for conjugate real exponents).

[L6]

The nonnegative integral is monotone and homogeneous (Monotonicity and nonnegative homogeneity of the nonnegative integral).

[L7]

The nonnegative integral is additive (Additivity of the nonnegative Lebesgue integral).

[L8]

The integral of a nonnegative function over a null set is zero (A nonnegative integral over a null set vanishes).

Proof

technique · For $1 < p < infinity$, normalize the nonzero norms and apply the published Young inequality pointwise before integrating. Treat the endpoint pairs $(1,\infty)$ and $(\infty,1)$ separately from the essential-bound definition

1.1L2L3L8given

Assume first $1 < p, q < \infty$ , and put $A : = ∥ f ∥_{p}$ and $B : = ∥ g ∥_{q}$ . If $A = 0$ or $B = 0$ , then the corresponding power integral is $0$ , so [L3] makes the corresponding function vanish outside a measurable null set. By [L8], $\int ∣ f g ∣ d μ = 0$ . Thus only the case $A, B > 0$ remains.

1.2L2L4L6L7L8given

For the endpoint pair $(p, q) = (1, \infty)$ , let $M : = ∥ g ∥_{\infty}$ . By [L4] there is a measurable null set $N$ outside which $∣ f g ∣ \leq M ∣ f ∣$ . On $N$ , [L8] gives zero integral for $∣ f g ∣ χ_{N}$ and $M ∣ f ∣ χ_{N}$ . Integrating the inequality on $X ∖ N$ with [L6] and [L7] therefore gives $\int ∣ f g ∣ d μ \leq M \int ∣ f ∣ d μ = ∥ g ∥_{\infty} ∥ f ∥_{1}$ .

2.1step 1.1L1L2L5L6L7algebra

In the remaining strict-exponent case, Young's inequality applied pointwise to $u = ∣ f ∣ / A$ and $v = ∣ g ∣ / B$ gives [step 1.1, L1, L2, L5, L6, L7, algebra] $\frac{∣ f ∣ ∣ g ∣}{A B} \leq \frac{∣ f ∣^{p}}{p A^{p}} + \frac{∣ g ∣^{q}}{q B^{q}} .$ Integrating and using additivity, monotonicity, homogeneity, and the definitions of $A$ and $B$ yields $\int ∣ f g ∣ d μ \leq \frac{B}{p A^{p - 1}} \int ∣ f ∣^{p} d μ + \frac{A}{q B^{q - 1}} \int ∣ g ∣^{q} d μ = \frac{A B}{p} + \frac{A B}{q} = A B .$

2.2step 1.2given

The case $(p, q) = (\infty, 1)$ is identical after exchanging $f$ and $g$ . [step 1.2, given] $\int ∣ f g ∣ d μ \leq ∥ f ∥_{\infty} ∥ g ∥_{1} .$

3.1step 2.1step 1.2step 2.2L2∎

Step 2.1 proves the strict-exponent case, and steps 1.2 and 2.2 prove the two endpoint cases. In every case the right-hand side is finite by [L2], so $f g$ is integrable.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-23 (gpt-6-sol)Open item page →

Equality in Holder's inequality for $1 < p < \infty$

Statement

Let $1 < p < \infty$ , let $q$ be its conjugate exponent, and let $f \in L^{p} (μ)$ and $g \in L^{q} (μ)$ . Then equality holds in Holder's inequality

$\int ∣ f g ∣ d μ = ∥ f ∥_{p} ∥ g ∥_{q}$

if and only if at least one of $f, g$ is zero almost everywhere, or there is a constant $c > 0$ such that

$∣ f ∣^{p} = c ∣ g ∣^{q} μ -almost everywhere .$

Facts & Assumptions

Given: A measure space, an exponent $1 < p < \infty$ , its conjugate exponent $q$ , and functions $f \in L^{p} (μ)$ , $g \in L^{q} (μ)$ .

[L1]

Holder's inequality for integrals has already been proved (Holder's inequality for integrals, including the endpoint cases).

[L2]

Young's inequality is the scalar step used in that proof (Young's inequality for conjugate real exponents).

[L3]

A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

[L4]

Membership in $L^{p} (μ)$ and $L^{q} (μ)$ means finiteness of the corresponding power integrals (The function space $L^{p} (μ)$ for $0 < p < \infty$ ).

[L5]

For a weight strictly between zero and one, equality in the two-point exponential convexity inequality holds exactly when its two inputs agree (The two-point convexity inequality for the exponential function).

Proof

technique · Trace where equality can occur in the normalized Young-inequality proof. Equality in Young forces the normalized powers $|f|^p$ and $|g|^q$ to be proportional almost everywhere, and conversely that proportionality makes the inequality an equality

1.1L1L3L4

If $∥ f ∥_{p} = 0$ or $∥ g ∥_{q} = 0$ , then the corresponding function is zero almost everywhere, and Holder's inequality becomes equality with both sides $0$ .

1.2L1L2L3

Assume now that $A : = ∥ f ∥_{p} > 0$ and $B : = ∥ g ∥_{q} > 0$ . The proof of [L1] integrated the nonnegative function [L1, L2, L3] $H : = \frac{∣ f ∣^{p}}{p A^{p}} + \frac{∣ g ∣^{q}}{q B^{q}} - \frac{∣ f g ∣}{A B} .$ If equality holds in Holder, then $\int H d μ = 0$ , so $H = 0$ almost everywhere. Thus equality holds in Young's inequality pointwise almost everywhere for $u = ∣ f ∣ / A$ and $v = ∣ g ∣ / B$ .

2.1L2L5step 1.2algebra

To check the equality condition omitted from [L2]'s Statement, first suppose $u, v > 0$ . Apply [L5] to $\log (u^{p})$ and $\log (v^{q})$ with weights $1 / p$ and $1 / q$ : its geometric side is $u v$ , its arithmetic side is $u^{p} / p + v^{q} / q$ , and equality holds exactly when $u^{p} = v^{q}$ . If one of $u, v$ is zero, equality in Young holds only when both vanish, again exactly when $u^{p} = v^{q}$ . Applying this criterion to step 1.2 gives [step 1.2, L2, L5, algebra] $\frac{∣ f ∣^{p}}{A^{p}} = \frac{∣ g ∣^{q}}{B^{q}} μ -almost everywhere,$ so $∣ f ∣^{p} = (A^{p} / B^{q}) ∣ g ∣^{q}$ almost everywhere.

3.1step 1.1step 2.1L1L4algebra

Conversely, if $∣ f ∣^{p} = c ∣ g ∣^{q}$ almost everywhere for some $c > 0$ and neither function vanishes almost everywhere, integration gives $A^{p} = c B^{q}$ . Thus the normalized powers agree almost everywhere, so the scalar criterion from step 2.1 makes Young an equality almost everywhere and the integrated Holder proof becomes an equality. If either function vanishes almost everywhere, step 1.1 already applies.

4.1step 1.1step 2.1step 3.1∎

Step 1.1 handles the zero-function case, step 2.1 proves the strict necessity, and step 3.1 proves sufficiency. These are exactly the alternatives in the Statement.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

Generalized Holder inequality puts products into $L^{r}$

Statement

Let $1 \leq p, q, r \leq \infty$ satisfy

$\frac{1}{r} = \frac{1}{p} + \frac{1}{q},$

with the convention $1 / \infty = 0$ . If $f$ and $g$ lie in the corresponding measurable-function spaces ( $L^{p} (μ)$ or $L^{\infty} (μ)$ according to whether the exponent is finite or infinite), then $f g$ lies in the corresponding space for $r$ and

$∥ f g ∥_{r} \leq ∥ f ∥_{p} ∥ g ∥_{q} .$

Facts & Assumptions

Given: Exponents $p, q, r$ with $1 / r = 1 / p + 1 / q$ and measurable functions $f, g$ in the spaces named in the Statement.

[L1]

Holder's inequality for integrals, including the endpoint cases, is available (Holder's inequality for integrals, including the endpoint cases).

[L2]

Conjugate exponents include the endpoint convention $1 / \infty = 0$ (Conjugate exponents, including the endpoint conventions).

[L3]

For $0 < s < \infty$ , $h \in L^{s} (μ)$ means $\int ∣ h ∣^{s} d μ < \infty$ , and $L^{\infty} (μ)$ means finite essential supremum (The function space $L^{p} (μ)$ for $0 < p < \infty$ , The space $L^{\infty} (μ)$ of essentially bounded measurable functions).

[L4]

If $h \in L^{\infty} (μ)$ , then $∣ h ∣ \leq ∥ h ∥_{\infty}$ almost everywhere (The essential supremum is attained as the least essential bound).

Proof

technique · Raise $|fg|$ to the $r$-th power and apply Holder to $|f|^r$ and $|g|^r$ with conjugate exponents $p/r$ and $q/r$

1.1L2L3L4given

If $r = \infty$ , then $1 / p = 1 / q = 0$ , so $p = q = \infty$ by [L2]. Hence $∣ f g ∣ \leq ∥ f ∥_{\infty} ∥ g ∥_{\infty}$ almost everywhere, and taking essential suprema gives $∥ f g ∥_{\infty} \leq ∥ f ∥_{\infty} ∥ g ∥_{\infty} .$ .

1.2L2L3L4given

If $p = \infty$ and $r < \infty$ , then $q = r$ . The pointwise bound and the definition of $∥ g ∥_{r}$ give $∥ f g ∥_{r}^{r} \leq ∥ f ∥_{\infty}^{r} ∥ g ∥_{r}^{r} .$ Indeed, [L4] gives $∣ f ∣ \leq ∥ f ∥_{\infty}$ almost everywhere, so $∣ f g ∣^{r} \leq ∥ f ∥_{\infty}^{r} ∣ g ∣^{r}$ almost everywhere. Taking $r$ -th roots yields the claim. The case $q = \infty$ is symmetric.

1.3L1L2L3givenalgebra

Assume now that $r < \infty$ and $p, q < \infty$ . Then $1 = \frac{r}{p} + \frac{r}{q},$ so the exponents $p / r$ and $q / r$ are conjugate. Because $∣ f ∣^{r} \in L^{p / r} (μ)$ and $∣ g ∣^{r} \in L^{q / r} (μ)$ , [L1] applied to these two functions gives $\int ∣ f g ∣^{r} d μ \leq {(\int ∣ f ∣^{p} d μ)}^{r / p} {(\int ∣ g ∣^{q} d μ)}^{r / q} = ∥ f ∥_{p}^{r} ∥ g ∥_{q}^{r} .$ .

2.1step 1.1step 1.2step 1.3∎

Step 1.1 covers $r = \infty$ , step 1.2 covers the one-infinite endpoint cases, and step 1.3 covers the fully finite case. [step 1.1, step 1.2, step 1.3] In each case $∥ f g ∥_{r} \leq ∥ f ∥_{p} ∥ g ∥_{q}$ , so $f g$ lies in the stated $r$ -space.

CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

Cauchy-Schwarz inequality for $L^{2}$

Statement

If $f, g \in L^{2} (μ)$ , then

$\int ∣ f g ∣ d μ \leq ∥ f ∥_{2} ∥ g ∥_{2} .$

Equality holds if and only if at least one of $f, g$ is zero almost everywhere, or there is a constant $c > 0$ with

$∣ f ∣^{2} = c ∣ g ∣^{2} μ -almost everywhere .$

Facts & Assumptions

Given: Functions $f, g \in L^{2} (μ)$ .

[L1]

Holder's inequality holds for conjugate exponents (Holder's inequality for integrals, including the endpoint cases).

[L2]

The strict-exponent equality criterion for Holder has already been proved (Equality in Holder's inequality for $1 < p < \infty$ ).

Proof

technique · Specialize Holder to $p=q=2$, and inherit the equality clause from the strict-exponent equality theorem

1.1L1

The exponent $2$ is conjugate to itself, so [L1] with $p = q = 2$ gives [L1] $\int ∣ f g ∣ d μ \leq ∥ f ∥_{2} ∥ g ∥_{2} .$

2.1L2step 1.1∎

Because $2 > 1$ , the equality clause is exactly the specialization of [L2] to [L2, step 1.1] $p = q = 2$ .

RemarkRemark: AI-adaptedProof: Not applicableaudited 2026-08-31Open item page →

Finite counting measure recovers finite Holder and implies the signed Cauchy-Schwarz inequality

Via $ℓ^{p}$ is the $L^{p}$ space of counting measure, the measure space ${0, \dots, n - 1}$ with counting measure identifies $L^{p}$ with the published $p$ -norm structure of The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ . Under that identification, Holder's inequality for integrals, including the endpoint cases becomes Holder's inequality for finite sums and conjugate real exponents for $p, q > 1$ . At $p = q = 2$ , Cauchy-Schwarz inequality for $L^{2}$ gives the stronger absolute-product estimate

$\sum_{k < n} ∣ x_{k} y_{k} ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2} .$

The real triangle inequality then gives $∣ \sum_{k < n} x_{k} y_{k} ∣ \leq \sum_{k < n} ∣ x_{k} y_{k} ∣$ , recovering the signed estimates in The Cauchy-Schwarz inequality for finite sums and Cauchy-Schwarz $∣ ⟨ x, y ⟩ ∣ \leq ∥ x ∥_{2} ∥ y ∥_{2}$ with its equality case, the triangle inequality for $∥ \cdot ∥_{2}$ , the parallelogram law and polarisation. Thus the integral and finite forms are compatible, but the absolute-product and signed left sides are not identical.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

Minkowski's inequality for integrals, including $p = \infty$

Statement

Let $(X, A, μ)$ be a measure space.

If $1 \leq p < \infty$ and $f, g \in L^{p} (μ)$ , then $∥ f + g ∥_{p} \leq ∥ f ∥_{p} + ∥ g ∥_{p} .$
If $f, g \in L^{\infty} (μ)$ , then $∥ f + g ∥_{\infty} \leq ∥ f ∥_{\infty} + ∥ g ∥_{\infty} .$

Facts & Assumptions

Given: A measure space and functions $f, g$ in the spaces named in the relevant clause of the Statement.

[L1]

Holder's inequality for integrals is available (Holder's inequality for integrals, including the endpoint cases).

[L2]

Membership in $L^{p} (μ)$ and $L^{\infty} (μ)$ is defined in The function space $L^{p} (μ)$ for $0 < p < \infty$ and The space $L^{\infty} (μ)$ of essentially bounded measurable functions.

[L3]

If $h \in L^{\infty} (μ)$ , then $∣ h ∣ \leq ∥ h ∥_{\infty}$ almost everywhere (The essential supremum is attained as the least essential bound).

[L4]

The nonnegative integral is monotone and homogeneous (Monotonicity and nonnegative homogeneity of the nonnegative integral).

[L5]

The nonnegative integral is additive (Additivity of the nonnegative Lebesgue integral).

[L6]

Holder's inequality for finite sums gives, for nonnegative reals $a, b$ , $a + b \leq 2^{1 / q} (a^{p} + b^{p})^{1 / p}$ when $q = p / (p - 1)$ (Holder's inequality for finite sums and conjugate real exponents).

Proof

technique · For $1 < p < infinity$, first use the two-term finite Holder inequality to show $|f + g|^p$ is integrable, then apply the standard Holder step to $|f + g| |f + g|^{p - 1}$. The cases $p = 1$ and $p = \infty$ are handled directly from subadditivity of absolute value and the essential-supremum bound

1.1L4L5given

If $p = 1$ , then $∣ f + g ∣ \leq ∣ f ∣ + ∣ g ∣$ pointwise, so [L4, L5, given] $∥ f + g ∥_{1} = \int ∣ f + g ∣ d μ \leq \int ∣ f ∣ d μ + \int ∣ g ∣ d μ = ∥ f ∥_{1} + ∥ g ∥_{1} .$

1.2L1L2L4L5L6givenalgebra

Assume $1 < p < \infty$ and let $q : = p / (p - 1)$ . Then [L1, L2, L4, L5, L6, given, algebra] $\int ∣ f + g ∣^{p} d μ \leq 2^{p - 1} (\int ∣ f ∣^{p} d μ + \int ∣ g ∣^{p} d μ) < \infty .$ Indeed, [L6] applied pointwise to the two-term families $(∣ f (x) ∣, ∣ g (x) ∣)$ and $(1, 1)$ gives $∣ f (x) ∣ + ∣ g (x) ∣ \leq 2^{1 / q} (∣ f (x) ∣^{p} + ∣ g (x) ∣^{p})^{1 / p},$ so $∣ f + g ∣^{p} \leq (∣ f ∣ + ∣ g ∣)^{p} \leq 2^{p - 1} (∣ f ∣^{p} + ∣ g ∣^{p})$ pointwise. Thus $f + g \in L^{p} (μ)$ . Put $C : = ∥ f + g ∥_{p}$ . If $C = 0$ , the claim is immediate. Otherwise $∣ f + g ∣^{p} = ∣ f + g ∣ ∣ f + g ∣^{p - 1} \leq ∣ f ∣ ∣ f + g ∣^{p - 1} + ∣ g ∣ ∣ f + g ∣^{p - 1} .$ Because $(p - 1) q = p$ , the function $∣ f + g ∣^{p - 1}$ lies in $L^{q} (μ)$ and has $q$ -norm $C^{p - 1}$ . Integrating and applying [L1] with conjugate exponents $p$ and $q$ to each term yields $C^{p} \leq ∥ f ∥_{p} {(\int ∣ f + g ∣^{(p - 1) q} d μ)}^{1 / q} + ∥ g ∥_{p} {(\int ∣ f + g ∣^{(p - 1) q} d μ)}^{1 / q} .$ Since $(p - 1) q = p$ , this becomes $C^{p} \leq (∥ f ∥_{p} + ∥ g ∥_{p}) C^{p - 1} .$ If $C > 0$ , divide by $C^{p - 1}$ to obtain the claim.

1.3L2L3given

For $p = \infty$ , let $M : = ∥ f ∥_{\infty}$ and $N : = ∥ g ∥_{\infty}$ . Then [L2, L3, given] $∣ f + g ∣ \leq ∣ f ∣ + ∣ g ∣ \leq M + N .$ Indeed, outside the union of the two null exceptional sets supplied by [L3], one has $∣ f ∣ \leq M$ and $∣ g ∣ \leq N$ . Therefore $∥ f + g ∥_{\infty} \leq M + N$ .

2.1step 1.1step 1.2step 1.3∎

Steps 1.1, 1.2, and 1.3 prove the $p = 1$ , $1 < p < \infty$ , and $p = \infty$ cases.

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

Equality in Minkowski's inequality for $1 < p < \infty$

Statement

Let $1 < p < \infty$ and let $f, g \in L^{p} (μ)$ . Then equality holds in Minkowski's inequality

$∥ f + g ∥_{p} = ∥ f ∥_{p} + ∥ g ∥_{p}$

if and only if at least one of $f, g$ is zero almost everywhere, or there is a constant $λ > 0$ such that

$f = λ g μ -almost everywhere .$

Facts & Assumptions

Given: An exponent $1 < p < \infty$ and functions $f, g \in L^{p} (μ)$ .

[L1]

Minkowski's inequality has already been proved (Minkowski's inequality for integrals, including $p = \infty$ ).

[L2]

The equality case in Holder has already been proved (Equality in Holder's inequality for $1 < p < \infty$ ).

[L3]

A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

[L4]

The nonnegative integral is additive (Additivity of the nonnegative Lebesgue integral).

Proof

technique · Examine the Holder step in the standard proof of Minkowski. Equality forces the nonnegative functions $|f|$ and $|g|$ to be proportional almost everywhere, and the pointwise triangle inequality then forces the same sign

1.1L1given

If at least one of $f, g$ is zero almost everywhere, then equality is immediate.

1.2L1given

If $f = λ g$ almost everywhere for some $λ > 0$ , then [L1, given] $∣ f + g ∣ = (λ + 1) ∣ g ∣$ almost everywhere, so $∥ f + g ∥_{p} = (λ + 1) ∥ g ∥_{p} = ∥ f ∥_{p} + ∥ g ∥_{p} .$

1.3L1L2L3L4

Conversely, assume equality in Minkowski and that neither $f$ nor $g$ is zero almost everywhere. [L1, L2, L3, L4] The proof of [L1] showed that equality in Minkowski can only occur when both inequalities $∣ f + g ∣ \leq ∣ f ∣ + ∣ g ∣$ and $\int ∣ f ∣ ∣ f + g ∣^{p - 1} d μ \leq ∥ f ∥_{p} ∥ f + g ∥_{p}^{p - 1}$ and its $g$ -analogue are equalities. The second and third equalities force the pairs $(∣ f ∣, ∣ f + g ∣^{p - 1})$ and $(∣ g ∣, ∣ f + g ∣^{p - 1})$ to satisfy Holder equality. By [L2], this makes $∣ f ∣$ and $∣ g ∣$ proportional almost everywhere. The first inequality then forces $(∣ f ∣ + ∣ g ∣) - ∣ f + g ∣$ to have integral $0$ ; [L3] and [L4] therefore give $∣ f + g ∣ = ∣ f ∣ + ∣ g ∣ μ -almost everywhere .$

2.1step 1.3algebra

Let $∣ f ∣ = c ∣ g ∣$ almost everywhere with $c > 0$ . Then [step 1.3, algebra] on the set where $g \neq 0$ , step 1.3 gives equality in the real triangle inequality for $f$ and $g$ , so they have the same sign there. Hence $f = c g$ almost everywhere on ${g \neq 0}$ , and on ${g = 0}$ both sides vanish. Thus $f = λ g$ almost everywhere for $λ = c > 0$ .

3.1step 1.1step 1.2step 1.3step 2.1∎

Steps 1.1 and 1.2 prove sufficiency, while steps 1.3 and 2.1 prove necessity.

RemarkRemark: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

On a finite counting space, Minkowski agrees with the published finite theorem for $p > 1$

Under the finite counting-measure identification of $ℓ^{p}$ is the $L^{p}$ space of counting measure, the integral Minkowski inequality of Minkowski's inequality for integrals, including $p = \infty$ becomes the published finite-sum theorem Minkowski's inequality for finite sums and real exponent p greater than one for real $p > 1$ on the same coordinates. The integral theorem's $p = 1$ and $p = \infty$ clauses also specialize to the corresponding elementary finite-sum inequalities, but those endpoints are outside the cited theorem's statement. This is an agreement seam, not a second construction.

PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31Open item page →

The essential supremum is attained as the least essential bound

Statement

Let $f : X \to R$ be measurable on a measure space $(X, A, μ)$ , and suppose $∥ f ∥_{\infty} < \infty$ . Then

$∣ f ∣ \leq ∥ f ∥_{\infty} μ -almost everywhere .$

Moreover, if $M \geq 0$ and $∣ f ∣ \leq M$ almost everywhere, then

$∥ f ∥_{\infty} \leq M .$

So $∥ f ∥_{\infty}$ is the least essential bound of $∣ f ∣$ .

Facts & Assumptions

Given: A measurable real-valued function $f$ with finite essential supremum $s : = ∥ f ∥_{\infty}$ .

[L1]

The essential supremum is the infimum of the essential bounds (The essential supremum of a measurable function with respect to a measure).

[L2]

Absolute values and threshold sets of measurable functions are measurable (Closure properties of measurable functions used by the integral).

[L3]

Countable subadditivity bounds the measure of a countable union (Finite and countable subadditivity of measures).

Proof

technique · Take the countable family of bad sets $\{|f|>\|f\|_\infty+1/n\}$; each is null by minimality of the infimum. Their union is null, giving $|f|\le\|f\|_\infty$ almost everywhere, and leastness is built into the definition

1.1L1L2given

For each $n \geq 1$ , the number $s + 1 / n$ is strictly larger than the infimum in [L1], so it is an essential bound. Therefore the measurable set $E_{n} : = {∣ f ∣ > s + 1 / n}$ has measure $0$ .

1.2L1

If $M \geq 0$ and $∣ f ∣ \leq M$ almost everywhere, then $M$ is one of the essential bounds in [L1], so the infimum $s$ satisfies $s \leq M$ . [L1].

2.1step 1.1L3algebra

Put $E : = ⋃_{n = 1}^{\infty} E_{n}$ . Then $E$ is measurable and $μ (E) \leq \sum_{n = 1}^{\infty} μ (E_{n}) = 0.$ If $x \notin E$ , then $∣ f (x) ∣ \leq s + 1 / n$ for every $n$ , hence $∣ f (x) ∣ \leq s$ . Therefore $∣ f ∣ \leq s$ almost everywhere.

3.1step 2.1step 1.2∎

Step 2.1 proves that $s$ itself is an essential bound, and step 1.2 proves that no smaller essential bound exists. Thus $s = ∥ f ∥_{\infty}$ is the least essential bound. [step 2.1, step 1.2].

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passverified 2026-09-23 (gpt-6-sol)Open item page →

$L^{p}$ and $L^{\infty}$ are vector spaces for $p \geq 1$

Statement

Let $(X, A, μ)$ be a measure space.

For each $1 \leq p < \infty$ , the class $L^{p} (μ)$ is a real vector space under pointwise addition and scalar multiplication.
The class $L^{\infty} (μ)$ is a real vector space under the same operations.

Facts & Assumptions

Given: A measure space $(X, A, μ)$ .

[L1]

$L^{p} (μ)$ and $L^{\infty} (μ)$ are the classes defined in The function space $L^{p} (μ)$ for $0 < p < \infty$ and The space $L^{\infty} (μ)$ of essentially bounded measurable functions.

[L2]

Sums, scalar multiples, and absolute values of measurable real-valued functions are measurable (Closure properties of measurable functions used by the integral).

[L3]

The nonnegative integral is monotone, additive, and homogeneous for positive finite scalars; the zero function has integral zero (Additivity of the nonnegative Lebesgue integral, Monotonicity and nonnegative homogeneity of the nonnegative integral).

[L4]

A finite essential supremum is an attained essential bound (The essential supremum is attained as the least essential bound).

[L5]

Countable unions of measurable null sets are measurable and null (Finite and countable subadditivity of measures).

[L6]

A vector space over $R$ means the structure defined in Vector space over a field.

Proof

technique · direct

1.1L1L2L3givenalgebra

Fix $1 \leq p < \infty$ and let $f, g \in L^{p} (μ)$ and $a \in R$ . Measurability of $f + g$ and $a f$ follows from [L2]. Pointwise, $∣ f + g ∣^{p} \leq (2 \max {∣ f ∣, ∣ g ∣})^{p} \leq 2^{p} (∣ f ∣^{p} + ∣ g ∣^{p}) .$ By [L3], $\int ∣ f + g ∣^{p} d μ \leq 2^{p} (\int ∣ f ∣^{p} d μ + \int ∣ g ∣^{p} d μ) < \infty .$ If $a \neq 0$ , positive homogeneity in [L3] gives $\int ∣ a f ∣^{p} d μ = ∣ a ∣^{p} \int ∣ f ∣^{p} d μ < \infty$ ; if $a = 0$ , the integral is zero without forming $0 \cdot \infty$ . Thus $f + g, a f \in L^{p} (μ)$ . The zero function belongs to the class, and additive inverses use $a = - 1$ .

1.2

Let $f, g \in L^{\infty} (μ)$ with $M : = ∥ f ∥_{\infty}$ and $N : = ∥ g ∥_{\infty}$ . There are measurable null sets $E_{f}, E_{g}$ such that $∣ f ∣ \leq M$ on $X ∖ E_{f}$ and $∣ g ∣ \leq N$ on $X ∖ E_{g}$ . With $E : = E_{f} \cup E_{g}$ , $E$ is measurable and null, and on $X ∖ E$ one has [L1, L2, L4, L5, given] $∣ f + g ∣ \leq ∣ f ∣ + ∣ g ∣ \leq M + N, ∣ a f ∣ = ∣ a ∣ ∣ f ∣ \leq ∣ a ∣ M .$ Thus $f + g$ and $a f$ are essentially bounded; measurability again comes from [L2].

2.1step 1.1L6

The pointwise addition and scalar-multiplication identities are inherited from real-valued functions. Hence [L6] makes $L^{p} (μ)$ a real vector space for every $1 \leq p < \infty$ .

3.1step 1.2L6∎

The pointwise identities are again inherited from real-valued functions, so [L6] makes $L^{\infty} (μ)$ a real vector space.

PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

Null functions form a linear subspace and are exactly the zero-seminorm class

Statement

Let $(X, A, μ)$ be a measure space.

For each $1 \leq p < \infty$ , the set $N_{p} (μ) : = L^{p} (μ) \cap N (μ)$ is a linear subspace of $L^{p} (μ)$ , and for $f \in L^{p} (μ)$ one has $f \in N (μ) ⟺ ∥ f ∥_{p} = 0.$
The set $N_{\infty} (μ) : = L^{\infty} (μ) \cap N (μ)$ is a linear subspace of $L^{\infty} (μ)$ , and for $f \in L^{\infty} (μ)$ one has $f \in N (μ) ⟺ ∥ f ∥_{\infty} = 0.$

Facts & Assumptions

Given: A measure space $(X, A, μ)$ .

[L1]

The null functions are those that vanish almost everywhere (The null subspace of measurable functions that vanish almost everywhere).

[L2]

$L^{p} (μ)$ and $L^{\infty} (μ)$ are vector spaces in the relevant ranges ( $L^{p}$ and $L^{\infty}$ are vector spaces for $p \geq 1$ ).

[L3]

A countable union of measurable null sets is null (Finite and countable subadditivity of measures).

[L4]

A linear subspace means the three closure conditions of Linear subspace of a vector space.

[L5]

For $1 \leq p < \infty$ , a nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

[L6]

If $∥ f ∥_{\infty} < \infty$ , then $∣ f ∣ \leq ∥ f ∥_{\infty}$ almost everywhere (The essential supremum is attained as the least essential bound).

Proof

technique · Use countable-union stability of null sets for addition and scalar multiplication. For $1 \le p < infinity$, the $p$-seminorm vanishes exactly when the integral of $|f|^p$ is zero; for $p = infinity$, vanishing means the essential supremum is zero

1.1L1L2L3L4

Fix $1 \leq p < \infty$ . If $f, g \in N_{p} (μ)$ , choose measurable null sets $E_{f}, E_{g}$ outside which $f = 0$ and $g = 0$ . Their union is null, and on its complement one has $f + g = 0$ and $a f = 0$ for every $a \in R$ . Because $L^{p} (μ)$ is a vector space, [L4] makes $N_{p} (μ)$ a linear subspace.

1.2

If $f \in N_{p} (μ)$ , then $∣ f ∣^{p} = 0$ almost everywhere, so [L1, L5] $\int ∣ f ∣^{p} d μ = 0,$ hence $∥ f ∥_{p} = 0$ . Conversely, if $∥ f ∥_{p} = 0$ , then the same theorem [L5] forces $∣ f ∣^{p} = 0$ almost everywhere and therefore $f = 0$ almost everywhere.

1.3L1L6

If $f \in N_{\infty} (μ)$ , then $0$ is an essential bound for $∣ f ∣$ , so $∥ f ∥_{\infty} = 0$ . Conversely, if $∥ f ∥_{\infty} = 0$ , then [L6] gives $∣ f ∣ \leq 0$ almost everywhere, hence $f = 0$ almost everywhere.

2.1L1L2L3L4

If $f, g \in N_{\infty} (μ)$ , the same null-set union argument as in step 1.1 shows that $f + g$ and $a f$ vanish almost everywhere, and [L2] places them in $L^{\infty} (μ)$ . Therefore [L4] makes $N_{\infty} (μ)$ a linear subspace.

3.1step 1.1step 1.2step 2.1step 1.3∎

Steps 1.1 and 2.1 prove the two subspace claims, and steps 1.2 and 1.3 identify the zero-seminorm class in every range.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$

Statement

Let $(X, A, μ)$ be a measure space.

For $1 \leq p < \infty$ , the rule $∥ [f] ∥_{p} : = ∥ f ∥_{p} ([f] \in L^{p} (μ))$ is well defined on the quotient classes.
For $p = \infty$ , the rule $∥ [f] ∥_{\infty} : = ∥ f ∥_{\infty} ([f] \in L^{\infty} (μ))$ is well defined.
In either case, with the quotient vector-space operations of Coset equality, well-defined quotient operations, and the canonical projection with kernel $W$ , the resulting pair $(L^{p} (μ), ∥ \cdot ∥_{p})$ is a normed space in the sense of A norm on a real vector space, the induced metric, and the dictionary with the metric axioms.

Facts & Assumptions

Given: A measure space $(X, A, μ)$ and an exponent $1 \leq p \leq \infty$ .

[L1]

The quotient spaces $L^{p} (μ)$ are those of The space $L^{p} (μ)$ as the quotient by null functions.

[L2]

The null representatives are exactly the zero-seminorm class (Null functions form a linear subspace and are exactly the zero-seminorm class).

[L3]

Minkowski's inequality supplies the triangle inequality for the representative seminorms (Minkowski's inequality for integrals, including $p = \infty$ ).

[L4]

The essential supremum is an attained essential bound (The essential supremum is attained as the least essential bound).

[L5]

The quotient operations are well defined and produce a vector space (Coset equality, well-defined quotient operations, and the canonical projection with kernel $W$ ).

[L6]

A normed space means a real vector space with separation, homogeneity, and triangle inequality (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

Proof

technique · Use the previous proposition to identify the null functions as the kernel of the seminorm. Hence the seminorm is constant on cosets and separates points on the quotient, while Minkowski and homogeneity descend from representatives

1.1

Suppose first $1 \leq p < \infty$ and $f - g \in N_{p} (μ)$ . Then $∥ f - g ∥_{p} = 0$ . Applying Minkowski twice yields [L2, L3] $∥ f ∥_{p} \leq ∥ g ∥_{p} + ∥ f - g ∥_{p} = ∥ g ∥_{p}, ∥ g ∥_{p} \leq ∥ f ∥_{p} + ∥ f - g ∥_{p} = ∥ f ∥_{p},$ so $∥ f ∥_{p} = ∥ g ∥_{p}$ . Thus $∥ [f] ∥_{p} : = ∥ f ∥_{p}$ is well defined.

1.2L2L4

For $p = \infty$ , if $f - g \in N_{\infty} (μ)$ then $f = g$ almost everywhere. Any essential bound for $f$ is therefore an essential bound for $g$ and conversely, so $∥ f ∥_{\infty} = ∥ g ∥_{\infty}$ . Thus $∥ [f] ∥_{\infty} : = ∥ f ∥_{\infty}$ is well defined.

1.3L2

For separation, $∥ [f] ∥_{p} = 0$ means $∥ f ∥_{p} = 0$ , and then [L2] gives $f \in N_{p} (μ)$ or $N_{\infty} (μ)$ . Hence $[f] = [0]$ . The converse is immediate because the zero representative has norm $0$ .

2.1step 1.1step 1.2L1L3L5

By [L1] and [L5], each quotient $L^{p} (μ)$ is already a real vector space. The representative functionals are homogeneous, and [L3] supplies the triangle inequality on representatives; steps 1.1 and 1.2 show that these formulas depend only on the class, so homogeneity and triangle inequality descend to the quotient.

3.1step 2.1step 1.3L6∎

Steps 2.1 and 1.3 verify the three norm axioms named in [L6]. Therefore $(L^{p} (μ), ∥ \cdot ∥_{p})$ is a normed space for every $1 \leq p \leq \infty$ .

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$

Statement

Assume countable choice (in particular, the Axiom of Choice suffices). Let $(X, A, μ)$ be a measure space and let $1 \leq p \leq \infty$ . Then $L^{p} (μ)$ , with the norm of The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ , is complete. Equivalently, the metric induced by that norm is a complete metric in the sense of Complete metric space: every Cauchy sequence converges in the space.

Moreover, if a sequence in $L^{p} (μ)$ converges in norm, then some subsequence admits measurable representatives converging almost everywhere in the sense of Convergence almost everywhere relative to a measure.

Facts & Assumptions

Given: A measure space, an exponent $1 \leq p \leq \infty$ , and countable choice, available under the Axiom of Choice (The Axiom of Choice). Its only use is to select a countable family of measurable representatives.

[L1]

$L^{p} (μ)$ is a normed space, so it has the norm metric (The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ , Complete metric space: every Cauchy sequence converges in the space).

[L2]

Minkowski's inequality holds in $L^{p}$ (Minkowski's inequality for integrals, including $p = \infty$ ).

[L3]

Monotone convergence and dominated convergence are available (Monotone convergence for the integral, Dominated convergence).

[L4]

Pointwise limits of measurable functions are measurable (Sequential suprema, infima, limsup, liminf, and pointwise limits of measurable functions are measurable).

[L5]

Sums and absolute values of measurable functions are measurable (Closure properties of measurable functions used by the integral).

[L6]

Countable unions of measurable null sets are measurable and null (Finite and countable subadditivity of measures).

[L7]

Finite essential suprema are attained almost everywhere (The essential supremum is attained as the least essential bound).

Proof

technique · For $1 \le p < infinity$, choose a rapidly Cauchy subsequence by least indices, sum the successive differences with monotone convergence and Minkowski, and recover the limit by dominated convergence. For $p = infinity$, union the exceptional null sets and take the pointwise limit outside them

1.1givenL2L3L4L5choose

Assume $1 \leq p < \infty$ and let $(u_{n})$ be Cauchy in $L^{p} (μ)$ . Choose by least indices a strictly increasing sequence $(n_{k})$ such that [L2, L3, L4, L5, given, choose] $∥ u_{n_{k + 1}} - u_{n_{k}} ∥_{p} < 2^{- k} (k \geq 0) .$ Use countable choice (the countable instance of AC) to choose measurable representatives $f_{k}$ of $u_{n_{k}}$ , and put $h_{k} : = ∣ f_{k + 1} - f_{k} ∣$ . Then each $h_{k}$ is measurable, belongs to $L^{p} (μ)$ , and satisfies $∥ h_{k} ∥_{p} < 2^{- k} .$ If $g_{m} : = \sum_{k < m} h_{k}$ , then $∥ g_{m} ∥_{p} \leq \sum_{k < m} ∥ h_{k} ∥_{p} \leq \sum_{k = 0}^{\infty} 2^{- k} < \infty,$ so each $g_{m}$ lies in $L^{p} (μ)$ . Monotone convergence then gives a measurable pointwise limit $g : = \sum_{k = 0}^{\infty} h_{k}$ with $g \in L^{p} (μ)$ .

1.2givenL4L5L6L7choose

Assume now $p = \infty$ and let $(u_{n})$ be Cauchy in $L^{\infty} (μ)$ . Choose least indices $n_{k}$ with [L4, L6, L7, given, choose] $∥ u_{n_{k + 1}} - u_{n_{k}} ∥_{\infty} < 2^{- k} .$ Use countable choice (the countable instance of AC) to choose measurable representatives $f_{k}$ of $u_{n_{k}}$ . By [L7], the explicitly defined measurable set $E_{k} : = {x : ∣ f_{k + 1} (x) - f_{k} (x) ∣ > 2^{- k}}$ is null, and $∣ f_{k + 1} - f_{k} ∣ \leq 2^{- k} on X ∖ E_{k} .$ With $E : = ⋃_{k} E_{k}$ , [L6] makes $E$ measurable and null, and for $x \in X ∖ E$ and $m > n$ , $∣ f_{m} (x) - f_{n} (x) ∣ \leq \sum_{k = n}^{m - 1} 2^{- k} .$ So $(f_{k} (x))$ is Cauchy in $R$ , hence converges to some value $f (x)$ . Set $f = 0$ on $E$ . The measurable functions equal to $f_{k}$ on $X ∖ E$ and zero on $E$ converge everywhere to $f$ , so [L4] makes $f$ measurable. Indeed, their inverse images of Borel sets are obtained by intersecting those of $f_{k}$ with $X ∖ E$ and adjoining $E$ exactly when the Borel set contains zero. No completeness of the measure is assumed.

2.1step 1.1L2L3L4

In the finite- $p$ case, $E : = {g = \infty}$ is measurable and null: for every $M > 0$ , $M^{p} μ (E) \leq \int g^{p} < \infty$ , hence $μ (E) = 0$ . On $X ∖ E$ the series $\sum_{k} ∣ f_{k + 1} - f_{k} ∣$ converges, so $f_{k}$ has a finite limit $f$ . Set $f = 0$ on $E$ . Replacing each $f_{k}$ by zero on $E$ gives measurable functions converging everywhere to $f$ , so [L4] gives measurability. Almost everywhere, $∣ f ∣ \leq ∣ f_{0} ∣ + g$ , so [L2] gives $f \in L^{p}$ , and [step 1.1, L2, L3, L4] $∣ f - f_{k} ∣ \leq \sum_{j \geq k} h_{j} \leq g .$ Dominated convergence in [L3] therefore gives $∥ f - f_{k} ∥_{p} \to 0$ . So the subsequence $u_{n_{k}}$ converges to $[f]$ , and its representatives converge to $f$ almost everywhere.

2.2step 1.2L1L2

For $x \in X ∖ E$ , the same tail estimate gives [step 1.2] $∣ f (x) - f_{n} (x) ∣ \leq \sum_{k = n}^{\infty} 2^{- k} \leq 2^{1 - n} .$ Taking $n = 0$ also gives $∣ f ∣ \leq ∣ f_{0} ∣ + 2$ almost everywhere, so $f \in L^{\infty}$ . Therefore $∥ u_{n_{k}} - [f] ∥_{\infty} \to 0$ . Given $ε > 0$ , choose $K$ with $∥ u_{n} - u_{m} ∥_{\infty} < ε / 2$ for $m, n \geq K$ , then choose $k$ with $n_{k} \geq K$ and $∥ u_{n_{k}} - [f] ∥_{\infty} < ε / 2$ . Hence $∥ u_{n} - [f] ∥_{\infty} \leq ∥ u_{n} - u_{n_{k}} ∥_{\infty} + ∥ u_{n_{k}} - [f] ∥_{\infty} < ε,$ so $u_{n} \to [f]$ in $L^{\infty} (μ)$ .

3.1step 2.1L1L2

Since $(u_{n})$ is Cauchy, for every $ε > 0$ there is $K$ with $∥ u_{n} - u_{m} ∥_{p} < ε / 2$ whenever $m, n \geq K$ . Choose $k$ with $n_{k} \geq K$ and $∥ u_{n_{k}} - [f] ∥_{p} < ε / 2$ from step 2.1. Then for every $n \geq K$ , [step 2.1, L2] $∥ u_{n} - [f] ∥_{p} \leq ∥ u_{n} - u_{n_{k}} ∥_{p} + ∥ u_{n_{k}} - [f] ∥_{p} < ε .$ So the whole sequence converges to $[f]$ . This proves completeness for $1 \leq p < \infty$ .

4.1step 2.1step 2.2step 3.1L1∎

Step 2.1 proves the almost-everywhere convergent subsequence clause in the finite- $p$ case, and step 2.2 gives the same for $p = \infty$ . Steps 3.1 and 2.2 prove completeness in every case, which by [L1] is exactly completeness of the norm metric.

CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-09-23 (gpt-6-sol)Open item page →

Assuming Countable Choice, $L^{p}$ -convergent sequences have almost-everywhere convergent subsequences

Statement

Assume Countable Choice. Let $1 \leq p \leq \infty$ . If $u_{n} \to u$ in $L^{p} (μ)$ , then some subsequence of $(u_{n})$ admits measurable representatives converging almost everywhere to a measurable representative of $u$ .

Facts & Assumptions

Given: Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) and a norm-convergent sequence $(u_{n})$ in $L^{p} (μ)$ .

[L1]

Under Countable Choice, Riesz-Fischer completeness states that every norm-convergent sequence in $L^{p} (μ)$ has an almost-everywhere convergent subsequence of representatives (Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ ). Applying that representative-selection clause in step 1.1 is the use of Countable Choice here.

Proof

technique · Choose a rapidly convergent subsequence from an $L^p$-convergent sequence and re-use the subsequence construction inside Riesz-Fischer

1.1L1given

The sequence $(u_{n})$ is Cauchy because it converges in norm. Applying [L1] to that Cauchy sequence gives an $L^{p}$ limit together with an almost-everywhere convergent subsequence of representatives. Because metric limits are unique, the limit supplied by [L1] must be the given $u$ .

2.1step 1.1∎

That subsequence is the required one.

CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31Open item page →

Convergence in $L^{p}$ implies convergence in measure

Statement

Let $1 \leq p \leq \infty$ , let $u_{n}, u \in L^{p} (μ)$ , and choose measurable representatives $f_{n}, f$ . If $∥ u_{n} - u ∥_{p} \to 0$ , then $f_{n} \to f$ in measure.

Facts & Assumptions

Given: Representatives $f_{n}, f$ of classes $u_{n}, u \in L^{p} (μ)$ with $∥ u_{n} - u ∥_{p} \to 0$ .

[L1]

Convergence in measure means the bad-set measures $μ ({∣ f_{n} - f ∣ > ε})$ tend to $0$ for every $ε > 0$ (Convergence in measure).

[L2]

The class norm agrees with the representative norm (The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ ).

[L3]

Chebyshev-Markov gives $μ ({h \geq t}) \leq t^{- 1} \int h d μ$ for nonnegative measurable $h$ (Chebyshev-Markov inequality for the integral).

[L4]

If $∥ g ∥_{\infty} < \infty$ , then $∣ g ∣ \leq ∥ g ∥_{\infty}$ almost everywhere (The essential supremum is attained as the least essential bound).

Proof

technique · For $1 \le p < infinity$, apply Chebyshev-Markov to $|f_n-f|^p$. For $p = infinity$, the essential-supremum bound makes the bad set null for all large $n$

1.1L3L2givenalgebra

Assume $1 \leq p < \infty$ and fix $ε > 0$ . Apply [L3] to [L2, L3, given, algebra] $h : = ∣ f_{n} - f ∣^{p}$ and $t : = ε^{p}$ : $μ ({∣ f_{n} - f ∣ > ε}) = μ ({∣ f_{n} - f ∣^{p} > ε^{p}}) \leq ε^{- p} \int ∣ f_{n} - f ∣^{p} d μ = ε^{- p} ∥ u_{n} - u ∥_{p}^{p} .$ By [L2], the right-hand side tends to $0$ .

1.2L2L4given

Assume $p = \infty$ and fix $ε > 0$ . Because $∥ u_{n} - u ∥_{\infty} \to 0$ , [L2, L4, given] for all large $n$ one has $∥ u_{n} - u ∥_{\infty} < ε$ . Then [L4] gives $∣ f_{n} - f ∣ \leq ε$ almost everywhere, so $μ ({∣ f_{n} - f ∣ > ε}) = 0.$ Hence the bad-set measures are eventually $0$ .

2.1step 1.1step 1.2L1∎

Step 1.1 proves the finite- $p$ case and step 1.2 proves the [step 1.1, step 1.2, L1] $p = \infty$ case, so [L1] gives convergence in measure.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

Finite-measure $L^{r}$ includes into $L^{p}$ for $p < r$

Statement

Let $(X, A, μ)$ be a measure space with $μ (X) < \infty$ .

If $1 \leq p < r < \infty$ and $f \in L^{r} (μ)$ , then $f \in L^{p} (μ)$ and $∥ f ∥_{p} \leq μ (X)^{1 / p - 1 / r} ∥ f ∥_{r} .$
If $1 \leq p < \infty$ and $f \in L^{\infty} (μ)$ , then $f \in L^{p} (μ)$ and $∥ f ∥_{p} \leq μ (X)^{1 / p} ∥ f ∥_{\infty} .$

Facts & Assumptions

Given: A finite measure space $(X, A, μ)$ .

[L1]

Holder's inequality for integrals is available (Holder's inequality for integrals, including the endpoint cases).

[L2]

Conjugate exponents are defined in Conjugate exponents, including the endpoint conventions.

[L3]

$L^{r} (μ)$ and $L^{\infty} (μ)$ are the measurable-function spaces of The function space $L^{p} (μ)$ for $0 < p < \infty$ and The space $L^{\infty} (μ)$ of essentially bounded measurable functions.

[L4]

If $h \in L^{\infty} (μ)$ , then $∣ h ∣ \leq ∥ h ∥_{\infty}$ almost everywhere (The essential supremum is attained as the least essential bound).

Proof

technique · Write $|f|^p$ as $|f|^p\cdot1$ and apply Holder with exponents $r/p$ and $r/(r-p)$. The finite total measure contributes the factor $\mu(X)^{1/p-1/r}$

1.1L1L2L3givenalgebra

Suppose $1 \leq p < r < \infty$ and put $a : = \frac{r}{p}, b : = \frac{r}{r - p} .$ Then $a, b \in (1, \infty)$ and $1 / a + 1 / b = 1$ , so [L2] makes them conjugate. Apply [L1] to the functions $∣ f ∣^{p}$ and $1$ : $\int ∣ f ∣^{p} d μ \leq {(\int ∣ f ∣^{p a} d μ)}^{1 / a} {(\int 1^{b} d μ)}^{1 / b} = {(\int ∣ f ∣^{r} d μ)}^{p / r} μ (X)^{1 - p / r} .$ Taking $p$ -th roots gives the claimed bound.

1.2L3L4givenalgebra

If $f \in L^{\infty} (μ)$ , then $\int ∣ f ∣^{p} d μ \leq ∥ f ∥_{\infty}^{p} μ (X) .$ Indeed, $∣ f ∣^{p} \leq ∥ f ∥_{\infty}^{p}$ almost everywhere by [L4]. Thus $f \in L^{p} (μ)$ and $∥ f ∥_{p} \leq μ (X)^{1 / p} ∥ f ∥_{\infty}$ .

2.1step 1.1step 1.2∎

Steps 1.1 and 1.2 prove the finite-measure inclusion laws. [step 1.1, step 1.2].

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31Open item page →

$ℓ^{p}$ includes into $ℓ^{r}$ for $p < r$

Statement

Let $1 \leq p < r \leq \infty$ . If $a = (a_{n})_{n \geq 0} \in ℓ^{p}$ , then $a \in ℓ^{r}$ . When $r < \infty$ one has

$∥ a ∥_{r} \leq ∥ a ∥_{p},$

and when $r = \infty$ one has

$∥ a ∥_{\infty} \leq ∥ a ∥_{p} .$

Facts & Assumptions

Given: A real sequence $a = (a_{n})_{n \geq 0}$ in $ℓ^{p}$ .

[L1]

$ℓ^{p}$ is $L^{p}$ of counting measure on $N$ ( $ℓ^{p}$ is the $L^{p}$ space of counting measure, Counting measure on an arbitrary set, Counting measure is a measure).

[L2]

Real powers obey the usual laws, and for fixed base $b > 1$ the map $x \mapsto b^{x}$ is strictly increasing (Real powers for positive bases, with the zero-base positive-exponent convention, The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents, The exponential function is strictly increasing).

Proof

technique · For counting measure, only finitely many terms can exceed $1$ when a sequence lies in $\ell^p$. Split the series at that finite set and compare $|a_n|^r$ to $|a_n|^p$ on the tail where $|a_n|\le1$

1.1L1

If $r = \infty$ , then for each $n$ , [L1] $∣ a_{n} ∣^{p} \leq \sum_{k = 0}^{\infty} ∣ a_{k} ∣^{p} = ∥ a ∥_{p}^{p},$ so taking $p$ -th roots gives $∣ a_{n} ∣ \leq ∥ a ∥_{p}$ . Hence $∥ a ∥_{\infty} \leq ∥ a ∥_{p}$ .

1.2L1L2

Assume $r < \infty$ . Because $\sum_{n} ∣ a_{n} ∣^{p} < \infty$ , only finitely many indices can satisfy $∣ a_{n} ∣ > 1$ ; otherwise the $p$ -series would dominate the divergent sum of infinitely many $1$ 's. Thus $∣ a_{n} ∣ \leq 1$ for all sufficiently large $n$ . Since $r > p$ , one has $∣ a_{n} ∣^{r} \leq ∣ a_{n} ∣^{p}$ on that tail, so $\sum_{n} ∣ a_{n} ∣^{r}$ converges. [L1, L2].

2.1step 1.1step 1.2L1algebra

For every $n$ , $\sum_{k < n} ∣ a_{k} ∣^{r} \leq (\sup_{k} ∣ a_{k} ∣^{r - p}) \sum_{k < n} ∣ a_{k} ∣^{p} \leq ∥ a ∥_{\infty}^{r - p} \sum_{k < n} ∣ a_{k} ∣^{p} .$ Using step 1.1, this becomes $\sum_{k < n} ∣ a_{k} ∣^{r} \leq ∥ a ∥_{p}^{r - p} \sum_{k < n} ∣ a_{k} ∣^{p} .$ Letting $n \to \infty$ yields $∥ a ∥_{r}^{r} \leq ∥ a ∥_{p}^{r}$ , hence $∥ a ∥_{r} \leq ∥ a ∥_{p}$ .

3.1step 1.1step 2.1∎

Step 1.1 proves the endpoint $r = \infty$ , and step 2.1 proves the finite- $r$ estimate. [step 1.1, step 2.1].

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31Open item page →

Lyapunov interpolation inequality for $L^{p}$ norms

Statement

Let $1 \leq p_{0} < p < p_{1} < \infty$ and let $θ \in (0, 1)$ satisfy

$\frac{1}{p} = \frac{θ}{p_{0}} + \frac{1 - θ}{p_{1}} .$

If $f \in L^{p_{0}} (μ) \cap L^{p_{1}} (μ)$ , then $f \in L^{p} (μ)$ and

$∥ f ∥_{p} \leq ∥ f ∥_{p_{0}}^{θ} ∥ f ∥_{p_{1}}^{1 - θ} .$

Facts & Assumptions

Given: Exponents $p_{0} < p < p_{1}$ and a function $f \in L^{p_{0}} (μ) \cap L^{p_{1}} (μ)$ .

[L1]

Holder's inequality for integrals is available (Holder's inequality for integrals, including the endpoint cases).

[L2]

Conjugate exponents are defined in Conjugate exponents, including the endpoint conventions.

[L3]

Membership in $L^{s} (μ)$ means finiteness of the $s$ -power integral (The function space $L^{p} (μ)$ for $0 < p < \infty$ ).

Proof

technique · If $1/p = \theta/p_0 + (1 - \theta)/p_1$, rewrite $|f|^p$ as the product $|f|^{\theta p}|f|^{(1-\theta)p}$ and apply Holder with conjugate exponents $p_0/(\theta p)$ and $p_1/((1-\theta)p)$

1.1L1L2L3givenalgebra

Put [L1, L2, L3, given, algebra] $a : = \frac{p_{0}}{θ p}, b : = \frac{p_{1}}{(1 - θ) p} .$ Then $\frac{1}{a} + \frac{1}{b} = \frac{θ p}{p_{0}} + \frac{(1 - θ) p}{p_{1}} = 1,$ so [L2] makes $a$ and $b$ conjugate exponents. Also ${(∣ f ∣^{θ p})}^{a} = ∣ f ∣^{p_{0}}, {(∣ f ∣^{(1 - θ) p})}^{b} = ∣ f ∣^{p_{1}} .$ Applying [L1] to the factors $∣ f ∣^{θ p}$ and $∣ f ∣^{(1 - θ) p}$ therefore yields $\int ∣ f ∣^{p} d μ \leq {(\int ∣ f ∣^{p_{0}} d μ)}^{θ p / p_{0}} {(\int ∣ f ∣^{p_{1}} d μ)}^{(1 - θ) p / p_{1}} = ∥ f ∥_{p_{0}}^{θ p} ∥ f ∥_{p_{1}}^{(1 - θ) p} .$

2.1step 1.1L3∎

Taking $p$ -th roots yields the Lyapunov interpolation inequality, and the right-hand side is finite by [L3], so $f \in L^{p} (μ)$ .

RemarkRemark: AI-adaptedProof: Not applicableaudited 2026-08-31Open item page →

Lyapunov inequality is equivalent to log-convexity of the reciprocal-exponent norm profile

Fix a measurable $f$ and an interval of exponents on which every $∥ f ∥_{p}$ is finite and positive. Write

$s : = \frac{1}{p}, s_{0} : = \frac{1}{p_{0}}, s_{1} : = \frac{1}{p_{1}} .$

Then Lyapunov interpolation inequality for $L^{p}$ norms says exactly that

$∥ f ∥_{p} \leq ∥ f ∥_{p_{0}}^{θ} ∥ f ∥_{p_{1}}^{1 - θ}$

whenever $s = θ s_{0} + (1 - θ) s_{1}$ . Equivalently,

$∥ f ∥_{1 / s} \leq ∥ f ∥_{1 / s_{0}}^{θ} ∥ f ∥_{1 / s_{1}}^{1 - θ} .$

So the map $s \mapsto ∥ f ∥_{1 / s}$ is log-convex in the sense of Log-convex positive functions on the reciprocal-exponent interval, and conversely that log-convexity is exactly the Lyapunov interpolation statement.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31Open item page →

$L^{p}$ norms converge to the essential supremum for essentially bounded $L^{r}$ functions

Statement

Let $0 < r < \infty$ , let $f \in L^{r} (μ) \cap L^{\infty} (μ)$ , and put $M : = ∥ f ∥_{\infty}$ . Then $f \in L^{p} (μ)$ for every finite $p \geq r$ and

$\lim_{p \to \infty} ∥ f ∥_{p} = M .$

Facts & Assumptions

Given: A real exponent $r > 0$ and a function $f \in L^{r} (μ) \cap L^{\infty} (μ)$ .

[L1]

If $M = ∥ f ∥_{\infty} < \infty$ , then $∣ f ∣ \leq M$ almost everywhere and $M$ is the least essential bound (The essential supremum is attained as the least essential bound).

[L2]

Membership in $L^{r} (μ)$ and $L^{\infty} (μ)$ is defined in The function space $L^{p} (μ)$ for $0 < p < \infty$ and The space $L^{\infty} (μ)$ of essentially bounded measurable functions.

[L3]

The nonnegative integral is monotone and homogeneous (Monotonicity and nonnegative homogeneity of the nonnegative integral).

Proof

technique · The upper bound is $\|f\|_p^p\le\|f\|_\infty^{p-r}\|f\|_r^r$. For the lower bound, every $\varepsilon$ below the essential supremum leaves a set of positive measure where $|f|$ exceeds $\|f\|_\infty-\varepsilon$, forcing the $p$-norm above that level as $p$ grows

1.1L1L3given

If $M = 0$ , then [L1] gives $∣ f ∣ = 0$ almost everywhere, so $∥ f ∥_{p}^{p} = \int ∣ f ∣^{p} d μ = 0$ for every finite $p \geq r$ . Hence $∥ f ∥_{p} = 0 = M$ for all such $p$ , and the conclusion follows in this case.

1.2L1L2L3givenalgebra

Assume from now on that $M > 0$ . For every finite $p \geq r$ , one has $∥ f ∥_{p}^{p} = \int ∣ f ∣^{p} d μ \leq M^{p - r} \int ∣ f ∣^{r} d μ = M^{p - r} ∥ f ∥_{r}^{r},$ because [L1] gives $∣ f ∣^{p} = ∣ f ∣^{r} ∣ f ∣^{p - r} \leq M^{p - r} ∣ f ∣^{r}$ almost everywhere. Thus $f \in L^{p} (μ)$ and $∥ f ∥_{p} \leq M^{1 - r / p} ∥ f ∥_{r}^{r / p} .$ In particular, ${lim sup}_{p \to \infty} ∥ f ∥_{p} \leq M .$

2.1L1L3step 1.2given

Fix $ε$ with $0 < ε < M$ . Because $M$ is the least essential bound, the set $E_{ε} : = {∣ f ∣ > M - ε}$ has positive measure. For every finite $p \geq r$ , step 1.2 gives $∥ f ∥_{p} < \infty$ , so $∥ f ∥_{p}^{p} = \int ∣ f ∣^{p} d μ \geq \int_{E_{ε}} ∣ f ∣^{p} d μ \geq (M - ε)^{p} μ (E_{ε})$ forces $μ (E_{ε}) < \infty$ . Therefore $∥ f ∥_{p} \geq (M - ε) μ (E_{ε})^{1 / p},$ and letting $p \to \infty$ gives ${lim inf}_{p \to \infty} ∥ f ∥_{p} \geq M - ε .$

3.1step 1.1step 1.2step 2.1∎

Because $0 < ε < M$ was arbitrary in the case $M > 0$ , step 2.1 yields ${lim inf}_{p \to \infty} ∥ f ∥_{p} \geq M$ . Combined with step 1.2, this proves $∥ f ∥_{p} \to M$ , while step 1.1 already handled the case $M = 0$ .

TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-31Open item page →

The parallelogram law in $L^{2}$

Statement

For $u, v \in L^{2} (μ)$ one has

$∥ u + v ∥_{2}^{2} + ∥ u - v ∥_{2}^{2} = 2 ∥ u ∥_{2}^{2} + 2 ∥ v ∥_{2}^{2} .$

Facts & Assumptions

Given: Classes $u, v \in L^{2} (μ)$ with measurable representatives $f, g$ .

[L1]

The $L^{2}$ norm is well defined on quotient classes (The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$ ).

[L2]

Products of two $L^{2}$ functions lie in $L^{1}$ (Generalized Holder inequality puts products into $L^{r}$ ).

[L3]

The Lebesgue integral is linear on $L^{1}$ (The Lebesgue integral is linear on $L^{1} (μ)$ ).

Proof

technique · Expand $|f + g|^2 + |f - g|^2$ pointwise to $2|f|^2 + 2|g|^2$ and integrate

1.1L2L3givenalgebra

Because $f, g \in L^{2} (μ)$ , step [L2] puts $f g$ in $L^{1} (μ)$ , so every term [L2, L3, given, algebra] in the algebraic expansions below is integrable. Pointwise, $∣ f + g ∣^{2} + ∣ f - g ∣^{2} = (f + g)^{2} + (f - g)^{2} = 2 f^{2} + 2 g^{2} .$ Integrating and using [L3] gives $\int ∣ f + g ∣^{2} d μ + \int ∣ f - g ∣^{2} d μ = 2 \int ∣ f ∣^{2} d μ + 2 \int ∣ g ∣^{2} d μ .$

2.1step 1.1L1∎

Rewriting the four integrals as $∥ u + v ∥_{2}^{2}$ , $∥ u - v ∥_{2}^{2}$ , [step 1.1, L1] $∥ u ∥_{2}^{2}$ , and $∥ v ∥_{2}^{2}$ is legitimate by [L1]. That yields the parallelogram identity.

PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31Open item page →

The $p$ -functional need not be a norm for $0 < p < 1$

Statement

Let $0 < p < 1$ . Then the functional

$[f] ⟼ {(\int ∣ f ∣^{p} d μ)}^{1 / p}$

on $L^{p} (μ)$ need not satisfy the triangle inequality. Consequently it is not a norm in general in the sense of A norm on a real vector space, the induced metric, and the dictionary with the metric axioms.

Facts & Assumptions

Given: A real exponent $0 < p < 1$ .

[L1]

Real powers are defined for positive bases and obey the exponent laws; for fixed base $2 > 1$ , the map $x \mapsto 2^{x}$ is strictly increasing (Real powers for positive bases, with the zero-base positive-exponent convention, The exponent, product, quotient, and iterated-power laws for positive real bases and real exponents, The exponential function is strictly increasing).

[L2]

A norm must satisfy the triangle inequality (A norm on a real vector space, the induced metric, and the dictionary with the metric axioms).

[L3]

Counting measure on a two-point set is a measure (Counting measure on an arbitrary set, Counting measure is a measure).

Proof

technique · Use two disjoint equal-mass indicators, so the triangle inequality becomes the scalar inequality $2^{1/p}\le2$, which fails because $1/p>1$

1.1L3given

Work on the two-point counting space ${0, 1}$ from [L3]. Let [L3, given] $e_{0} : = χ_{{0}}$ and $e_{1} : = χ_{{1}}$ . Then $∥ e_{0} ∥_{p} = ∥ e_{1} ∥_{p} = 1, ∥ e_{0} + e_{1} ∥_{p} = (1 + 1)^{1 / p} = 2^{1 / p} .$

2.1L1step 1.1

Because $0 < p < 1$ , one has $1 / p > 1$ . Strict monotonicity in [L1] therefore [L1, step 1.1] gives $2^{1 / p} > 2^{1} = 2.$ So $∥ e_{0} + e_{1} ∥_{p} > ∥ e_{0} ∥_{p} + ∥ e_{1} ∥_{p} .$

3.1L2step 2.1∎

The triangle inequality from [L2] fails on this concrete measure space, so [L2, step 2.1] the $p$ -functional is not a norm in general for $0 < p < 1$ .

TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-09-24 (gpt-6-sol)Open item page →

The $L^{p}$ distance for $0 < p < 1$ is a complete translation-invariant metric

Statement

Assume the Axiom of Choice. Let $0 < p < 1$ and let $L^{p} (μ)$ denote the set of almost-everywhere classes of functions in $L^{p} (μ)$ . Define

$d_{p} ([f], [g]) : = \int ∣ f - g ∣^{p} d μ .$

Then $d_{p}$ is a translation-invariant metric on $L^{p} (μ)$ , and $(L^{p} (μ), d_{p})$ is complete.

Facts & Assumptions

Given: The Axiom of Choice, a measure space $(X, A, μ)$ and an exponent $0 < p < 1$ .

[A1]

The countable instance of the Axiom of Choice selects representatives of a countable sequence of nonempty almost-everywhere equivalence classes (The Axiom of Choice).

[L1]

The class notation $L^{p} (μ)$ means almost-everywhere equivalence classes of $L^{p} (μ)$ representatives (The space $L^{p} (μ)$ as the quotient by null functions).

[L2]

A metric and a complete metric space are defined in Metric space: $d (x, y) = 0$ iff $x = y$ , symmetry, and the triangle inequality; pseudometric and ultrametric and Complete metric space: every Cauchy sequence converges in the space.

[L3]

Monotone convergence, dominated convergence, and measurability of pointwise limits are available (Monotone convergence for the integral, Dominated convergence, Sequential suprema, infima, limsup, liminf, and pointwise limits of measurable functions are measurable).

[L4]

Zero nonnegative integral means zero almost everywhere (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere).

[L5]

Sums, scalar multiples, absolute values, and pointwise limits of measurable functions are measurable (Closure properties of measurable functions used by the integral).

[L6]

Countable unions of measurable null sets are measurable and null (Finite and countable subadditivity of measures).

[L7]

For $0 < p < 1$ and nonnegative reals $a, b$ , $(a + b)^{p} \leq a^{p} + b^{p} .$ Indeed, if $a + b = 0$ the claim is trivial. Otherwise set $s : = a + b > 0$ , $u : = a / s$ , and $v : = b / s$ . Then $u, v \in [0, 1]$ and $u + v = 1$ . If $0 < u \leq 1$ , then $\log u \leq 0$ , so $p \log u \geq \log u$ because $0 < p < 1$ ; strict increase of the exponential and the definition of real power therefore give $u^{p} \geq u$ . The same holds for $v$ . Hence $a^{p} + b^{p} = s^{p} (u^{p} + v^{p}) \geq s^{p} (u + v) = s^{p} = (a + b)^{p} .$ (Real powers for positive bases, with the zero-base positive-exponent convention, Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm, The exponential function is strictly increasing) [given]

Proof

technique · Because $(a+b)^p\le a^p+b^p$ for $0<p<1$, $d_p([f],[g])=\int|f-g|^p\,d\mu$ defines a metric on quotient classes and is translation invariant. Completeness follows by repeating the Riesz-Fischer telescoping argument without taking $p$-th roots

1.1L1L5L6L7

If $f \sim f^{'}$ and $g \sim g^{'}$ , then $f - f^{'}$ and $g - g^{'}$ vanish almost everywhere. Outside the union of those two null sets, one has $f - g = f^{'} - g^{'}$ , so $∣ f - g ∣^{p} = ∣ f^{'} - g^{'} ∣^{p}$ almost everywhere. Hence $d_{p} ([f], [g])$ is well defined. The same union-of-null-sets argument shows that addition and scalar multiplication descend to the quotient classes, and the inequality in [L7] shows that $L^{p} (μ)$ is closed under those operations.

1.2L7L2L4

Symmetry of $d_{p}$ is immediate. If $d_{p} ([f], [g]) = 0$ , then $\int ∣ f - g ∣^{p} d μ = 0$ , so $f = g$ almost everywhere and hence $[f] = [g]$ . For the triangle inequality, the pointwise inequality from [L7] gives [L2, L4, L7] $∣ f - h ∣^{p} = ∣ (f - g) + (g - h) ∣^{p} \leq ∣ f - g ∣^{p} + ∣ g - h ∣^{p},$ and integrating yields $d_{p} ([f], [h]) \leq d_{p} ([f], [g]) + d_{p} ([g], [h]) .$ Thus [L2] makes $d_{p}$ a metric.

1.3L3L5givenchooseA1

Let $(u_{n})$ be Cauchy in $d_{p}$ . Choose by least indices a subsequence $(u_{n_{k}})$ with [L3, L5, given, choose] $d_{p} (u_{n_{k + 1}}, u_{n_{k}}) < 2^{- k} .$ By [A1], choose a sequence of representatives $f_{k}$ of $u_{n_{k}}$ and define $h_{k} : = ∣ f_{k + 1} - f_{k} ∣^{p}, g_{m} : = \sum_{j < m} h_{j} .$ Each $h_{k}$ is measurable and integrable, and [L3] gives a measurable pointwise limit $g : = \sum_{j = 0}^{\infty} h_{j}$ with $\int g d μ = \lim_{m \to \infty} \int g_{m} d μ \leq \sum_{j = 0}^{\infty} 2^{- j} < \infty .$ Hence $g < \infty$ almost everywhere. This is the only use of Choice in the argument.

2.1step 1.1

Translation invariance is pointwise: [step 1.1] $d_{p} ([f] + [u], [g] + [u]) = \int ∣ (f + u) - (g + u) ∣^{p} d μ = d_{p} ([f], [g]) .$

2.2step 1.3L3L7L5

Put $E : = {x : g (x) = \infty}$ , a measurable null set by step 1.3. Fix $x \notin E$ . Then $\sum_{j} ∣ f_{j + 1} (x) - f_{j} (x) ∣^{p} < \infty$ , so the terms tend to $0$ . Thus $∣ f_{j + 1} (x) - f_{j} (x) ∣ \leq 1$ for all large $j$ , and then [step 1.3, L3, L7] $∣ f_{j + 1} (x) - f_{j} (x) ∣ \leq ∣ f_{j + 1} (x) - f_{j} (x) ∣^{p} = h_{j} (x) .$ So the real series $\sum_{j} ∣ f_{j + 1} (x) - f_{j} (x) ∣$ converges by comparison with $\sum_{j} h_{j} (x)$ , which makes $(f_{k} (x))$ converge to some real value $f (x)$ . Define $f = 0$ on $E$ . The measurable functions equal to $f_{k}$ on $X ∖ E$ and zero on $E$ converge pointwise to $f$ , so [L3] and [L5] make $f$ measurable. Moreover, $∣ f ∣^{p} \leq ∣ f_{0} ∣^{p} + g$ on $X ∖ E$ by [L7], and $f = 0$ on $E$ . Since $f_{0} \in L^{p} (μ)$ and $g$ is integrable, $f \in L^{p} (μ)$ , so $[f]$ is a valid element of $L^{p} (μ)$ . Also $∣ f (x) - f_{k} (x) ∣^{p} \leq {(\sum_{j \geq k} ∣ f_{j + 1} (x) - f_{j} (x) ∣)}^{p} \leq \sum_{j \geq k} h_{j} (x) .$ Integrating and using monotone convergence on the tails yields $d_{p} (u_{n_{k}}, [f]) \leq \sum_{j \geq k} d_{p} (u_{n_{j + 1}}, u_{n_{j}}) \leq \sum_{j \geq k} 2^{- j} \to 0.$

3.1step 1.2step 2.2

Because $(u_{n})$ is Cauchy, given $ε > 0$ choose $K$ with $d_{p} (u_{n}, u_{m}) < ε / 2$ for $m, n \geq K$ , then choose $k$ with $n_{k} \geq K$ and $d_{p} (u_{n_{k}}, [f]) < ε / 2$ from step 2.2. The triangle inequality from step 1.2 gives $d_{p} (u_{n}, [f]) < ε$ for all $n \geq K$ . Hence $(L^{p} (μ), d_{p})$ is complete.

4.1step 1.2step 2.1step 3.1∎

Steps 1.2 and 2.1 prove that $d_{p}$ is a translation-invariant metric, and step 3.1 proves completeness.

TheoremStatement: AI-adaptedProof: AI-generatedprecheck passverified 2026-09-24 (gpt-6-sol)Open item page →

The $p$ -power triangle inequality for nonnegative functions when $0 < p < 1$

Statement

Let $0 < p < 1$ and let $f, g \in L^{p} (μ)$ be nonnegative. Then

$∥ f + g ∥_{p}^{p} \leq ∥ f ∥_{p}^{p} + ∥ g ∥_{p}^{p} .$

Equivalently,

$\int (f + g)^{p} d μ \leq \int f^{p} d μ + \int g^{p} d μ .$

Facts & Assumptions

Given: An exponent $0 < p < 1$ and nonnegative functions $f, g \in L^{p} (μ)$ .

[L1]

For $0 < p < 1$ and $0 < u \leq 1$ , $p \log u \geq \log u$ , hence $u^{p} \geq u$ by the definition of real power and strict increase of the exponential. The same inequality holds at $u = 0$ (Real powers for positive bases, with the zero-base positive-exponent convention, Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm, The exponential function is strictly increasing).

[L2]

The nonnegative integral is monotone and additive (Monotonicity and nonnegative homogeneity of the nonnegative integral, Additivity of the nonnegative Lebesgue integral).

Proof

technique · For nonnegative numbers $a,b$ and $0<p<1$ one has $(a+b)^p\le a^p+b^p$. Apply this pointwise to $f$ and $g$ and integrate

1.1L1givenalgebra

For nonnegative reals $a, b$ with $a + b > 0$ , set $s = a + b$ , $u = a / s$ and $v = b / s$ . Then $u, v \in [0, 1]$ , $u + v = 1$ , and [L1] gives $a^{p} + b^{p} = s^{p} (u^{p} + v^{p}) \geq s^{p} = (a + b)^{p}$ ; for $a + b = 0$ equality is immediate. Apply this scalar inequality pointwise to $f, g$ to obtain [L1, given, algebra] $(f + g)^{p} \leq f^{p} + g^{p} .$

2.1L2step 1.1∎

Integrating and using monotonicity and additivity from [L2] yields [step 1.1, L2] $\int (f + g)^{p} d μ \leq \int f^{p} d μ + \int g^{p} d μ .$ This is exactly the displayed $∥ \cdot ∥_{p}^{p}$ inequality.

RemarkRemark: AI-adaptedProof: Not applicablejudge pass (gpt-5.6-terra)audited 2026-08-31Open item page →

$L^{p}$ completeness and the Banach-property wording

Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$ proves the mathematics this page needs: with the descended norm, $L^{p} (μ)$ is complete for $1 \leq p \leq \infty$ . Later functional-analysis pages supply the vocabulary seam by naming a complete normed space a Banach space. The terminology is deferred; the completeness proof is not.

5 · Examples, counterexamples and false statements

None yet.

The LpLp Spaces Holder Minkowski and Riesz Fischer

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Conjugate exponents, including the endpoint conventions

Definition

The essential supremum of a measurable function with respect to a measure

Definition

The function space Lp(μ)Lp(μ) for 0<p<∞0<p<∞

Definition

The space L∞(μ)L∞(μ) of essentially bounded measurable functions

Definition

The null subspace of measurable functions that vanish almost everywhere

Definition

The space Lp(μ)Lp(μ) as the quotient by null functions

Definition

Elements of LpLp are equivalence classes, so pointwise statements require a representative

ℓpℓp is the LpLp space of counting measure

Remark

Holder's inequality for integrals, including the endpoint cases

Statement

Facts & Assumptions

Proof

Equality in Holder's inequality for 1<p<∞1<p<∞

Statement

Facts & Assumptions

Proof

Generalized Holder inequality puts products into LrLr

Statement

Facts & Assumptions

Proof

Cauchy-Schwarz inequality for L2L2

Statement

Facts & Assumptions

Proof

Finite counting measure recovers finite Holder and implies the signed Cauchy-Schwarz inequality

Minkowski's inequality for integrals, including p=∞p=∞

Statement

Facts & Assumptions

Proof

Equality in Minkowski's inequality for 1<p<∞1<p<∞

Statement

Facts & Assumptions

Proof

On a finite counting space, Minkowski agrees with the published finite theorem for p>1p>1

The essential supremum is attained as the least essential bound

Statement

Facts & Assumptions

Proof

LpLp and L∞L∞ are vector spaces for p≥1p≥1

Statement

Facts & Assumptions

Proof

Null functions form a linear subspace and are exactly the zero-seminorm class

Statement

Facts & Assumptions

Proof

The LpLp norm descends to the quotient and makes LpLp a normed space for 1≤p≤∞1≤p≤∞

Statement

Facts & Assumptions

Proof

Riesz-Fischer completeness of LpLp for 1≤p≤∞1≤p≤∞

Statement

Facts & Assumptions

Proof

Assuming Countable Choice, LpLp-convergent sequences have almost-everywhere convergent subsequences

Statement

Facts & Assumptions

Proof

Convergence in LpLp implies convergence in measure

Statement

Facts & Assumptions

Proof

Finite-measure LrLr includes into LpLp for p<rp<r

Statement

Facts & Assumptions

Proof

ℓpℓp includes into ℓrℓr for p<rp<r

Statement

The $L^{p}$ Spaces Holder Minkowski and Riesz Fischer

The function space $L^{p} (μ)$ for $0 < p < \infty$

The space $L^{\infty} (μ)$ of essentially bounded measurable functions

The space $L^{p} (μ)$ as the quotient by null functions

Elements of $L^{p}$ are equivalence classes, so pointwise statements require a representative

$ℓ^{p}$ is the $L^{p}$ space of counting measure

Equality in Holder's inequality for $1 < p < \infty$

Generalized Holder inequality puts products into $L^{r}$

Cauchy-Schwarz inequality for $L^{2}$

Minkowski's inequality for integrals, including $p = \infty$

Equality in Minkowski's inequality for $1 < p < \infty$

On a finite counting space, Minkowski agrees with the published finite theorem for $p > 1$

$L^{p}$ and $L^{\infty}$ are vector spaces for $p \geq 1$

The $L^{p}$ norm descends to the quotient and makes $L^{p}$ a normed space for $1 \leq p \leq \infty$

Riesz-Fischer completeness of $L^{p}$ for $1 \leq p \leq \infty$

Assuming Countable Choice, $L^{p}$ -convergent sequences have almost-everywhere convergent subsequences

Convergence in $L^{p}$ implies convergence in measure

Finite-measure $L^{r}$ includes into $L^{p}$ for $p < r$

$ℓ^{p}$ includes into $ℓ^{r}$ for $p < r$

Lyapunov interpolation inequality for $L^{p}$ norms

$L^{p}$ norms converge to the essential supremum for essentially bounded $L^{r}$ functions

The parallelogram law in $L^{2}$

The $p$ -functional need not be a norm for $0 < p < 1$

The $L^{p}$ distance for $0 < p < 1$ is a complete translation-invariant metric

The $p$ -power triangle inequality for nonnegative functions when $0 < p < 1$

$L^{p}$ completeness and the Banach-property wording