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Riesz Potentials and the Hardy–Littlewood–Sobolev Inequality

1 · Prerequisites

2 · Summary

This page develops the unit-normalized Riesz potential $I_{α} f (x) = \int_{R^{n}} ∣ x - y ∣^{α - n} f (y) d y (0 < α < n)$ on complex Euclidean Lebesgue spaces and proves the strict-range Hardy–Littlewood–Sobolev fractional integration theorem. The kernel is the unit normalization $c_{n, α} = 1$ used by the cited sources; no Fourier multiplier identity is asserted for $I_{α}$ .

The definition fixes exactly where the pointwise integral is absolutely meaningful and claims no all- $L^{p}$ existence. The near/far lemma then splits the kernel at a radius $R$ : the near part is controlled by $R^{α} M f (x)$ through the centered maximal function, while the far part is controlled by Hölder's inequality and the polar-coordinate computation of the radial weight, whose finiteness is exactly the strict condition $p < n / α$ ; the same lemma proves local integrability of $L^{p}$ representatives and independence of the measurable representative at every convergent point. Hedberg's pointwise inequality balances the two bounds at $R = (∥ f ∥_{p} / M f (x))^{p / n}$ and gives $∣ I_{α} f (x) ∣ \leq C (M f (x))^{1 - θ} ∥ f ∥_{p}^{θ}$ with $θ = α p / n$ .

The theorem takes $1 / q = 1 / p - α / n$ , so $q > p$ , and proves that the defining integral converges absolutely almost everywhere for every complex $L^{p}$ input, that the resulting classes form a bounded linear map $L^{p} \to L^{q}$ with $∥ I_{α} f ∥_{q} \leq C_{n, α, p} ∥ f ∥_{p}$ , and that on the dense smooth core the map is the pointwise integral, whose unique bounded dense-core extension is the same almost-everywhere integral operator. Countable Choice is declared on every item consuming the maximal-function, Tonelli, polar, measurability, density, completeness or extension interfaces, and the endpoint remark is recorded, not proved, and supplies no argument.

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Riesz potential of order alpha

Definition

Assume the Axiom of Countable Choice for the Euclidean Lebesgue framework (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Fix an integer $n \geq 1$ and a real order $0 < α < n$ , and work on $R^{n}$ with Lebesgue measure $λ_{n}$ on the Lebesgue sigma-algebra (Lebesgue measurable sets, the family $L (R^{n})$ , and the restricted set function $λ_{n}$ ).

The kernel. Put $K_{α} (z) : = ∣ z ∣^{α - n} (z \neq 0), K_{α} (0) : = 0.$ Since $α - n < 0$ , the kernel is strictly positive and continuous on $R^{n} ∖ {0}$ and has a singularity at the origin. The assigned value $K_{α} (0) = 0$ is a normalization convention: it changes the integrand $R^{n} ∋ y \mapsto K_{α} (x - y) f (y)$ only at the single point $y = x$ , which is the diagonal point of the domain; every statement in this pair is unchanged if another finite value is assigned instead, and the value at the origin is never used as a bound on the kernel everywhere.

The potential. Let $f : R^{n} \to C$ be measurable, with real and imaginary parts measurable in the sense of Integrable real and complex functions, and their integrals. The Riesz potential of order $α$ of $f$ is defined at a point $x \in R^{n}$ precisely when the nonnegative integral is finite, $\int_{R^{n}} K_{α} (x - y) ∣ f (y) ∣ d λ_{n} (y) < \infty,$ and at every such point it is $I_{α} f (x) : = \int_{R^{n}} K_{α} (x - y) f (y) d λ_{n} (y) .$ The integral displayed in the definition is the Lebesgue integral of the complex function $y \mapsto K_{α} (x - y) f (y)$ , which is absolutely convergent exactly at the points where the first display holds; the value $I_{α} f (x)$ is then a complex number.

Scope of the definition. The set of points at which $I_{α} f$ is defined may be empty, all of $R^{n}$ , or anything in between, and this definition asserts nothing about which case occurs: in particular it makes no claim that $I_{α} f$ exists at every point, at almost every point, or for every $f$ belonging to any Lebesgue space. The strict-range theorem of this pair proves almost-everywhere absolute existence for every $f \in L^{p} (R^{n}; C)$ when $1 < p < n / α$ .

Normalization. The unit normalization is fixed once and for all by $c_{n, α} = 1$ in $I_{α} f (x) = c_{n, α} \int K_{α} (x - y) f (y) d y$ , the convention of both cited sources. A different positive constant rescales every potential and every estimate of this pair by that constant; no constant carrying a Fourier multiplier identity is asserted, and the identification of $I_{α}$ with a power of $- Δ$ is not used here.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-10-02Open item page →

Near and far bounds for a Riesz potential

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ , $0 < α < n$ , $1 < p < n / α$ , and let $f$ be an element of the complex Lebesgue space $L^{p} (R^{n}; C)$ (Complex Lp classes and Euclidean test-function conventions). Every measurable representative of $f$ is locally integrable (A locally integrable function on $R^{n}$ ).

At every $x$ with $M f (x) < \infty$ , where $M$ is the centered Hardy-Littlewood maximal operator (The centered and uncentered Hardy-Littlewood maximal functions), and for every $R > 0$ , the two absolute integrals $N_{R} (x) : = \int_{∣ x - y ∣ < R} ∣ x - y ∣^{α - n} ∣ f (y) ∣ d y, F_{R} (x) : = \int_{∣ x - y ∣ \geq R} ∣ x - y ∣^{α - n} ∣ f (y) ∣ d y$ are finite and satisfy $N_{R} (x) \leq C_{n, α} R^{α} M f (x), F_{R} (x) \leq C_{n, α, p} R^{α - n / p} ∥ f ∥_{p} .$ The far bound holds at every $x \in R^{n}$ . Each convergent integral is independent of the measurable representative of the class $f$ : if two representatives agree almost everywhere, then at every point $x$ the integrals $N_{R}$ and $F_{R}$ coincide, and at every point where both are finite the total potential $I_{α} f (x)$ of Riesz potential of order alpha is defined and likewise independent of the representative.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $0 < α < n$ , $1 < p < n / α$ , and a class $f \in L^{p} (R^{n}; C)$ with a fixed measurable representative, also written $f$ .

[F1]

The unit-normalized Riesz potential is $I_{α} f (x) = \int K_{α} (x - y) f (y) d y$ at exactly those points where $\int K_{α} (x - y) ∣ f (y) ∣ d y < \infty$ , with $K_{α} (z) = ∣ z ∣^{α - n}$ for $z \neq 0$ and $K_{α} (0) = 0$ . (Riesz potential of order alpha)

[F2]

Complex $L^{p}$ classes, the seminorm $N_{p} (g) = (\int ∣ g ∣^{p})^{1 / p}$ , the set quotient by almost-everywhere equality, the convention that a complex function is measurable when its real and imaginary parts are, and the Euclidean conventions for $C_{c}^{\infty} (R^{n}; C)$ ; local integrability means finite absolute integral over every Euclidean ball. (Complex Lp classes and Euclidean test-function conventions, A locally integrable function on $R^{n}$ )

[F3]

Under Countable Choice the centered maximal function of a locally integrable $f$ is $M f (x) = \sup_{r > 0} λ (B (x, r))^{- 1} \int_{B (x, r)} ∣ f ∣$ , with values in $[0, \infty]$ ; every finite value bounds every ball average. (The centered and uncentered Hardy-Littlewood maximal functions)

[F4]

For conjugate exponents $p, p^{'} \in (1, \infty)$ and measurable representatives $g \in L^{p} (R^{n}; C)$ , $h \in L^{p^{'}} (R^{n}; C)$ , $\int ∣ g h ∣ \leq ∥ g ∥_{p} ∥ h ∥_{p^{'}}$ and $∣ \int g h ∣ \leq ∥ g ∥_{p} ∥ h ∥_{p^{'}}$ . (Complex Holder, Minkowski, and the quotient norm)

[F5]

Every ball $B (x, r)$ is Lebesgue measurable with $0 < λ (B (x, r)) < \infty$ ; every box between its open and closed forms is Lebesgue measurable with its usual volume; Lebesgue measure is monotone on measurable sets. (Euclidean balls have positive finite Lebesgue measure, A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included, Measures are monotone)

[F6]

Under Countable Choice a $C^{1}$ diffeomorphism $T : U \to V$ satisfies $\int_{V} h d λ_{n} = \int_{U} h \circ T ∣ \det D T ∣ d λ_{n}$ for every nonnegative Lebesgue measurable $h$ ; an affine map $y \mapsto y - x$ is a $C^{1}$ diffeomorphism of $R^{n}$ with derivative the identity, whose determinant is $1$ . (A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions, $C^{k}$ Euclidean maps and diffeomorphisms, The determinant of a triangular matrix is the product of its diagonal entries)

[F7]

Under Countable Choice, polar coordinates express the integral of a nonnegative Borel function $h$ on $R^{n}$ as $\int_{0}^{\infty} \int_{S^{n - 1}} h (r ω) r^{n - 1} d σ (ω) d r$ with $σ$ a finite Borel measure on $S^{n - 1}$ . (Polar coordinates decompose Lebesgue measure into r^{n-1} dr d sigma)

[F8]

Continuous maps on Euclidean space are Borel measurable; the composition of a measurable map with a Borel measurable function of its codomain is measurable; sums, products, scalar multiples and absolute values of measurable real functions are measurable; every Borel subset of $R^{n}$ is Lebesgue measurable under Countable Choice, so a Borel measurable function into $[0, \infty]$ is Lebesgue measurable. (Continuous functions on Euclidean spaces are Borel measurable, Composition with a Borel measurable outer map preserves measurability, Arithmetic and lattice operations preserve measurability whenever they are defined, Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable, Borel measurable and Lebesgue measurable functions on $R^{n}$ )

[F9]

For nonnegative measurable functions the Lebesgue integral is additive, monotone, homogeneous for nonnegative scalars, and computes the integral of an increasing pointwise limit as the limit of the integrals; the integral of the zero function is zero. (Additivity of the nonnegative Lebesgue integral, Monotone convergence for the integral, Monotonicity and nonnegative homogeneity of the nonnegative integral)

[F10]

A nonnegative measurable function has integral zero if and only if it vanishes almost everywhere; two integrable real or complex functions that agree almost everywhere have equal integrals over every measurable set. (A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere, Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree)

[F11]

Countable Choice says that every sequence of nonempty sets has a choice function; it is the choice principle assumed by the maximal-function, polar, change-of-variables and measurability interfaces used below, and no other choice principle is invoked. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

[F12]

$B (x, r) = {y : ∥ y - x ∥_{2} < r}$ and $∥ v ∥_{2}^{2} = \sum_{i} v_{i}^{2}$ , so $∥ v ∥_{2} < r$ forces $∣ v_{i} ∣ < r$ for every coordinate; the published Euclidean metric is induced by this norm. (Open ball, closed ball and sphere in a metric space, The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ , Each $∥ \cdot ∥_{p}$ is a norm on $R^{n}$ , and the induced metrics are exactly $d_{1}$ , $d_{2}$ and $d_{\infty}$ of the published metric-spaces page)

Proof

technique · direct; split the kernel integral at $R$, bound the near part by the maximal function on dyadic shells and the far part by Hölder after translating, then transfer to representatives

1.1F2F3F4F5given

Local integrability. Let $f$ be a measurable representative of the class with $N_{p} (f) = ∥ f ∥_{p} < \infty$ , and let $B (x, r)$ be any Euclidean ball. By [F4] with the conjugate pair $p, p^{'}$ , whose second exponent is finite because $p > 1$ , $\int_{B (x, r)} ∣ f ∣ \leq ∥ f \cdot 1_{B (x, r)} ∥_{p} ∥ 1_{B (x, r)} ∥_{p^{'}} \leq ∥ f ∥_{p} λ (B (x, r))^{1 / p^{'}} < \infty,$ the last inequality using $λ (B (x, r)) < \infty$ from [F5]. Hence every measurable representative is locally integrable, and [F3] defines $M f (x) \in [0, \infty]$ at every $x$ .

1.2F3F5F12algebra

Ball bounds. For every $x$ and $r > 0$ , the inclusion $B (x, r) \subseteq x + (- r, r)^{n}$ holds by [F12], and the open cube is Lebesgue measurable with measure $(2 r)^{n}$ by [F5]; monotonicity in [F5] gives $λ (B (x, r)) \leq (2 r)^{n} .$ Combining with the averaging inequality of [F3], for every $r > 0$ , $\int_{B (x, r)} ∣ f ∣ \leq λ (B (x, r)) M f (x), λ (B (x, r))^{- 1} \int_{B (x, r)} ∣ f ∣ \leq M f (x) .$

1.3F2F8

Measurability of the integrands. Fix $x$ . The map $y \mapsto x - y$ is continuous, hence Borel measurable, and $K_{α}$ is Borel measurable because it is continuous off the origin and takes the finite value $0$ there; by the composition clause of [F8] the map $y \mapsto K_{α} (x - y)$ is Borel measurable, hence Lebesgue measurable by the Borel-containment clause of [F8]. The modulus $∣ f ∣$ and the components of $f$ are measurable by [F2], so the product clause of [F8] makes $y \mapsto K_{α} (x - y) ∣ f (y) ∣$ and each component of $y \mapsto K_{α} (x - y) f (y)$ Lebesgue measurable.

1.4F6F7F8givenalgebra

Far kernel integral. Fix $x$ and $R > 0$ , and define $h_{R} (z) : = ∣ z ∣^{(α - n) p^{'}} 1_{{∣ z ∣ \geq R}}$ . The function $z \mapsto ∣ z ∣$ is continuous and $t \mapsto t^{(α - n) p^{'}} 1_{{t \geq R}}$ is Borel, so $h_{R}$ is Borel, nonnegative and hence Lebesgue measurable by [F8]. The affine map $T (y) : = y - x$ has $T (T^{- 1})$ equal to the identity with inverse $y \mapsto y + x$ , identity derivative and determinant $1$ , so it is a $C^{1}$ diffeomorphism of $R^{n}$ by [F6]; applying the change-of-variables formula of [F6] to $h_{R}$ gives $\int_{R^{n}} ∣ y - x ∣^{(α - n) p^{'}} 1_{{∣ y - x ∣ \geq R}} d y = \int_{R^{n}} h_{R} (z) d z .$ The right-hand integral is radial and $h_{R}$ is Borel, so the polar formula [F7] computes it as $σ (S^{n - 1}) \int_{R}^{\infty} r^{(α - n) p^{'} + n - 1} d r = σ (S^{n - 1}) \frac{R^{(α - n) p^{'} + n}}{(n - α) p^{'} - n},$ where the antiderivative is evaluated at the convergent upper end because the exponent $(α - n) p^{'} + n - 1 < - 1$ ; that inequality is equivalent to $(n - α) p^{'} > n$ , which in turn is equivalent to $p < n / α$ , the hypothesis.

2.1F1F3F5F9step 1.2step 1.3algebra

Near shell estimate. Fix $x$ with $M f (x) < \infty$ and $R > 0$ . For $j \geq 0$ put $S_{j} : = {y : 2^{- j - 1} R \leq ∣ x - y ∣ < 2^{- j} R} .$ The sets $S_{j}$ are pairwise disjoint Lebesgue measurable sets with union $B (x, R) ∖ {x}$ : a point $y$ lies in exactly one shell according to the dyadic size of $∣ x - y ∣ \in (0, R)$ , and the excluded point is exactly $x$ . On $S_{j}$ one has $∣ x - y ∣ < 2^{- j} R$ , so the ball $B (x, 2^{- j} R)$ contains $S_{j}$ , and since $α - n < 0$ reverses the inequality at the positive lower endpoint $2^{- j - 1} R \leq ∣ x - y ∣$ , $K_{α} (x - y) = ∣ x - y ∣^{α - n} \leq (2^{- j - 1} R)^{α - n} (y \in S_{j}) .$ Therefore, by monotonicity and the averaging bound of step 1.2, $\int_{S_{j}} K_{α} (x - y) ∣ f (y) ∣ d y \leq (2^{- j - 1} R)^{α - n} \int_{B (x, 2^{- j} R)} ∣ f ∣ \leq (2^{- j - 1} R)^{α - n} (2 \cdot 2^{- j} R)^{n} M f (x) = 2^{2 n - α} R^{α} 2^{- j α} M f (x) .$

2.2F4step 1.4givenalgebra

Far bound. The $p^{'}$ -th root of the value in step 1.4 is $C^{'} R^{α - n / p}, C^{'} : = σ (S^{n - 1})^{1 / p^{'}} ((n - α) p^{'} - n)^{- 1 / p^{'}},$ using $(α - n) p^{'} + n = (α - n + n / p^{'}) p^{'}$ and $α - n + n / p^{'} = α - n / p$ ; the constant $C^{'}$ is finite and positive because $σ (S^{n - 1})$ is finite by [F7] and $(n - α) p^{'} - n > 0$ . Hence $F_{R} (x) = \int_{R^{n}} ∣ y - x ∣^{α - n} 1_{{∣ y - x ∣ \geq R}} ∣ f (y) ∣ d y \leq (\int_{R^{n}} (∣ y - x ∣^{α - n} 1_{{∣ y - x ∣ \geq R}})^{p^{'}} d y)^{1 / p^{'}} ∥ f ∥_{p} \leq C^{'} R^{α - n / p} ∥ f ∥_{p} < \infty,$ where the first inequality is Hölder [F4] applied to the pair $∣ f ∣$ and the radial weight, and the weight's exact $L^{p^{'}}$ norm is the quantity computed in step 1.4. The estimate uses no hypothesis on $M f (x)$ , so it holds at every $x$ and every $R > 0$ , and in particular proves finiteness of $F_{R} (x)$ everywhere.

2.3F9F10step 1.3

Representative independence of the absolute integrals. Let $f$ and $g$ be measurable representatives of the same class, so that $f = g$ almost everywhere, and fix $x$ . The two nonnegative measurable integrands $h_{f} : = K_{α} (x - \cdot) ∣ f ∣$ and $h_{g} : = K_{α} (x - \cdot) ∣ g ∣$ of step 1.3 are equal off the null set ${f \neq g}$ ; hence $∣ h_{f} - h_{g} ∣ = 0$ almost everywhere and [F10] gives $\int ∣ h_{f} - h_{g} ∣ = 0$ . Since $h_{f} \leq h_{g} + ∣ h_{f} - h_{g} ∣$ and $h_{g} \leq h_{f} + ∣ h_{f} - h_{g} ∣$ pointwise, additivity and monotonicity in [F9] give $\int h_{f} \leq \int h_{g} + 0, \int h_{g} \leq \int h_{f} + 0,$ so the two extended nonnegative integrals are equal; in particular one is finite if and only if the other is. Applying this to the restrictions ${∣ x - y ∣ < R}$ and ${∣ x - y ∣ \geq R}$ (each restriction has the same form with the additional indicator) shows that $N_{R} (x)$ and $F_{R} (x)$ are independent of the representative, as functions of $x$ .

3.1F1F9step 2.1algebra

Near bound and finiteness. Since $K_{α} (0) = 0$ by [F1], the integrand $K_{α} (x - \cdot) ∣ f ∣ \cdot 1_{{x}}$ vanishes identically, so the pointwise identity $K_{α} (x - y) ∣ f (y) ∣ 1_{B (x, R)} (y) = \sum_{j \geq 0} K_{α} (x - y) ∣ f (y) ∣ 1_{S_{j}} (y)$ holds; the partial sums increase to the left-hand side, so [F9] (additivity followed by monotone convergence) gives $N_{R} (x) = \int_{B (x, R)} K_{α} (x - \cdot) ∣ f ∣ = \sum_{j \geq 0} \int_{S_{j}} K_{α} (x - \cdot) ∣ f ∣ \leq 2^{2 n - α} R^{α} M f (x) \sum_{j \geq 0} 2^{- j α} = \frac{2^{2 n - α}}{1 - 2^{- α}} R^{α} M f (x) < \infty,$ because $α > 0$ makes the geometric series converge and $M f (x) < \infty$ . This proves the near bound and the finiteness of $N_{R} (x)$ with $C_{n, α} = 2^{2 n - α} / (1 - 2^{- α})$ , a constant depending only on $n$ and $α$ .

3.2F1F9F10step 2.3

Representative independence of the complex integrals and of the potential. Keep the notation of step 2.3 and suppose now that the common absolute integral is finite at $x$ . Then the complex functions $K_{α} (x - \cdot) f$ and $K_{α} (x - \cdot) g$ are both integrable, and they agree almost everywhere, so [F10] applied to their real and imaginary parts gives $\int K_{α} (x - \cdot) f = \int K_{α} (x - \cdot) g$ . Consequently the convergence set of the defining integral and its value at every convergent point depend only on the class $f$ , and if $N_{R} (x)$ and $F_{R} (x)$ are both finite, then additivity in [F9] applied on the complementary measurable sets $B (x, R)$ and its complement gives $\int K_{α} (x - \cdot) ∣ f ∣ = N_{R} (x) + F_{R} (x) < \infty,$ so $I_{α} f (x)$ is defined by [F1] and is representative-independent.

4.1F11step 1.1step 3.1step 2.2step 2.3step 3.2∎

Conclusion. The local-integrability assertion is step 1.1; the finiteness and the bound for $N_{R}$ are step 3.1; the finiteness and the bound for $F_{R}$ at every point are step 2.2; and representative independence of the convergent integrals and of the total potential is steps 2.3 and 3.2. Countable Choice is used exactly through the maximal-function interface [F3], the ball, box, change-of-variables and polar interfaces [F5]-[F7], the measurability interfaces [F8] and the integral-lattice facts [F9]-[F10], all of which are stated under Countable Choice; no full Axiom of Choice is used. No estimate is asserted at a point with $M f (x) = \infty$ , and no endpoint case $p = 1$ or $p = n / α$ is claimed.

LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Hedberg pointwise inequality for Riesz potentials

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ , $0 < α < n$ , $1 < p < n / α$ and put $θ : = α p / n \in (0, 1)$ . Let $f$ be an element of $L^{p} (R^{n}; C)$ (Complex Lp classes and Euclidean test-function conventions) and let $x$ be a point with $M f (x) < \infty$ , where $M$ denotes the centered Hardy-Littlewood maximal operator (The centered and uncentered Hardy-Littlewood maximal functions). Then the defining integral of the Riesz potential (Riesz potential of order alpha) converges absolutely at $x$ and $∣ I_{α} f (x) ∣ \leq C_{n, α, p} (M f (x))^{1 - θ} ∥ f ∥_{p}^{θ} .$ If $∥ f ∥_{p} = 0$ , or if $M f (x) = 0$ , then $f = 0$ almost everywhere, $M f = 0$ , and $I_{α} f (x) = 0$ at the stated point, so no zero or infinity power with an undefined value is used: only the exact powers $θ \in (0, 1)$ and $1 - θ \in (0, 1)$ of the finite nonnegative numbers $M f (x)$ and $∥ f ∥_{p}$ occur.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $0 < α < n$ , $1 < p < n / α$ , $θ = α p / n$ , a class $f \in L^{p} (R^{n}; C)$ with a fixed measurable representative, and a point $x$ with $M f (x) < \infty$ .

[F1]

The unit Riesz potential is $I_{α} f (x) = \int K_{α} (x - y) f (y) d y$ at every point where $\int K_{α} (x - y) ∣ f (y) ∣ d y < \infty$ , with $K_{α} (z) = ∣ z ∣^{α - n}$ for $z \neq 0$ and $K_{α} (0) = 0$ . (Riesz potential of order alpha)

[F2]

Complex $L^{p}$ classes are quotients by almost-everywhere equality, with norm $N_{p} (g) = (\int ∣ g ∣^{p})^{1 / p}$ , and $∥ g ∥_{p} = 0$ exactly for the zero class; local integrability of the representatives is a consequence of $L^{p}$ membership for finite $p$ and finite measure balls. (Complex Lp classes and Euclidean test-function conventions, A locally integrable function on $R^{n}$ , Complex Holder, Minkowski, and the quotient norm)

[F3]

The centered maximal function of a locally integrable function satisfies $M f (x) = \sup_{r > 0} λ (B (x, r))^{- 1} \int_{B (x, r)} ∣ f ∣$ with values in $[0, \infty]$ , so every ball average is at most $M f (x)$ . (The centered and uncentered Hardy-Littlewood maximal functions)

[F4]

Near and far splitting: at every point $y$ with $M f (y) < \infty$ and every $R > 0$ , the near and far integrals $N_{R} (y), F_{R} (y)$ are finite, $N_{R} (y) \leq C_{n, α} R^{α} M f (y)$ , $F_{R} (y) \leq C_{n, α, p} R^{α - n / p} ∥ f ∥_{p}$ , the far bound holds everywhere, and where both are finite the potential $I_{α} f (y)$ is defined by the total absolute integral and depends only on the class $f$ . (Near and far bounds for a Riesz potential)

[F5]

For integrable complex $g$ , $∣ \int g d μ ∣ \leq \int ∣ g ∣ d μ$ . (The modulus of an integral is bounded by the integral of the modulus)

[F6]

The nonnegative Lebesgue integral is additive over complementary measurable sets, monotone, and homogeneous for nonnegative scalars; a nonnegative measurable function has integral zero if and only if it vanishes almost everywhere. (Additivity of the nonnegative Lebesgue integral, Monotonicity and nonnegative homogeneity of the nonnegative integral, A nonnegative measurable function has integral $0$ exactly when it vanishes almost everywhere)

[F7]

A countable union of Lebesgue null sets is null, so a function vanishing almost everywhere on every ball $B (x, k)$ , $k \geq 1$ , vanishes almost everywhere on $R^{n}$ . (Finite and countable subadditivity of measures)

[F8]

Countable Choice is the choice principle assumed by the maximal-function and splitting interfaces used here. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

Proof

technique · direct; dispose of the degenerate identically-zero cases, then balance the near and far bounds of the splitting lemma at the optimal radius

1.1F1F2F3F6givenalgebra

The degenerate case. Suppose first that $∥ f ∥_{p} = 0$ . Then $f = 0$ almost everywhere by the definiteness clause of [F2]; for every $x$ the nonnegative integrand $K_{α} (x - \cdot) ∣ f ∣$ vanishes almost everywhere, so [F6] gives $\int K_{α} (x - y) ∣ f (y) ∣ d y = 0 < \infty$ , the potential is defined at $x$ by [F1] and $I_{α} f (x) = 0$ . Also every ball average in [F3] is the integral of a function vanishing almost everywhere, hence is $0$ , so $M f (x) = 0$ , and the asserted inequality reads $0 \leq C_{n, α, p} \cdot 0^{1 - θ} \cdot 0^{θ} = 0$ .

2.1F2F3F6F7step 1.1

The case $M f (x) = 0$ . If $M f (x) = 0$ , then every ball average of $∣ f ∣$ is at most $0$ , so $\int_{B (x, k)} ∣ f ∣ = 0$ for every integer $k \geq 1$ ; the nonnegative function $∣ f ∣$ therefore vanishes almost everywhere on each ball $B (x, k)$ by [F6], and the balls $B (x, k)$ cover $R^{n}$ , so $∣ f ∣ = 0$ almost everywhere by [F7]. Hence $f$ is the zero class, $∥ f ∥_{p} = 0$ , and step 1.1 applies. Thus in the remaining case both $M : = M f (x)$ and $F : = ∥ f ∥_{p}$ are strictly positive, and both are finite by hypothesis and by [F2].

3.1F4F5F6step 2.1algebra

The balanced estimate. Assume $0 < M < \infty$ and $0 < F < \infty$ and put $R : = (F / M)^{p / n} > 0$ . Step 2.1 and [F2] give every representative $f$ locally integrable, so $M$ is defined and the splitting lemma [F4] applies at $x$ with this radius: $I_{α} f (x)$ is defined and $∣ I_{α} f (x) ∣ \leq \int K_{α} (x - y) ∣ f (y) ∣ d y = N_{R} (x) + F_{R} (x) \leq C_{n, α} R^{α} M + D_{n, α, p} R^{α - n / p} F,$ where $D_{n, α, p}$ is the far constant of [F4], renamed here to avoid a clash with the constant defined below, the first inequality is [F5], and the equality of the total integral with the sum of the near and far integrals is additivity in [F6] applied on the complementary sets $B (x, R)$ and ${y : ∣ x - y ∣ \geq R}$ . Since $θ = α p / n$ gives $α = θ n / p$ and $α - n / p = (θ - 1) n / p$ , one has $R^{α} = (F / M)^{α p / n} = (F / M)^{θ}, R^{α - n / p} = (F / M)^{(α - n / p) p / n} = (F / M)^{θ - 1},$ so $R^{α} M = F^{θ} M^{1 - θ}$ and $R^{α - n / p} F = F^{θ} M^{1 - θ}$ ; hence $∣ I_{α} f (x) ∣ \leq (C_{n, α} + D_{n, α, p}) M^{1 - θ} F^{θ}$ .

4.1step 1.1step 2.1step 3.1algebra

Conclusion of the estimate. Setting $C_{n, α, p} : = C_{n, α} + D_{n, α, p}$ , with $D_{n, α, p}$ the far constant of [F4], step 3.1 gives the asserted bound in the nondegenerate case; together with steps 1.1 and 2.1 every case is covered, the exponential factors are the exact positive powers $θ \in (0, 1)$ and $1 - θ \in (0, 1)$ of finite nonnegative quantities, and no expression $0^{0}$ or $\infty^{0}$ occurs. Absolute convergence at $x$ is the finiteness of $N_{R} (x) + F_{R} (x)$ from step 3.1.

5.1F8step 1.1step 2.1step 3.1step 4.1∎

Choice accounting. The argument uses Countable Choice only through the maximal-function interface [F3] and the splitting lemma [F4], both of which are stated under Countable Choice, and through the measure and integral facts [F6]-[F7] of the Euclidean Lebesgue framework; no full Axiom of Choice and no choice over an uncountable family is invoked. The hypothesis $M f (x) < \infty$ is used only at the single point $x$ , and the conclusion is pointwise at that point.

TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02Open item page →

Hardy–Littlewood–Sobolev fractional integration inequality

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $n \geq 1$ , $0 < α < n$ and $1 < p < n / α$ , and set $1 / q : = 1 / p - α / n .$ Then $q$ is finite and $q > p$ . For every $f$ in the complex Lebesgue space $L^{p} (R^{n}; C)$ the unit-normalized Riesz integral $I_{α} f$ of Riesz potential of order alpha exists absolutely for almost every $x \in R^{n}$ , and the resulting almost-everywhere defined function determines an element of $L^{q} (R^{n}; C)$ . The assignment is independent of the measurable representative of the class and defines a bounded linear map $I_{α} : L^{p} (R^{n}; C) \to L^{q} (R^{n}; C), ∥ I_{α} f ∥_{q} \leq C_{n, α, p} ∥ f ∥_{p},$ with a constant depending only on $n$ , $α$ and $p$ . On the dense subspace $C_{c}^{\infty} (R^{n}; C)$ the map is the pointwise integral $x \mapsto \int K_{α} (x - y) f (y) d y$ , which is finite at every point there, and its unique bounded dense-core extension is this same almost-everywhere integral operator.

Facts & Assumptions

Given: Countable Choice, $n \geq 1$ , $0 < α < n$ , $1 < p < n / α$ , the exponent $q$ with $1 / q = 1 / p - α / n$ , and a class $f \in L^{p} (R^{n}; C)$ with a fixed measurable representative, again written $f$ .

[F1]

The unit Riesz potential is $I_{α} f (x) = \int K_{α} (x - y) f (y) d y$ at exactly those points where $\int K_{α} (x - y) ∣ f (y) ∣ d y < \infty$ , with $K_{α} (z) = ∣ z ∣^{α - n}$ for $z \neq 0$ and $K_{α} (0) = 0$ ; where the absolute integral is infinite no value is assigned. (Riesz potential of order alpha)

[F2]

Complex $L^{p}$ classes are quotients by almost-everywhere equality carrying the well-defined norm $∥ g ∥_{p} = (\int ∣ g ∣^{p})^{1 / p}$ and the complex vector operations; a complex function is measurable when its real and imaginary parts are; integration is componentwise, with $\int u = \int u^{+} - \int u^{-}$ for integrable real $u$ ; $C_{c}^{\infty} (R^{n}; C)$ consists of bounded measurable functions; local integrability means finite absolute integral over every Euclidean ball. (Complex Lp classes and Euclidean test-function conventions, Complex Holder, Minkowski, and the quotient norm, Integrable real and complex functions, and their integrals, A locally integrable function on $R^{n}$ )

[F3]

Near/far splitting: for $f \in L^{p}$ with $1 < p < n / α$ every measurable representative is locally integrable, and at every $x$ with $M f (x) < \infty$ and every $R > 0$ the integrals $N_{R} (x) = \int_{∣ x - y ∣ < R} K_{α} (x - y) ∣ f (y) ∣ d y$ and $F_{R} (x) = \int_{∣ x - y ∣ \geq R} K_{α} (x - y) ∣ f (y) ∣ d y$ are finite with $N_{R} (x) \leq C_{n, α} R^{α} M f (x)$ and $F_{R} (x) \leq C_{n, α, p} R^{α - n / p} ∥ f ∥_{p}$ ; the far bound holds at every $x$ ; and if two representatives agree almost everywhere then, at every $x$ and every $R > 0$ , their integrals $N_{R}$ and $F_{R}$ coincide, and the total potential is defined on the common finite set and agrees for the two representatives. (Near and far bounds for a Riesz potential)

[F4]

Hedberg's pointwise inequality: with $θ = α p / n \in (0, 1)$ , at every $x$ with $M f (x) < \infty$ the defining integral of $I_{α} f$ converges absolutely and $∣ I_{α} f (x) ∣ \leq C_{n, α, p} (M f (x))^{1 - θ} ∥ f ∥_{p}^{θ}$ . (Hedberg pointwise inequality for Riesz potentials)

[F5]

The centered maximal function is $M f (x) = M (∣ f ∣) (x) = \sup_{r > 0} λ (B (x, r))^{- 1} \int_{B (x, r)} ∣ f ∣$ with values in $[0, \infty]$ ; for $1 < p < \infty$ there is $C_{n, p}$ with $∥ M h ∥_{p} \leq C_{n, p} ∥ h ∥_{p}$ for every real $h \in L^{p} (R^{n})$ ; and $M h$ is Borel measurable whenever $h \in L_{l o c}^{1} (R^{n})$ . (The centered and uncentered Hardy-Littlewood maximal functions, The centered maximal operator is bounded on $L^{p} (R^{n})$ for $1 < p < \infty$ , The centered Hardy-Littlewood maximal function is Borel measurable)

[F6]

A nonnegative measurable function with finite integral is finite almost everywhere. (A nonnegative measurable function with finite integral is finite almost everywhere)

[F7]

Extended-real measurability is equivalent to measurability of all strict superlevel sets: $h$ is measurable exactly when ${h > a}$ is measurable for every real $a$ . (Threshold characterisations of real-valued and extended-real-valued measurability)

[F8]

Tonelli: for sigma-finite measure spaces $(X, A, μ)$ and $(Y, B, ν)$ and a product-measurable $H : X \times Y \to [0, \infty]$ , the partial integral $x \mapsto \int_{Y} H (x, y) d ν (y)$ is measurable and the three iterated integrals agree. Euclidean Lebesgue measure on $L (R^{n})$ is sigma-finite, and every bounded measurable set has finite measure. (Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure)

[F9]

Product measurability toolkit: $B (R^{n}) \otimes B (R^{n}) = B (R^{2 n})$ under the usual identification; every Borel subset of $R^{2 n}$ is Lebesgue measurable and $B (R^{n}) \subseteq L (R^{n})$ , so $B (R^{n}) \otimes B (R^{n}) \subseteq L (R^{n}) \otimes L (R^{n})$ ; continuous Euclidean maps are Borel; the composition of measurable maps is measurable, and composition of a measurable map with a Borel map on its codomain preserves measurability; coordinate projections are measurable; sums, products, scalar multiples, absolute values and positive and negative parts of measurable extended-real functions are measurable. (The Borel product of R^m and R^n is the Borel sigma-algebra of R^{m+n}, Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable, Continuous functions on Euclidean spaces are Borel measurable, Composition with a Borel measurable outer map preserves measurability, Arithmetic and lattice operations preserve measurability whenever they are defined, Borel measurable and Lebesgue measurable functions on $R^{n}$ , A measurable function between measurable spaces)

[F10]

Integral rules: the Lebesgue integral is complex-linear on $L^{1}$ ; the nonnegative integral is monotone and additive; a nonnegative measurable function has integral zero exactly when it vanishes almost everywhere; integrable functions equal almost everywhere have equal integrals over every measurable set; a nonnegative measurable function has zero integral over every null set; and finite unions of null sets are null. (The Lebesgue integral is linear on $L^{1} (μ)$ , Monotonicity and nonnegative homogeneity of the nonnegative integral, Additivity of the nonnegative Lebesgue integral, Two integrable functions are equal almost everywhere exactly when all of their indefinite integrals agree, A nonnegative integral over a null set vanishes, Finite and countable subadditivity of measures)

[F11]

Density and completeness: under Countable Choice $C_{c}^{\infty} (R^{n}; C)$ is dense in the Euclidean Lebesgue space $L^{p}$ for $1 \leq p < \infty$ , and $L^{r} (μ; C)$ is complete for every $1 \leq r \leq \infty$ , so $L^{q}$ is a Banach space for its quotient norm. (Complex finite-simple and smooth compact-support density for finite p, Complex Lp completeness and almost-everywhere subsequences, Banach space)

[F12]

Extension and continuity: a bounded linear map $T : D \to Y$ on a dense normed subspace $D$ of a normed space $X$ with Banach target $Y$ has a unique bounded linear extension $\tilde{T} : X \to Y$ with $∥ \tilde{T} ∥ = ∥ T ∥$ ; bounded linear maps are continuous; limits of convergent sequences in a metric space are unique. (A bounded linear map from a dense normed subspace into a Banach space extends uniquely with the same norm, Normed subspace, Interior, closure, boundary, limit point, isolated point and dense subset of a metric space, For a linear operator, boundedness, continuity at 0, continuity, and Lipschitz continuity are equivalent, A sequence in a metric space has at most one limit)

[F13]

Countable Choice is the choice principle assumed by the maximal-function, Tonelli, density, completeness and extension interfaces used below. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

Proof

technique · direct; the exponent algebra fixes $q$; Tonelli makes the integral measurable; the maximal bound makes $Mf$ finite almost everywhere and Hedberg converts that into almost-everywhere absolute convergence and an $L^q$ estimate; classes, linearity and the dense smooth core are then handled separately

1.1givenalgebra

The exponents. Put $θ : = α p / n$ , which lies in $(0, 1)$ because $0 < α p < n$ . The exponent $q$ of the statement satisfies $1 / q = 1 / p - α / n = (n - α p) / (p n) = (1 - θ) / p$ , so $q = p / (1 - θ)$ ; hence $q < \infty$ , $q > p$ , and $(1 - θ) q = p$ , while $θ q = α p q / n = q - p$ .

1.2F2F9

Joint product measurability. Identify $R^{n} \times R^{n}$ with $R^{2 n}$ ; by [F9] its Borel sigma-algebra is $B (R^{n}) \otimes B (R^{n})$ , which is contained in $L (R^{n}) \otimes L (R^{n})$ . The difference map $(x, y) \mapsto x - y$ is continuous, hence Borel, so its composition with the Borel function $K_{α}$ is Borel by [F9] and therefore product measurable. The second coordinate projection is measurable into the Lebesgue sigma-algebra since the inverse image of every Lebesgue set $A$ is $R^{n} \times A$ ; composing it with measurable $f$ makes $(x, y) \mapsto f (y)$ product measurable. The lattice and product clauses of [F9] then make $(x, y) \mapsto ∣ f (y) ∣$ and $(x, y) \mapsto (Re f)^{\pm} (y)$ , $(Im f)^{\pm} (y)$ product measurable. Products of measurable finite-valued functions are measurable by [F9], so the five functions $(x, y) \mapsto K_{α} (x - y) ∣ f (y) ∣$ and $K_{α} (x - y) (Re f)^{\pm} (y)$ , $K_{α} (x - y) (Im f)^{\pm} (y)$ are product measurable and take values in $[0, \infty)$ .

1.3F5F6F7givenalgebra

The maximal function is finite almost everywhere. The function $∣ f ∣$ is real, measurable and lies in $L^{p}$ , so the maximal bound of [F5] applied to $∣ f ∣$ gives $∥ M f ∥_{p} = ∥ M (∣ f ∣) ∥_{p} \leq C_{n, p} ∥ ∣ f ∣ ∥_{p} = C_{n, p} ∥ f ∥_{p} < \infty$ . The power $(M f)^{p}$ is nonnegative and measurable by the threshold criterion [F7]: for $a < 0$ its superlevel set is all of $R^{n}$ , and for $a \geq 0$ it equals ${M f > a^{1 / p}}$ . Applying [F6] to $(M f)^{p}$ gives $(M f)^{p} < \infty$ almost everywhere, that is, $M f < \infty$ almost everywhere.

2.1F1F2F8step 1.2algebra

Tonelli and measurability of the integral. By [F8] the measure space $(R^{n}, L (R^{n}), λ_{n})$ is sigma-finite, so Tonelli applies to each product-measurable function of step 1.2: the functions $G (x) : = \int K_{α} (x - y) ∣ f (y) ∣ d y, G_{1} (x) : = \int K_{α} (x - y) (Re f)^{+} (y) d y, \dots, G_{4} (x) : = \int K_{α} (x - y) (Im f)^{-} (y) d y$ are measurable $[0, \infty]$ -valued functions of $x$ . The set $E : = {G < \infty} = ⋃_{m \geq 1} {G \leq m}$ is measurable, and $0 \leq G_{j} \leq G$ on all of $R^{n}$ for $j = 1, \dots, 4$ ; hence on $E$ the four numbers $G_{j} (x)$ are finite and $h (x) : = (G_{1} (x) - G_{2} (x)) + i (G_{3} (x) - G_{4} (x))$ is a well-defined complex number. Since each $G_{j}$ is measurable and $E$ is measurable, the product ${\tilde{I}}_{α} f : = h \cdot 1_{E}$ (with value $0$ off $E$ ) is a measurable complex-valued function; and at every $x \in E$ it equals $\int K_{α} (x - y) f (y) d y$ by componentwise integration [F2] and the definition [F1], because $K_{α} (x - \cdot) ∣ f ∣$ has finite integral there.

3.1F1F3F4step 1.3step 2.1

Almost-everywhere absolute convergence and the pointwise bound. By [F3] every representative of $f$ is locally integrable, so $M f$ is defined everywhere; by the Hedberg inequality [F4], at every $x$ with $M f (x) < \infty$ the defining integral converges absolutely and $∣ I_{α} f (x) ∣ \leq C_{n, α, p}^{H} (M f (x))^{1 - θ} ∥ f ∥_{p}^{θ}$ , where $C_{n, α, p}^{H}$ denotes the constant of [F4], renamed to avoid a clash with the constant claimed in the Statement. Let $S : = {M f < \infty}$ ; step 1.3 makes $S$ conull, and $S \subseteq E$ by the definition of $E$ in step 2.1. Hence ${\tilde{I}}_{α} f = h \cdot 1_{E}$ agrees with the pointwise potential $I_{α} f$ of [F1] at every point of $S$ and differs from it only on the null set $R^{n} ∖ S$ ; in particular the defining integral converges absolutely almost everywhere, and ${\tilde{I}}_{α} f$ is a measurable representative of the almost-everywhere defined integral. At every point of $S$ the displayed inequality is exactly the Hedberg bound. At a point with $M f (x) = + \infty$ : if $∥ f ∥_{p} > 0$ then the right-hand side is $+ \infty$ , because $1 - θ \in (0, 1)$ and $θ \in (0, 1)$ are applied to $+ \infty$ and to the strictly positive number $∥ f ∥_{p}$ , so the inequality holds trivially against the finite value $∣ {\tilde{I}}_{α} f (x) ∣$ ; and if $∥ f ∥_{p} = 0$ then $f$ is the zero class with $M f = 0$ everywhere by [F4], so the case $M f (x) = + \infty$ cannot occur. In every case the pointwise bound holds at every $x \in R^{n}$ in the form $∣ {\tilde{I}}_{α} f (x) ∣ \leq C_{n, α, p}^{H} (M f (x))^{1 - θ} ∥ f ∥_{p}^{θ} .$

4.1F10step 1.1step 1.3step 3.1algebra

The $L^{q}$ estimate. Raising the bound of step 3.1 to the $q$ -th power and using $(1 - θ) q = p$ from step 1.1 gives, at every $x$ , $∣ {\tilde{I}}_{α} f (x) ∣^{q} \leq (C_{n, α, p}^{H})^{q} (M f (x))^{p} ∥ f ∥_{p}^{θ q},$ an inequality between nonnegative measurable functions. Monotonicity of the integral [F10], the identity $∥ M f ∥_{p}^{p} = \int (M f)^{p}$ , and the maximal bound of step 1.3 give $∥ {\tilde{I}}_{α} f ∥_{q}^{q} = \int ∣ {\tilde{I}}_{α} f ∣^{q} \leq (C_{n, α, p}^{H})^{q} ∥ f ∥_{p}^{θ q} ∥ M f ∥_{p}^{p} \leq (C_{n, α, p}^{H})^{q} C_{n, p}^{p} ∥ f ∥_{p}^{θ q + p} = (C_{n, α, p}^{H})^{q} C_{n, p}^{p} ∥ f ∥_{p}^{q},$ where $θ q + p = q$ by step 1.1. Hence $∥ {\tilde{I}}_{α} f ∥_{q} \leq C_{n, α, p} ∥ f ∥_{p}$ with $C_{n, α, p} : = C_{n, α, p}^{H} C_{n, p}^{p / q} = C_{n, α, p}^{H} C_{n, p}^{1 - θ} < \infty$ , a constant depending only on $n, α, p$ .

5.1F3F10step 1.3step 2.1step 3.1step 4.1

Representative independence. Let $f$ and $g$ be measurable representatives of the same class, so that $f = g$ almost everywhere; then $∣ f ∣ = ∣ g ∣$ almost everywhere, and both $∣ f ∣ 1_{B}$ and $∣ g ∣ 1_{B}$ are integrable for every ball $B$ because the representatives are locally integrable by [F3] and balls have finite measure, so the almost-everywhere-equality clause of [F10] gives $\int_{B} ∣ f ∣ = \int_{B} ∣ g ∣$ for every ball $B$ . Hence $M f = M g$ as extended-real functions and the two conull sets coincide: writing $S : = {M f < \infty} = {M g < \infty}$ , step 1.3 makes $S$ conull. At every $x \in S$ and every $R > 0$ the splitting lemma [F3] gives that $N_{R}$ and $F_{R}$ are finite for each of the two representatives, and its representative-independence clause gives that the total potential is defined at $x$ and takes the same value for $f$ and for $g$ ; thus $I_{α} f (x) = I_{α} g (x)$ for every $x \in S$ . By step 3.1 both ${\tilde{I}}_{α} f$ and ${\tilde{I}}_{α} g$ agree with these potentials at every point of $S$ ; since $S$ is conull, the two measurable functions of step 2.1 define the same class in $L^{q}$ . Therefore the class $[{\tilde{I}}_{α} f] \in L^{q}$ depends only on the class $[f] \in L^{p}$ , and step 4.1 gives the bound $∥ {\tilde{I}}_{α} [f] ∥_{q} \leq C_{n, α, p} ∥ [f] ∥_{p}$ for this well-defined assignment.

6.1F10step 1.3step 3.1step 5.1algebra

Linearity. Let $f, g$ be measurable representatives of classes in $L^{p}$ and let $c \in C$ ; then $f + g$ and $c f$ are measurable representatives of the corresponding classes. At every point $x$ of $S_{f} \cap S_{g} \cap S_{f + g} \cap S_{c f}$ , all of the complex functions $K_{α} (x - \cdot) f$ , $K_{α} (x - \cdot) g$ , $K_{α} (x - \cdot) (f + g)$ and $K_{α} (x - \cdot) (c f)$ are integrable, and the linearity of the Lebesgue integral on $L^{1}$ [F10] gives $\int K_{α} (x - y) (f + g) (y) d y = \int K_{α} (x - y) f (y) d y + \int K_{α} (x - y) g (y) d y$ and $\int K_{α} (x - y) (c f) (y) d y = c \int K_{α} (x - y) f (y) d y$ . The four sets $S_{f}, S_{g}, S_{f + g}, S_{c f}$ are conull by step 1.3 and their intersection is conull by the null-union clause of [F10]; on that intersection, where each ${\tilde{I}}_{α}$ is the corresponding integral by step 3.1, the a.e.-equal functions ${\tilde{I}}_{α} (f + g)$ and ${\tilde{I}}_{α} f + {\tilde{I}}_{α} g$ define the same class in $L^{q}$ , and likewise ${\tilde{I}}_{α} (c f)$ and $c {\tilde{I}}_{α} f$ . So the assignment of step 5.1 is complex-linear.

7.1F1F2F3F8step 4.1step 5.1step 6.1

The smooth core. Let $g \in C_{c}^{\infty} (R^{n}; C)$ . Then $∣ g ∣ \leq ∥ g ∥_{\infty} < \infty$ everywhere, so every ball average of $∣ g ∣$ is at most $∥ g ∥_{\infty}$ and $M g (x) \leq ∥ g ∥_{\infty}$ for every $x$ ; also $g \in L^{p}$ because it is bounded and supported in a bounded measurable set of finite measure by [F8]. By the near/far bounds [F3] the near and far integrals of $g$ are finite at every $x$ , so the defining absolute integral is finite everywhere and [F1] defines $I_{α} g$ everywhere as the pointwise integral $x \mapsto \int K_{α} (x - y) g (y) d y$ . Hence $S = R^{n}$ and $E = R^{n}$ , and ${\tilde{I}}_{α} g$ equals this pointwise integral at every point. Let $D \subseteq L^{p} (R^{n}; C)$ be the image of $C_{c}^{\infty} (R^{n}; C)$ ; the assignment of steps 4.1, 5.1 and 6.1 restricts to a bounded linear map $T : D \to L^{q}$ which is exactly the pointwise-integral map, with $∥ T ∥$ at most the constant of step 4.1.

8.1F11F12step 7.1

Density and the abstract extension. By [F11] the subspace $D$ is dense in the normed space $L^{p}$ and $L^{q}$ is complete, hence a Banach space; the map $T$ of step 7.1 is bounded and linear. The extension theorem [F12] therefore produces a unique bounded linear map $\tilde{T} : L^{p} (R^{n}; C) \to L^{q} (R^{n}; C)$ with $\tilde{T} ∣_{D} = T$ and $∥ \tilde{T} ∥ = ∥ T ∥$ .

9.1F11F12step 4.1step 7.1step 8.1algebra

Identification of the extension with the integral operator. Let $I : L^{p} \to L^{q}$ denote the bounded linear almost-everywhere integral map of steps 4.1, 5.1 and 6.1. Let $x \in L^{p}$ . By density [F11] there are $d_{k} \in D$ with $∥ d_{k} - x ∥_{p} \to 0$ . Both $I$ and $\tilde{T}$ are bounded linear, hence continuous on the normed space $L^{p}$ by [F12], and they agree on $D$ because $I ∣_{D} = T = \tilde{T} ∣_{D}$ by steps 7.1 and 8.1. Therefore $I (x) = \lim_{k} I (d_{k}) = \lim_{k} T (d_{k}) = \lim_{k} \tilde{T} (d_{k}) = \tilde{T} (x),$ the two outer equalities by continuity and the middle one because $d_{k} \in D$ ; limits in the normed space $L^{q}$ are unique by [F12]. Hence the unique bounded dense-core extension of the pointwise-integral map on $C_{c}^{\infty}$ is precisely the almost-everywhere integral operator $I$ , and it satisfies the bound $∥ I (f) ∥_{q} \leq C_{n, α, p} ∥ f ∥_{p}$ of step 4.1.

10.1F13step 1.1step 4.1step 5.1step 6.1step 7.1step 8.1step 9.1∎

Conclusion. For $1 < p < n / α$ and the exponent $q$ of the statement, steps 1.1, 4.1, 5.1 and 6.1 prove that the defining integral of $I_{α} f$ converges absolutely almost everywhere for every $f \in L^{p}$ and that its class obeys the $L^{q}$ bound with a constant depending only on $n, α, p$ , and that this gives a well-defined bounded linear map on the quotient classes; and steps 7.1, 8.1 and 9.1 prove that on the dense smooth core the map is the pointwise integral and that its unique bounded dense-core extension is this same almost-everywhere integral operator. Countable Choice is spent exactly through the maximal-function, Tonelli and sigma-finiteness, density, completeness and extension interfaces [F5], [F8], [F11], [F12]; no full Axiom of Choice is invoked.

RemarkRemark: Literature-sourcedProof: Not applicable‡ sources checked 2026-10-02‡ not proved hereOpen item page →

‡ Recorded, not proved here. This statement is included so the library can refer to it honestly, with a citation to the literature. It is not proved anywhere in this library: the track that would prove it has not been developed here yet.

Endpoint bounds require separate formulations

Statement

Assume the Axiom of Countable Choice for the Lebesgue conventions of Riesz potential of order alpha. Recorded orientation, not proved here. Let $0 < α < n$ , let $I_{α}$ be the unit-normalized Riesz potential on $R^{n}$ of Riesz potential of order alpha, and let the strict-range theorem of this pair be Hardy–Littlewood–Sobolev fractional integration inequality, whose hypothesis is $1 < p < n / α$ . In the notation of Sublinear operators and weak or strong type $(p, q)$ bounds, the following endpoint claims are recorded from the cited source but are not proved, used, or reproduced in this library:

Lower endpoint. $I_{α}$ is of weak type $(1, n / (n - α))$ , and it is not of strong type $(1, n / (n - α))$ . Consequently the hypothesis $1 < p$ of the strong theorem cannot be relaxed to $p = 1$ : no constant bounds $∥ I_{α} f ∥_{n / (n - α)}$ by $∥ f ∥_{1}$ .
Upper endpoint. At $p = n / α$ the raw potential is not of strong type $(n / α, \infty)$ : there are $f \in L^{n / α} (R^{n})$ for which $I_{α} f$ is not essentially bounded (indeed it may fail to be finite on a set of positive measure).
Critical mean oscillation. For $f \in L^{n / α} (R^{n})$ with compact support, the potential $I_{α} f$ is finite almost everywhere and its mean-oscillation seminorm modulo additive constants is bounded by $C ∥ f ∥_{n / α}$ . For general $f \in L^{n / α} (R^{n})$ use the renormalized potential ${\tilde{I}}_{α} f (x) : = \int_{R^{n}} [K_{α} (x - y) - 1_{{∣ y ∣ \geq 1}} K_{α} (- y)] f (y) d y,$ with subtraction inside the integral. It is finite almost everywhere and locally integrable, and satisfies the same mean-oscillation bound. If $C_{f} : = \int_{∣ y ∣ \geq 1} K_{α} (- y) f (y) d y$ is absolutely convergent, as it is for compactly supported critical data, then ${\tilde{I}}_{α} f = I_{α} f - C_{f}$ wherever the raw potential is defined. In general the raw integral may diverge everywhere, so no finite additive constant relating it to the renormalized potential is asserted.

Recorded orientation

These are orientation facts about the boundary of the strict-range theorem, recorded with their exact hypotheses and not established here. The library does not currently define the weak $L^{q}$ space or the space $B M O$ of functions of bounded mean oscillation, so clauses 1 and 3 are quoted from the source in the source's own vocabulary; clause 1's weak-type inequality is the $p = 1$ case of the weak $(p, q)$ estimate stated in the proof of the source's Theorem 1, and clause 3 is the source's Theorem 4 together with the remark that follows it. None of these endpoint claims is a proof supplier for this pair: the strict range $1 < p < n / α$ retains the hypothesis of Hardy–Littlewood–Sobolev fractional integration inequality, and no item of the pair lists this remark among its dependencies. The companion page's two counterexamples exhibit the failures of clause 2 and of the strong part of clause 1 directly, in $L^{p_{0}}$ and $L^{1}$ respectively, without proving the weak-type or mean-oscillation bounds recorded above.

5 · Examples, counterexamples and false statements

None yet.

Riesz Potentials and the Hardy–Littlewood–Sobolev Inequality

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

Riesz potential of order alpha

Definition

Near and far bounds for a Riesz potential

Statement

Facts & Assumptions

Proof

Hedberg pointwise inequality for Riesz potentials

Statement

Facts & Assumptions

Proof

Hardy–Littlewood–Sobolev fractional integration inequality

Statement

Facts & Assumptions

Proof

Endpoint bounds require separate formulations

Statement

Recorded orientation

5 · Examples, counterexamples and false statements

Sources