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For $1<p<\infty$, let $q=p/(p-1)$ and partition the coordinates into consecutive finite blocks $B_k$ of lengths $m_k\to \infty$. The displayed map defines a norm-one operator on the l-p sequence space. By Hölder's inequality, $|(Tx{)}_k{|}^p\le {\sum }_{j\in B_k}|x_j{|}^p$, giving the upper norm bound. For the unit vector $e_j$ with $j\in B_k$, $T{e}_j=m_k^{-1/q}{e}_k$, so its norm tends to zero as $j\to \infty$.

Nevertheless, $x^{(k)}=m_k^{-1/p}{1}_{B_k}$ has norm one and maps to $e_k$. Those images are pairwise distance $2^{1/p}$ apart. Thus $T$ is not a compact operator, and has norm exactly one. Vanishing images of a chosen basis do not alone test compactness on ${\ell }^p$ for $p>1$. The map is the adjoint of the isometric block embedding $(Uy{)}_j=m_k^{-1/q}{y}_k$, $j\in B_k$, on ${\ell }^q$.

Let $1<p<\infty$ and $q=p/(p-1)$. For a nonzero bounded linear functional $\ell$ on an Lp space, choose $f_0$ with $\ell (f_0)=1$ and a closest point $h$ in its closed kernel. Put $z=f_0-h$. The variational condition is $\int |z{|}^{p-2}\overline{z}k=0$ for every kernel element $k$. Since $f-\ell (f)z$ lies in the kernel, $\ell (f)=\int fg$ with $g=|z{|}^{p-2}\overline{z}/\Vert z{\Vert }_p^p\in L^q$. Hölder's inequality gives the upper norm bound; testing against $|g{|}^{q-2}\overline{g}$ gives equality. Thus the representation is an isometric linear isomorphism when the pairing is bilinear.

On a bounded open set, let $Tv\in W_0^{1,2}$ be the weak solution of $\Delta Tv=v$. The energy estimate bounds $T:L^2\to W_0^{1,2}$. For $p>n$, put $q=2p/(p-2)<2n/(n-2)$. The zero-boundary Rellich-Kondrachov compactness theorem gives compact embedding into $L^q$, and Hölder's inequality gives $\Vert wz{\Vert }_2\le \Vert w{\Vert }_p\Vert z{\Vert }_q$. Hence $Kv=wTv$ is compact on $L^2$, without a regularity assumption on the boundary. Writing $v=\Delta u$ makes $\Delta u-wu=f$ equivalent to $(I-K)v=f$. The Fredholm alternative for a compact operator then identifies unique solvability with triviality of the homogeneous kernel.

For $1<p,q<\infty$ and nonzero factors in their respective Lp spaces, equality in the product-integral bound holds precisely whenIn other words, the normalized powers match. Zero factors give the trivial equality separately. For real vectors, equality for the signed finite sum also needs the nonzero coordinate products to have a common sign; otherwise the preceding triangle inequality loses information.

For $f(t)=t$ and $g(t)=1-t$ on $[0,1]$, with $p=3$, $q=3/2$,Changing only $g$ to $t^2$ gives $f^3=g^{3/2}=t^3$, and both sides become $1/4$. The factors need not be equal: their conjugate powers are what must agree after normalization.

The proof outline makes the equality condition natural. Normalize $u=|f|/\Vert f{\Vert }_p$ and $v=|g|/\Vert g{\Vert }_q$. The Young inequality gives $uv\le u^p/p+v^q/q$; integration yields $\int uv\le 1/p+1/q=1$. Equality requires $u^p=v^q$ almost everywhere. The endpoint proof uses the essential supremum directly, so it does not need division by $p-1$ at $p=1$.

Take $p=3$, $q=3/2$, and vectors $x=(1,2,-1)$ and $y=(2,-1,3)$. Their coordinate products are $(2,-2,-3)$, soTheir Lp norms give the larger boundThus cancellation explains $3<7$, and Hölder's inequality explains the remaining gap between $7$ and the norm product. Small signed sums need not imply small individual coordinate products.

Now keep $x$ and choose $y=(1,4,-1)$. The coordinate products are $(1,8,1)$, with no cancellation, andThe matching powers are $|x_i{|}^3=|y_i{|}^{3/2}$, namely $(1,8,1)$.

Suppose a linear configuration count has a Fourier-integral representation with $k$ factors. If one has a uniform Fourier bound for the difference between two weights and compatible $L^{k-1}$ bounds for each weight, expanding the difference one factor at a time and applying Hölder's inequality bounds the change in the count.

Each nonnegative integrable $f_j$ depends on all coordinates except the $j$th. In dimension three, Cauchy-Schwarz inequality first in the third coordinate and then in the remaining pair proves the displayed estimate. Induction integrates the last coordinate by Hölder's inequality on the first $d-1$ factors, then applies Hölder's inequality to the remaining factor and the dimension-$(d-1)$ product. Taking indicators of coordinate projections of a Euclidean body gives the volume version of the Loomis--Whitney inequality.

Let $1<p<\infty$ and let $f$ be a real or complex measurable function in Lp space $L^p((0,\infty ),dx)$, with Lebesgue measure $dx$. The Hardy operator, denoted here by $H$, takes the running averageHölder's inequality ensures ${\int }_0^x|f(t)|dt\le x^{1-1/p}\Vert f{\Vert }_p<\infty$, so every such average exists. Hardy's integral inequality statesThe real signed and complex extensions of the nonnegative statement follow from the triangle inequality $|Hf|\le H|f|$. Changing $f$ on a null set changes neither average nor Lp norm.^[ref]

Averaging over an expanding interval spreads a localized input across the whole positive half-line. Hardy's integral inequality bounds the total Lp norm of that spread. The Hardy operator is consequently a bounded linear operator on this Lp space. The factor $p/(p-1)$ bounds its operator norm; its $p$th power is the constant in the displayed integral bound. These are different constants: at $p=2$, the L2 norm bound is two and the squared L2 norm bound is four.

If $1<p<n/\alpha$, the pointwise Hedberg inequality isSplit the integral at radius $r$. Dyadic annuli bound the near part by $C{r}^{\alpha }Mf(x)$, while Hölder's inequality bounds the far part by $C\Vert f{\Vert }_p{r}^{\alpha -n/p}$. Choosing $r^{n/p}=\Vert f{\Vert }_p/Mf(x)$ balances the terms.

Hölder's inequality controls a product using separate information about its two factors. If one factor is large on part of a domain, the other factor's complementary Lp norm limits how much their product can accumulate there. The same statement bounds finite sums, integrals and expected values; no independence is required in the probabilistic version.

Let $p,q\in [1,\infty ]$ be conjugate exponents, meaningOn a measure space, if $f\in L^p$ and $g\in L^q$, then $fg\in L^1$ andHere the Lp norms are finite; at exponent $\infty$ the norm uses the essential supremum. For a finite vector, integration with counting measure givesThe first step is the triangle inequality; the second is Hölder's inequality.^[ref]

For finite vectors, the endpoint case $p=1$, $q=\infty$ is $|x^{T}y|\le \Vert x{\Vert }_1\Vert y{\Vert }_{\infty }$: bound every $|y_i|$ by its maximum, then sum the $|x_i|$. Swapping the factors gives the other endpoint. The choice $p=q=2$ gives the Cauchy-Schwarz inequality.

An Lp norm measures the size of a measurable function by averaging a power of its absolute value with respect to a specified measure. For $1\le p<\infty$,when the integral is finite. The exponent $p$ controls how strongly large values influence the result. The Lp space identifies functions equal almost everywhere, so this becomes a norm on the corresponding equivalence classes. For $p=\infty$, use the essential supremum of $|f|$, as defined by the L-infinity norm. ^[ref].

For a two-coordinate vector $(3,4)$, treat the coordinates as a function on two inputs with counting measure. Then $\Vert (3,4){\Vert }_1=3+4=7$, $\Vert (3,4){\Vert }_2=\sqrt{9+16}=5$, and $\Vert (3,4){\Vert }_{\infty }=4$. The middle case is the Euclidean norm; the first adds the coordinate magnitudes, and the last takes their largest value. These are different answers to different questions about size, all satisfying the triangle inequality.

If instead each input has probability $1/2$, the same values give $\Vert f{\Vert }_1=3.5$ and $\Vert f{\Vert }_2=\sqrt{12.5}$, while $\Vert f{\Vert }_{\infty }=4$ remains unchanged. Thus the measure is not an optional label. On a probability space, the finite Lp norms are nondecreasing in $p$ whenever they are finite, by Hölder's inequality; the unnormalized two-coordinate example above decreases instead. Statements comparing different exponents must specify their measure convention.

Why are these sizes useful? If $e$ is an error function, its L1 norm measures aggregate absolute error, its L2 norm measures the square root of aggregate quadratic loss, and its L-infinity norm bounds error almost everywhere. Concentrating a large error on a set of small positive measure can give a modest finite Lp norm but a large L-infinity norm. The three-point example and diagram in L-infinity norm show this explicitly, including why a null set behaves differently from a rare event.

For measurable functions $f,g$ with values in the real numbers or complex numbers on the same measure space $(X,\mathcal{A},\mu )$, and $1\le p\le \infty$, membership in the same Lp space givesThus adding two functions cannot make their combined Lp norm exceed the sum of their separate Lp norms. The measure need not be finite or a probability measure. The hypotheses require finite Lp norms; $p=\infty$ uses the essential supremum of the absolute value, not an ordinary pointwise supremum. For complex numbers, the absolute value is the complex modulus. As usual, Lp spaces identify functions equal almost everywhere. ^[ref]

The Minkowski inequality makes the Lp norm a workable notion of size: it supplies the triangle inequality needed to turn $d_p(f,g)=\Vert f-g{\Vert }_p$ into a metric on Lp space. Applied to $(f-h)+(h-g)$, it says that travelling through an intermediate function $h$ cannot shorten the direct distance measured by this metric. Applied to two error functions, it gives a bound on the total error even when the errors reinforce one another; Minkowski inequality for error aggregation gives explicit examples.

The geometry is already visible in the Euclidean plane. Under counting measure on two coordinates, the closed unit balls for the L1 norm, Euclidean norm, and L-infinity norm are respectively $|x|+|y|\le 1$, $x^2+y^2\le 1$, and $\max (|x|,|y|)\le 1$. Every one is a convex set: if $u,v$ have Lp norm at most one, the Minkowski inequality and absolute homogeneity of a norm give $\Vert tu+(1-t)v{\Vert }_p\le t+(1-t)=1$ for $0\le t\le 1$. The Euclidean norm has no flat boundary segments, whereas the other two norms do. This is the distinction captured by a strictly convex normed space and developed in equality in Minkowski inequality.

A short Hölder's inequality proof outline handles $1<p<\infty$. First $|f+g{|}^p\le 2^{p-1}(|f{|}^p+|g{|}^p)$, so $f+g$ has finite Lp norm before any division. With conjugate exponents $p$ and $q=p/(p-1)$, set $h=|f+g{|}^{p-1}$; then $\Vert h{\Vert }_q=\Vert f+g{\Vert }_p^{p-1}$. The scalar triangle inequality followed by Hölder's inequality givesDivide if $\Vert f+g{\Vert }_p>0$; otherwise the desired bound is immediate. For $p=1$, integrate $|f+g|\le |f|+|g|$. For $p=\infty$, the almost everywhere bounds $|f|\le \Vert f{\Vert }_{\infty }$ and $|g|\le \Vert g{\Vert }_{\infty }$ give the result by the essential supremum. The same scalar triangle inequality works for real numbers and complex numbers.

Choose the real zero-trace Sobolev spaceThe Poincaré inequality makes this an equivalent Sobolev norm. Since the dimension is five, the Sobolev conjugate exponent is $4^{\ast }=5\cdot 4/(5-4)=20$. The Sobolev embedding theorem gives $X\hookrightarrow L^{20}(\Omega )$ continuously. Boundedness of $\Omega$ and Lp inclusion on a finite measure space then give $X\hookrightarrow L^8(\Omega )$ continuously, say $\Vert u{\Vert }_8\le C_8\Vert u{\Vert }_X$.

For $f\in Y$, the weak formulation isThe zero Dirichlet boundary condition is imposed through the Sobolev trace. Hölder's inequality verifies every term:Thus $|\nabla u{|}^2\nabla u\in L^{4/3}(\Omega ;{\mathbb{R}}^5)$, while $u^7\in L^{8/7}(\Omega )$, and all three pairings are finite bounded linear functionals in $v$. The choice $Y=L^{8/7}(\Omega )$ is sufficient; it is not intended to be the largest possible space of data.

We prove existence by the direct method in the calculus of variations, making the step with weakly compact sets explicit. For $X=W_0^{1,4}(\Omega )$ and $f\in L^{8/7}(\Omega )$, introduce the functionalIt is finite by part (b)(i). Hölder's inequality and the Sobolev embedding theorem giveThus $E$ is a coercive function, bounded below, and every minimizing sequence is bounded in $X$.

The space $X$ is a reflexive Banach space: it is isomorphic to a closed linear subspace of the finite product of the reflexive spaces $L^4(\Omega )$ recording the function and its weak derivatives. A bounded sequence in a reflexive Banach space has a weakly convergent subsequence. Hence, after passing to a subsequence, a minimizing sequence satisfies $u_j\rightharpoonup u_{\ast }$ in $X$. The continuous linear maps $u\mapsto \nabla u$ into $L^4$ and $u\mapsto u$ into $L^8$ preserve weak convergence, soThe norm in any Banach space has weak lower semicontinuity, since it is the supremum of absolute values of bounded linear functionals of dual norm at most one, by the Hahn-Banach theorem. Therefore the two nonnegative norm powers in $E$ have weak lower semicontinuity. Also $u\mapsto {\int }_{\Omega }fudx$ belongs to $X^{\ast }$ and is continuous for weak convergence. It follows thatFor any $v\in X$, the real function $t\mapsto E(u_{\ast }+tv)$ has a minimum at $t=0$. Its derivative there isThis differentiation is justified directly by finite polynomial expansions: the fourth power of the Euclidean norm of $\nabla u_{\ast }+t\nabla v$ and the eighth power of $u_{\ast }+tv$ have integrable coefficients by Hölder's inequality. Thus $A(u_{\ast })=0$, giving existence of a weak solution.

If $u,w$ are two weak solutions, testing the difference by $u-w$ and using part (b)(iii) givesHenceNo sign assumption on $f$ or on the weak solution is required.

Take the functionalThe factor $1/8$ is essential: the derivative of $|s{|}^8/8=s^8/8$ is $s^7$, while the gradient with respect to $\xi$ of $|\xi {|}^4/4$ is $|\xi {|}^2\xi$. As in part (c), finite polynomial expansions and Hölder's inequality justify differentiation along every direction $v\in X$, giving $DJ(u)[v]=⟨A(u),v⟩$.

We prove the global and strict minimum directly. The strictly convex functions $\Phi (\xi )=|\xi {|}^4/4$ and $\Psi (s)=|s{|}^8/8$ satisfy their tangent inequalitieswith strict inequality when the respective arguments differ. For example, strict convexity of $\Phi$ follows from strict operator monotonicity of its gradient, proved by the midpoint identity in part (b)(iii); strict convexity of $\Psi$ follows because its derivative $s^7$ is strictly increasing.

Apply these inequalities pointwise to $\nabla u_{\ast },\nabla v$ and $u_{\ast },v$, then integrate. The weak formulation at $u_{\ast }$ with test function $v-u_{\ast }$ givesIf $v\ne u_{\ast }$ in $X$, the Poincaré inequality implies $\nabla v\ne \nabla u_{\ast }$ on a set of positive Lebesgue measure. The first tangent inequality is strict on that set and the second is nonnegative everywhere. Their integrable remainders therefore have strictly positive total integral. HenceThus $u_{\ast }$ is the unique global minimizer of $J$. Conversely, every minimizer has zero directional derivative in every direction and so satisfies $A(u)=0$, which also identifies it with the unique weak solution from part (c).

Let $1<p<\infty$ and $q=p/(p-1)$. The dual of ${\ell }^p$ is ${\ell }^q$. For every $a\in {\ell }^q$, $⟨a,e_n⟩=a_n\to 0$, so $e_n\rightharpoonup 0$ weakly. A compact operator sends a weakly convergent sequence to a norm-convergent sequence with the corresponding limit. Indeed, any failure of $\Vert T{e}_n\Vert \to 0$ would give a subsequence bounded away from zero, with a further norm-convergent subsequence by compactness; weak continuity of $T$ forces that norm limit to be zero, a contradiction. The condition is therefore necessary.

For failure of sufficiency, partition the positive integers into consecutive blocksand define the block averaging compactness counterexampleBy Hölder's inequality, $|(Tx{)}_k{|}^p\le {\sum }_{j\in B_k}|x_j{|}^p$. Summing over disjoint blocks shows $\Vert Tx{\Vert }_p\le \Vert x{\Vert }_p$, so $T$ is bounded. If $j\in B_k$, then $T{e}_j=k^{-1/q}{e}_k$; as $j\to \infty$, also $k\to \infty$, so $\Vert T{e}_j{\Vert }_p\to 0$.

However, the vectors $x^{(k)}=k^{-1/p}{1}_{B_k}$ have norm one and satisfy $T{x}^{(k)}=e_k$. Distinct output vectors have distance $2^{1/p}$, so this bounded sequence of inputs has no norm-convergent image subsequence. Thus $T$ is not compact.

To connect with the hint, define $(Uy{)}_j=k^{-1/q}{y}_k$ for $j\in B_k$. Then $U:{\ell }^q\to {\ell }^q$ is an isometry, and the dual pairing calculation gives $T=U^{\ast }$. The vectors $U{e}_k$ have disjoint supports and norm one. The direct estimates above establish the counterexample without relying only on the hint.

A real-valued random variable is a measurable function $Y:(\Omega ,\mathcal{F})\to (\mathbb{R},\mathcal{B}(\mathbb{R}))$: equivalently, $\{Y\le t\}\in \mathcal{F}$ for every real $t$.

For $p=1$ the asserted inequality is equality. For $p>1$, let $q=p/(p-1)$. If $\mathbb{E}{Y}^p=\infty$, the inequality is automatic. Otherwise Hölder's inequality on the probability space givesThus $\boxed{\Vert Y{\Vert }_1\le \Vert Y{\Vert }_p}$, using that the total measure is one.

Use the independent product measure on the finite coordinate spaces. For $A\ne \varnothing$, let $V_A(x)$ be the convex hull of its mismatch vectors $(1_{x_i\ne y_i}{)}_{i=1}^n$, $y\in A$. The distance in the question equalsTo verify this, let $v_{\ast }$ be the nearest point. Any unit vector $\alpha$ satisfies ${\min }_{v\in V_A}\alpha \cdot v\le \alpha \cdot v_{\ast }\le \Vert v_{\ast }{\Vert }_2$. Conversely the projection property $v_{\ast }\cdot (v-v_{\ast })\ge 0$ gives equality for $\alpha =v_{\ast }/\Vert v_{\ast }{\Vert }_2$ when $v_{\ast }\ne 0$. If $v_{\ast }=0$, nonnegative weights give supremum zero. Linear functionals have the same minimum on the hull as on its original vectors, so this is exactly the quantified definition of the Talagrand convex distance.

We prove the stronger section induction for the Talagrand exponential moment,Zero-probability atoms may be discarded, replacing $A$ by its supported part: this preserves its probability and can only increase its distance. Thus assume all coordinate atoms have positive mass. The induction starts with zero coordinates, when a positive-probability event is the singleton space and its distance is zero.

Split the last coordinate into values $j$, with probabilities $q_j$, and write $A_j$ for the corresponding section in the first $n-1$ coordinates. Put $a_j=\mathrm{Pr}(A_j)$, choose a largest section $A_s$ of probability $a>0$, and set $r_j=a_j/a\in [0,1]$. For a point $(x^{\prime },j)$, mix nearest mismatch vectors from $A_j$ and $A_s$ with weights $\lambda ,1-\lambda$. Their final coordinates are zero and, if $j\ne s$, one. Convexity of the squared norm givesFor an empty $A_j$, use $\lambda =0$ directly. If $j=s$, the final-coordinate term is an unnecessary nonnegative allowance, so the bound still holds.

Exponentiate, integrate in $x^{\prime }$, and use Hölder's inequality and the induction hypothesis. For $r_j>0$ this givesWe supply the scalar estimate for Talagrand product induction used in the last step. Set $L=-\log r$. For $0\le L\le 1/2$, choose $1-\lambda =2L$, making the expression $e^{L-L^2}$. The functionhas $h(0)=h^{\prime }(0)=0$ andThus $e^{L-L^2}\le 2-e^{-L}$. For $L\ge 1/2$, choose $\lambda =0$, giving $e^{1/4}$; the already proved boundary case and monotonicity of $2-e^{-L}$ give the same bound. For $r=0$, the direct empty-section bound is $e^{1/4}/a\le 2/a$.

Now average over $j$. With $m={\sum }_j{q}_j{r}_j=\mathrm{Pr}(A)/a\in (0,1]$, the exponential integral is at most $(2-m)/a$. Since $m(2-m)\le 1$, this is at most $1/(am)=1/\mathrm{Pr}(A)$, closing the induction.

Finally apply the Markov inequality to the nonnegative exponential. For the distance-tail event ${\overline{A}}_t$,This proves Talagrand's convex distance inequality for the finite product. If $\mathrm{Pr}(A)=0$, the required probability-product inequality is immediate, including the empty-event case with distance defined as infinity; no finite exponential-moment assertion is needed there.

Write $F=f_2$. Split the integral in the assumed tail bound according to whether $g\le a/2$ or $g>a/2$:The finite-superlevel hypothesis permits subtraction of the first term. With $a=2b$, this becomesThus $F/2$ satisfies part (i), which proves finiteness and the preliminary estimateNow integrate the original tail bound against $p{a}^{p-2}$. The layer cake representation and Tonelli theorem givewhere the second step is Hölder's inequality. The preliminary estimate justified the finiteness needed here. If $\Vert F{\Vert }_p=0$ the result is immediate; otherwise divide by its $(p-1)$st power. This proves the Lp bound from a tail domination inequality:The printed strict signs cannot hold when $g=F=0$: both sides are zero. The universally valid formulation therefore has non-strict inequalities. If $\Vert g{\Vert }_p>0$, the sharper estimate is actually strict, as follows.

Suppose equality held and $\Vert F{\Vert }_p>0$. Equality in Hölder's inequality and the norm relation force $g=F/q$ almost everywhere; equality in the preceding integrated tail inequality forces equality in that tail inequality for almost every $a>0$. Put $H(a)=\mu \{F>a\}$ and $J(a)={\int }_a^{\infty }H(t)dt$. The layer-cake identity on $\{F>a\}$ gives ${\int }_{\{F>a\}}F=aH(a)+J(a)$, so equality impliesBecause $F\in L^p$ and $p>1$, $J$ is finite for positive $a$, locally absolutely continuous, and $J^{\prime }=-H$ almost everywhere. It follows that $J^{\prime }=-(p-1)J/a$, whose solutions on $(0,\infty )$ are $J(a)=c{a}^{1-p}$. If $c>0$, then $H(a)=c(p-1)a^{-p}$ almost everywhere, and $p{\int }_0^{\infty }{a}^{p-1}H(a)da$ diverges. If $c=0$, then $F=0$ almost everywhere. Both contradict the assumed nonzero finite equality case. If $F=0$ but $g\ne 0$, strictness is immediate.

Thus both requested strict estimates hold when $\Vert g{\Vert }_p>0$: the coarse constant is strictly larger than the sharp one since $q^{1/q}\le e^{1/e}<2$, so $q^p<2^{p}q$. The zero-data exception is the only qualification needed here.

Let $Mf$ be the uncentered maximal function of $|f|$, defined by its averages over balls. First establish the strong $L^p$ estimate needed below. The uncentered maximal weak-type inequality gives the weak $L^1$ bound with $C_d=3^d$, and plainly $Mf\le \Vert f{\Vert }_{\infty }$ for bounded inputs. For $f\in L^p$, split it into its parts on $\{|f|>a/2\}$ and $\{|f|\le a/2\}$. The first part is in $L^1$, since $|f|1_{\{|f|>a/2\}}\le (a/2{)}^{1-p}|f{|}^p$, and the second part contributes at most $a/2$ to the maximal function. Sublinearity therefore givesIntegrating this bound with layer cake representation and Tonelli theorem yieldsThis proves the strong Lp bound from weak L1 and L-infinity bounds directly. In particular $Mf$ is finite almost everywhere.

Now split the positive Riesz potential of $|f|$ at a radius $r$. In the near region, decompose into annuli $2^{-j-1}r<|x-y|\le 2^{-j}r$. On the $j$th annulus,The geometric series converges because $\alpha >0$. In the far region, Hölder's inequality, with $p^{\prime }=p/(p-1)$, givesThe radial integral is finite exactly because $p<d/\alpha$. Combining these bounds, including the fixed Riesz normalization constant, givesFor $0<Mf(x)<\infty$, choose $r=(\Vert f{\Vert }_p/Mf(x){)}^{p/d}$. If $Mf(x)=0$, every ball average vanishes and $f=0$ almost everywhere, so the assertion is trivial. The zero-norm case is likewise trivial. The resulting Hedberg inequality isThis proves absolute convergence of the defining integral for almost every $x$, also for signed or complex $f$. Put $\theta =\alpha p/d\in (0,1)$. The proposed exponent satisfies $q=p/(1-\theta )$, so $(1-\theta )q=p$. Raising the pointwise estimate to the $q$th power and using the proved maximal bound givesThus the Hardy–Littlewood–Sobolev inequality follows from the estimates just derived, rather than being quoted in place of its proof.

For uniqueness of the target exponent, suppose such a uniform estimate holds with index $r$ instead. Choose a nonzero nonnegative bump function $f$ and let $f_t(x)=f(tx)$ for $t>0$. Homogeneity of the Riesz potential givesThe assumed estimate at $t=1$ makes the nonzero output's $L^r$ norm finite. For all $t>0$ it must therefore satisfywith $d/r=0$ when $r=\infty$. Letting $t$ tend to zero or infinity forces the power of $t$ to vanish:Hence no other target index is possible.

For the Sobolev embedding, take $d>1$ and $1<p<d$, so the preceding result is available with $\alpha =1$. For $u\in C_c^{\infty }({\mathbb{R}}^d)$, the vector field $(x-y)/|x-y{|}^d$ has zero divergence in $y$ away from $x$. Integrating by parts outside the ball $|x-y|\le \epsilon$, its inner boundary contributes $|{\mathbb{S}}^{d-1}|$ times the spherical average of $u$ about $x$. The outer boundary contributes zero because $u$ has compact support. Letting $\epsilon \downarrow 0$ provesThe absolute value is at most $C{I}_1(|\nabla u|)(x)$. ThusFor a general member of the first-order Sobolev space $W^{1,p}({\mathbb{R}}^d)=L_1^p({\mathbb{R}}^d)$, approximate in the Sobolev norm by compactly supported smooth functions. This density of smooth functions in a Sobolev space follows from cutoffs tending to one and subsequent mollification: the cutoff-gradient term tends to zero in $L^p$, and a mollifier approximates both the function and each weak derivative. The displayed estimate makes the approximants Cauchy in $L^{q_1}$. Their $L^p$ limit is the original function, so subsequences converging almost everywhere identify the $L^{q_1}$ limit with it. This proves continuous embedding. The range $1<p<d$ is essential for this consequence; the formula is not a general critical $p=d$ embedding into $L^{\infty }$.