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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.

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.

For conjugate exponents $p,q\in [1,\infty ]$, Hölder's inequality states $\Vert fg{\Vert }_1\le \Vert f{\Vert }_p\Vert g{\Vert }_q$. For finite vectors, the case $p=1$, $q=\infty$ is $|x^{T}y|\le \Vert x{\Vert }_1\Vert y{\Vert }_{\infty }$.

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.

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 $L^{\infty }$.

For $d\ge 2$, write ${\hat{x}}_j$ for the $(d-1)$ coordinates other than $x_j$, and setBy Tonelli theorem, $\Vert S_j{\Vert }_1=\Vert s_j{\Vert }_1$. The key estimate is the functional Loomis-Whitney inequality:Here is a proof in every dimension. In dimension two, the left-hand side separates into a product of two one-dimensional integrals, so Tonelli theorem gives equality. For $d\ge 3$, first integrate $x_d$. Apply Hölder's inequality with $d-1$ factors to $S_1,\dots ,S_{d-1}$, which may depend on $x_d$, and leave the factor $S_d$ outside that integral. WriteEach $H_j$ depends on the $d-2$ coordinates other than $j,d$. With $x^{\prime }=(x_1,\dots ,x_{d-1})$, the full integral is at mostPut $G(x^{\prime })={\prod }_{j<d}{H}_j^{1/(d-2)}$. A second application of Hölder's inequality, with exponents $d-1$ and $(d-1)/(d-2)$, bounds this expression byBy induction in dimension $d-1$, $\int G\le {\prod }_{j<d}(\int H_j{)}^{1/(d-2)}$. Since $\int H_j=\int S_j$, substitution proves the claimed functional inequality. Nonnegative integrands justify all integrations; truncation gives the extended-valued version if necessary.

The pointwise hypotheses imply $f^d\le {\prod }_j{S}_j$, soApply the proved functional inequality and take the $(d-1)/d$ power to obtainFor the useful case all the right-hand norms are finite. If one of them is zero, its line integral vanishes for almost every transverse coordinate and forces $f=0$ almost everywhere; there is no division by that factor. In dimension one the corresponding direct assertion is $\Vert f{\Vert }_{\infty }\le \Vert s_1{\Vert }_1$.

For odd $p$, write $e_p(t)=e^{2\pi it/p}$ and use the discrete paraboloidHere surface integration means $p^{-2}{\sum }_P$, whereas ambient norms below use unnormalized counting measure on ${\mathbb{F}}_p^3$. With the negative-phase transform convention,When $x_3=0$, additive-orthogonality of characters makes this one at $x=0$ and zero at all other points of that plane. When $x_3\ne 0$, completing the two squares givesThe Quadratic Gauss sum identity is $G(t{)}^2=\chi (-1)p$, where $\chi$ is the Legendre symbol. For completeness, counting the preimages of a square gives $G(t)=\chi (t)G(1)$ for nonzero $t$. Also $|G(1){|}^2={\sum }_{u,v}{e}_p(u^2-v^2)=p$, by the invertible change $(u,v)\mapsto (u-v,u+v)$ and orthogonality of characters. Since $\overline{G(1)}=G(-1)=\chi (-1)G(1)$, the asserted square identity follows. Thus for $p\equiv 3(\mathrm{mod}4)$,Using positive phase instead changes the sign inside the character, not the magnitude or any norm estimate.

Define $Eg(x)=p^{-2}{\sum }_{\xi \in P}g(\xi )e_p(x\cdot \xi )$, and define $R^{\ast }(2\to q)$ as the smallest constant forThe adjoint relative to these measures is $E^{\ast }h(\xi )={\sum }_{x}h(x)e_p(-x\cdot \xi )$. Consequently $E{E}^{\ast }h=h\ast \hat{\sigma }(-\cdot )$, where convolution is the unnormalized ambient sum. Put $K=\hat{\sigma }(-\cdot )-1_{\{0\}}$. Its supremum is at most $p^{-1}$, so convolution by $K$ has ${\ell }^1\to {\ell }^{\infty }$ norm at most $p^{-1}$.

Under the unnormalized ambient Fourier transform on a finite group, the convolution multiplier of $\hat{\sigma }(-\cdot )$ is $p^3\sigma =p{1}_P$, up to the harmless reflection convention. Thus the multiplier of $K$ is $p{1}_P-1$, with maximum absolute value at most $p$. The Plancherel theorem gives ${\ell }^2\to {\ell }^2$ norm at most $p$. The Riesz-Thorin interpolation theorem between these two bounds gives ${\ell }^{4/3}\to {\ell }^4$ norm at most one for convolution by $K$. Convolution by $1_{\{0\}}$ is the identity, whose ${\ell }^{4/3}\to {\ell }^4$ norm is at most one for counting measure. Hence $\Vert E{E}^{\ast }h{\Vert }_4\le 2\Vert h{\Vert }_{4/3}$.

By Lp duality and Hölder's inequality, $\Vert E^{\ast }h{\Vert }_{L^2(\sigma )}^2=⟨E{E}^{\ast }h,h⟩\le 2\Vert h{\Vert }_{4/3}^2$. Taking adjoints proves the finite-field paraboloid fourth-moment extension estimateIn fact this argument works for both residue classes of odd primes.

If $p\equiv 1(\mathrm{mod}4)$, choose $i\in {\mathbb{F}}_p$ with $i^2=-1$. The complete line $\ell =\{(t,it,0):t\in {\mathbb{F}}_p\}$ lies in $P$. For $g=1_{\ell }$ the normalized surface $L^2$ norm is $p^{-1/2}$, and orthogonality of characters gives $Eg(x)=p^{-1}{1}_{\{x_1+i{x}_2=0\}}$. That annihilator plane has $p^2$ points, so the isotropic-line obstruction to finite-field restriction givesThis tends to infinity along these primes for every $q<4$, as required.

For $x\in \Omega$ and nonempty $A$, letThe Talagrand convex distance isEquivalently,For completeness, let $v_{\ast }$ be a closest point of the finite convex hull to zero. If $v_{\ast }\ne 0$, minimality gives $v_{\ast }\cdot (v-v_{\ast })\ge 0$ for every $v\in V$. Taking $\alpha =v_{\ast }/\Vert v_{\ast }\Vert$ proves that the supremum is at least $\Vert v_{\ast }\Vert$. Conversely each permitted $\alpha$ has ${\inf }_{v\in V}\alpha \cdot v\le \alpha \cdot v_{\ast }\le \Vert v_{\ast }\Vert$ by Cauchy-Schwarz. The weights are nonnegative because the hull has nonnegative coordinates. If $v_{\ast }=0$, both sides are zero. This proves the dual definition used below.

We prove Talagrand's convex distance inequality by induction on the number of factors. Write $\Omega ={\Omega }^{\prime }\times X_n$, with measure $\mu \otimes \nu$, and $F_A=d_T(\cdot ,A{)}^2$. For a last-coordinate value $z$, let $A_z=\{y^{\prime }:(y^{\prime },z)\in A\}$. Choose a section $B=A_b$ of largest probability $m=\mu (B)>0$, and put $a_z=\mu (A_z)/m\in [0,1]$. The total event probability is $p=\mathrm{Pr}(A)=m{\sum }_z\nu (z)a_z$.

If $A_z$ has positive probability, combine mismatch vectors from $A_z$ and from $B$. A weight $t$ on the second section contributes at most $t$ in the last coordinate. Convexity of the squared Euclidean norm gives, for every $0\le t\le 1$,The last-coordinate contribution is actually zero if $z=b$, so the displayed upper bound remains valid. Hölder's inequality and the induction hypothesis yieldFor a zero-probability section use $t=1$ directly, so no division by zero or exponent of an infinite quantity is needed.

The scalar estimate for Talagrand product induction isHere is its proof. For $a>0$ set $L=-\log a$. If $L\le 1/2$, take $t=2L$; the logarithm of the left expression is $L-L^2$. Put $h(L)=\log (2-e^{-L})-L+L^2$. Then $h(0)=h^{\prime }(0)=0$ andsince $(2u-1{)}^2-u=(4u-1)(u-1)\ge 0$ for $u=e^L\ge 1$. Thus $h\ge 0$. If $L\ge 1/2$, choose $t=1$. The already proved boundary case gives $e^{1/4}\le 2-e^{-1/2}\le 2-a$. The same direct choice covers $a=0$.

Average the resulting section bounds over $z$:where the last inequality is equivalent to $(m-p{)}^2\ge 0$. The induction starts on the zero-coordinate singleton, where every positive-probability event is the whole space and the moment is one. HenceWe assume $\mathrm{Pr}(A)>0$; a zero-probability event has no finite right-hand bound to prove. Markov's inequality also gives

For the subsequence application, let $L(x)$ be the longest increasing subsequence length for independent sequence coordinates. A witness of length $\ell$ uses only $\ell$ positions. If $L(x)=\ell \ge b$ and $y$ satisfies $L(y)\le a<b$, at least $\ell -a$ witness positions must change: all unchanged witness positions remain in increasing order. Put weights $1/\sqrt{\ell }$ on the witness and zero elsewhere. The dual distance definition giveswhere the latter expression increases with $\ell$ for $a\ge 0$. Therefore the increasing-subsequence certificate concentration bound isIf the lower event has probability zero the conclusion is immediate.

Let $m$ be an integer median, so $\mathrm{Pr}(L\le m),\mathrm{Pr}(L\ge m)\ge 1/2$. Taking the two thresholds in the bound gives, for $t>0$,For $t>m$ the lower event is empty. In a nonempty sequence $m\ge 1$. Integration of these tails gives an explicit mean-median estimate: use $m+t\le 2m$ for $t\le m$ and $m+t\le 2t$ for $t\ge m$, obtainingWriting $\mu =\mathbb{E}L$, the median property also gives $m\le 2\mu$. Consequently, for a universal constant $C$, all $u\ge 0$ satisfyThus the subsequence length is concentrated near its mean on the scale $\sqrt{\mu }$, rather than the generic coordinate-exposure scale $\sqrt{n}$. This argument applies directly to finite-valued independent sequences. Independent continuous samples follow by refining finite coordinate discretizations; the strict-order pattern stabilizes almost surely when ties have probability zero. A uniform random permutation is the rank sequence of independent continuous samples, so it has the same subsequence-length concentration. Arbitrarily dependent random sequences are not covered by the product-space assumption.

The uniform covers theorem says that if a multiset $\mathcal{C}$ of coordinate subsets covers every coordinate exactly $k$ times, then a Euclidean body $S\subseteq {\mathbb{R}}^n$ satisfieswhere $S_A$ is its coordinate projection of a Euclidean body. Work with bounded Borel bodies, whose projections are Lebesgue measurable sets; an empty-coordinate projection has measure one.

Prove the theorem by induction on dimension, with the one-dimensional case immediate. Slice at the last coordinate, writing $S_t\subseteq {\mathbb{R}}^{n-1}$. The subsets $A\setminus \{n\}$ still cover each remaining coordinate $k$ times, so induction givesFor $A$ not containing $n$, the projected slice is contained in $S_A$, giving a constant bound. For $A$ containing $n$, its measure is the slice measure $f_A(t)$ of $S_A$. Exactly $k$ members of $\mathcal{C}$ contain $n$, counted with multiplicity. Integrate and apply Hölder's inequality to those $k$ factors:Fubini's theorem identifies $\int f_A=|S_A|$, completing the proof. Exact covering is needed for real volumes, which may be smaller than one.