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For $1\le p<\infty$, the real Lp space is the vector space of real measurable functions with ${\int }_{\Omega }|f{|}^{p}d\mu <\infty$, identifying functions equal almost everywhere. Its Lp norm is $\Vert f{\Vert }_p=(\int |f{|}^{p}d\mu {)}^{1/p}$. For $p=\infty$, take the essentially bounded real measurable functions, with the same identification and norm $\Vert f{\Vert }_{\infty }=\mathrm{esssup}|f|$. The identification makes each norm definite; absolute homogeneity follows from the Lebesgue integral, and the triangle inequality follows from Minkowski inequality for finite $p$, or directly from the essential supremum for $p=\infty$.

Here is a completeness proof valid on any measure space. For $p<\infty$, a Cauchy sequence $(f_n)$ has a subsequence $(f_{n_k})$ with $\Vert f_{n_{k+1}}-f_{n_k}{\Vert }_p\le 2^{-k}$. Choose measurable function representatives and put $h_k=f_{n_{k+1}}-f_{n_k}$. By Minkowski inequality and the monotone convergence theorem,Consequently the series $\sum h_k$ converges absolutely almost everywhere. Define $f=f_{n_1}+\sum h_k$ there, and define it to be zero on the measurable exceptional null set. Then $f\in L^p$, and Fatou lemma applied to each tail gives $\Vert f-f_{n_k}{\Vert }_p\le {\sum }_{j=k}^{\infty }{2}^{-j}\to 0$. The original Cauchy sequence also converges in Lp norm, by the triangle inequality. For $p=\infty$, choose the same subsequence using the essential supremum norm. Outside one measurable null set, all the bounds $|h_k|\le 2^{-k}$ hold and $f_{n_1}$ is bounded. The series then converges uniformly there, with an essentially bounded measurable function limit and the same tail estimate in essential supremum norm. Thus all these spaces are Banach spaces.

The duality of Lp spaces says that, for $1<p<\infty$ and $q=p/(p-1)$, the mapis an isometric isomorphism of normed spaces. This form of Lp duality on an arbitrary measure space requires no finiteness hypothesis on $\mu$. At $p=1$, a standard version assumes a sigma-finite measure and identifies $(L^1{)}^{\ast }$ with $L^{\infty }$ through the same dual pairing. That endpoint assertion must not be made without a suitable measure-space hypothesis.

We first prove the required duality of Lp spaces for a finite measure. Let $F\in (L^p{)}^{\ast }$ and set $\nu (E)=F(1_E)$. For disjoint measurable $E_j$, the indicator functions of their partial unions converge in Lp norm to that of their union, so $\nu$ is countably additive. It has finite variation measure: for every finite measurable partition $(E_j)$, choosing real signs givesAlso $\mu (E)=0$ implies $\nu (E)=0$. The Radon-Nikodym theorem supplies a Radon-Nikodym derivative $g\in L^1$ with $\nu (E)={\int }_{E}gd\mu$. Linearity gives $F(f)=\int fgd\mu$ for simple functions. Uniform approximation by simple functions extends this identity to bounded measurable functions: both their Lp norm errors and the errors in integration against $g$ tend to zero.

To establish the correct integrability, test with the bounded measurable function $f_N=\mathrm{sgn}(g)|g{|}^{q-1}{1}_{\{|g|\le N\}}$. Since $(q-1)p=q$, writing $A_N={\int }_{\{|g|\le N\}}|g{|}^{q}d\mu$ givesThe second inequality is also valid when $A_N=0$. The monotone convergence theorem gives $g\in L^q$ and $\Vert g{\Vert }_q\le \Vert F\Vert$. Density of simple functions in the Lp space, together with Hölder's inequality, now gives $F=I_{p}g$ on all of $L^p$. Conversely Hölder's inequality gives $\Vert I_{p}g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, testing againstgives $\Vert f{\Vert }_p=1$ and $(I_{p}g)(f)=\Vert g{\Vert }_q$, so $\boxed{\Vert I_{p}g\Vert =\Vert g{\Vert }_q}$. This also proves uniqueness of the representing Radon-Nikodym derivative.

For completeness, the passage to an arbitrary measure space can be made without losing a hypothesis in the reflexive Banach space argument below. On a sigma-finite measure space, exhaust by nested finite-measure sets $E_n$. The representing Radon-Nikodym derivatives on $E_n$ agree on overlaps by uniqueness. Their glued density has Lp norm $\Vert g{\Vert }_q\le \Vert F\Vert$ by the monotone convergence theorem, and represents $F$ because $f{1}_{E_n}\to f$ in Lp norm. Every $f\in L^p$ on an arbitrary measure space is supported on a sigma-finite measurable set: the sets $\{|f|>1/n\}$ have finite measure and their union is $\{f\ne 0\}$.

Use support localization of an Lp functional as follows. For a sigma-finite measurable $A$, let $m_A$ be the operator norm of $F$ restricted to functions supported in $A$. The support observation gives ${\sup }_A{m}_A=\Vert F\Vert$. Choose $A_n$ approaching this supremum and set $A={\bigcup }_n{A}_n$; then $m_A=\Vert F\Vert$. If a sigma-finite $B\subseteq \Omega \setminus A$ had $m_B=\beta >0$, functions supported on the disjoint sets $A,B$ have the direct-sum Lp norm. Optimizing their two scalar coefficients by Hölder's inequality, and using functions approaching the two restriction operator norms, would givea contradiction. Thus $F$ vanishes on functions supported outside $A$. The density on $A$, extended by zero, represents $F$ globally. The case $F=0$ simply uses $g=0$. This proves the stated Lp duality on an arbitrary measure space.

Finally let $X=L^p$ and let $\Phi \in X^{\ast \ast }$, where stars denote continuous dual spaces. Compose with $I_p$ to obtain the bounded linear functional $g\mapsto \Phi (I_{p}g)$ on $L^q$. Applying duality of Lp spaces with the exponents reversed gives $f\in L^p$ with $\Phi (I_{p}g)=\int fgd\mu$. For the canonical embedding into the bidual $J_{X}f$, its value on $I_{p}g$ is also $\int fgd\mu$. Since $I_p$ is onto, $\Phi =J_{X}f$. Its norm is $\Vert f{\Vert }_p$ by the same dual pairing norm identity. Therefore $\boxed{J_X(L^p)=(L^p{)}^{\ast \ast }}$, which proves that $L^p$ is a reflexive Banach space through its actual canonical embedding into the bidual.

Hölder's inequality states that for conjugate exponents $1/p+1/q=1$, $f\in L^p$ and $g\in L^q$ imply $fg\in L^1$ andIt includes $p=1,q=\infty$ (and the reversed case), with the essential-supremum norm for $L^{\infty }$.

The Radon-Nikodym theorem for positive measures says: if $\mu$ and $\nu$ are sigma-finite measures on the same measurable space, and $\nu$ is absolutely continuous with respect to $\mu$, then there is a nonnegative measurable function $g$, unique $\mu$-almost everywhere, such thatHere sigma-finiteness means that the space is a countable union of measurable sets of finite measure; absolute continuity of measures, written $\nu \ll \mu$, means that $\mu (E)=0$ implies $\nu (E)=0$. The function $g=d\nu /d\mu$ is the Radon-Nikodym derivative. For a finite signed or complex measure $\nu$ of finite total variation norm of a measure, absolutely continuous with respect to a sigma-finite $\mu$, the corresponding density belongs to $L^1(\mu )$. This follows by applying the positive theorem to the positive and negative parts of the real and imaginary parts of $\nu$.

For every measure space and $1<p<\infty$, $(L^p{)}^{\ast }$ is isometrically $L^q$, where $1/p+1/q=1$. We use the complex-linear pairingWith the convention $\int h\overline{g}$, the same identification is conjugate-linear in $g$. By Hölder's inequality, ${\Phi }_g$ is a bounded linear functional with $\Vert {\Phi }_g\Vert \le \Vert g{\Vert }_q$. If $g\ne 0$, setinterpreting the numerator as zero where $g=0$. The identity $(q-1)p=q$ gives $\Vert h{\Vert }_p=1$ and ${\Phi }_g(h)=\Vert g{\Vert }_q$. ThusIt remains to represent an arbitrary $\Phi \in (L^p{)}^{\ast }$, rather than merely produce functionals from $L^q$.

We prove Lp duality on an arbitrary measure space without imposing sigma-finiteness on $\mu$. Write $C=\Vert \Phi \Vert$. For every measurable set $E$ with $\mu (E)<\infty$, define a complex measure on $E$ byIt is countably additive: for disjoint $F_j\subseteq E$, the partial sums of their indicator functions tend in $L^p$ to $1_{\bigcup F_j}$, because the measure of the omitted tail tends to zero. It is absolutely continuous with respect to $\mu {|}_E$, since indicator functions of null sets represent zero in $L^p$.

Its total variation norm of a measure is finite. For any finite measurable partition $E={\bigcup }_j{E}_j$, choose scalars $c_j$ of modulus $1$ with $c_j{\nu }_E(E_j)=|{\nu }_E(E_j)|$. ThenTaking the supremum over partitions gives the variation bound. Since $\mu {|}_E$ is finite, the Radon-Nikodym theorem supplies $g_E\in L^1(E)$ with ${\nu }_E(F)={\int }_F{g}_{E}d\mu$. By uniform approximation with simple functions,for every bounded measurable $h$ supported in $E$.

To improve $g_E$ from $L^1$ to $L^q$, test with the bounded functionIf $I_N={\int }_{E\cap \{|g_E|\le N\}}|g_E{|}^{q}d\mu$, thenThus $I_N^{1/q}\le C$ when $I_N>0$, and the same bound is trivial when it is zero. The monotone convergence theorem givesIf $E,F$ both have finite measure, the densities $g_E,g_F$ agree almost everywhere on $E\cap F$: their integrals over every measurable subset of the intersection equal the same functional value. This is uniqueness in the Radon-Nikodym theorem.

We now perform support localization of an Lp functional. SetChoose finite-measure sets $E_n$ whose displayed integrals tend to $M$, and let $D_n=E_1\cup \cdots \cup E_n$, $S={\bigcup }_n{D}_n$. If $M=0$, take $E_n=\varnothing$. Compatibility allows us to define a measurable function $g$ on $S$ by taking $g_{D_n}$ on the disjoint measurable sets $D_n\setminus D_{n-1}$, and put $g=0$ off $S$. It agrees almost everywhere with $g_{D_n}$ on every $D_n$. Moreover,Indeed each integral is at most $M$, and it is at least the integral over $E_n$, which tends to $M$.

For any finite-measure set $F\subseteq \Omega \setminus S$, compatibility on the disjoint union $D_n\cup F$ givesLetting $n\to \infty$ forces $g_F=0$ almost everywhere. For an arbitrary finite-measure set $E$, compatibility on $E\cap D_n$ and the preceding conclusion on $E\setminus S$ show that $g_E=g$ almost everywhere on $E$. Consequently $\Phi (h)=\int hgd\mu$ for every simple function supported on a finite-measure set.

Those simple functions are dense in $L^p$ even for this arbitrary measure space. To see the needed finite-support property, for $h\in L^p$ the sets $\{|h|>1/n\}$ have finite measure, bounded by $n^p\Vert h{\Vert }_p^p$. First truncate $h$ to such sets and to bounded values, then approximate by simple functions; the discarded $L^p$ integral tends to zero. Continuity of $\Phi$ and Hölder's inequality therefore extend the representation to every $h\in L^p$.

We have constructed $g\in L^q$ with $\Phi ={\Phi }_g$, and the previously proved norm identity gives $\Vert g{\Vert }_q=\Vert \Phi \Vert$. It also proves uniqueness: if ${\Phi }_g={\Phi }_{\tilde{g}}$, then $\Vert g-\tilde{g}{\Vert }_q=0$. This completes the isometric duality of Lp spaces.

The dominated sequence actually converges to zero in norm, and hence weakly. The assumptions give $|f_n{|}^p\le |f{|}^p\in L^1$ and $|f_n{|}^p\to 0$ almost everywhere. The dominated convergence theorem yieldsFor every $\Phi \in (L^p{)}^{\ast }$, $|\Phi (f_n)|\le \Vert \Phi \Vert \Vert f_n{\Vert }_p\to 0$. Therefore

Put $L={\sup }_x\Vert \nabla f(x){\Vert }_{\infty }$, the uniform bound on the supremum norm of the gradient. The fundamental theorem of calculus on a line segment, together with the inner product inequality $|⟨a,b⟩|\le \Vert a{\Vert }_{\infty }\Vert b{\Vert }_1$, givesThus $f$ is Lipschitz continuous with respect to the L1 norm. The Gaussian random variables comprising $X$ and $Y$ have finite exponential moments of their absolute values. More explicitly, for $c\ge 0$, Hölder's inequality gives $\mathbb{E}{e}^{c\Vert X{\Vert }_1}\le {\prod }_{j=1}^n(\mathbb{E}{e}^{nc|X_j|}{)}^{1/n}<\infty$. Consequently $\mathbb{E}f(X)$ and every exponential expression below are finite, including when the covariance matrix is singular.

Use the convex function $\Psi (v)=e^{\lambda |v|}$. Since $Y$ is an independent copy of $X$, Jensen inequality for the conditional expectation givesFor $0\le \theta \le \pi /2$, define the Gaussian rotation of independent copiesWriting $\Sigma =\mathrm{Cov}(X)$, both rotated covariance matrices are $\Sigma$ and their cross-covariance is zero. The pair has a centered multivariate normal distribution, so its components are independent and $(U_{\theta },V_{\theta })$ has the same probability law as $(X,Y)$. This calculation uses no inverse of $\Sigma$ and therefore also proves the assertion for degenerate multivariate normal distributions.

The path $\theta \mapsto U_{\theta }$ runs from $Y$ to $X$. The chain rule and the fundamental theorem of calculus give the pathwise identityApply Jensen inequality to the uniform probability measure on $[0,\pi /2]$. Then use Tonelli theorem and the Gaussian rotation of independent copies to obtainThis proves the Gaussian rotation interpolation inequality in the required form:The stated almost-sure domination now gives an exponential moment of an absolute standard normal variable bound. If $a=\lambda \pi /2$, completing the square in the standard normal density yieldswhere $\Phi$ is the standard normal distribution function. Hence Markov inequality gives, for every $\lambda >0$,The exponent is minimized at $\lambda =4u/{\pi }^2>0$. Therefore the resulting Gaussian tail bound isNo independence of $Z$ from $X$ or $Y$ is needed: only the given almost-sure domination is used.

Let $q$ be conjugate to $p$, so $1/q=1-1/p$, including $q=1$ for $p=\infty$. Hölder's inequality implies, for every measurable finite-measure set $E$,This proves integrability on $A$ and gives uniform integrability there. Choose a set $B\subseteq A$ with $\lambda (A\setminus B)\le \delta$ on which convergence is uniform, using Egorov theorem. ThenFirst take $n\to \infty$ and then $\delta \downarrow 0$. Thus even the local $L^1$ difference tends to zero, and in particularFor $1<p<\infty$, simple functions supported on finite-measure Borel sets are dense in $L^q(\mathbb{R})$. The displayed convergence proves pairing convergence for such functions. Approximate any $h\in L^q$ by a simple $h_0$ and useBy the duality of Lp spaces, these are all continuous linear functionals on $L^p$, so $f_n$ converges weakly to $f$. It need not converge strongly: $f_n=1_{[n,n+1]}$ converges pointwise to zero but has $L^p$ norm one. For $p=\infty$ the pairing argument gives weak-star convergence against $L^1$, rather than asserting weak convergence against every functional in the much larger full dual.

For conjugate exponents $1<p,q<\infty$, $1/p+1/q=1$, Hölder's inequality is $\boxed{\mathbb{E}|XY|\le \Vert X{\Vert }_p\Vert Y{\Vert }_q}$. If either finite norm is zero, the corresponding variable is zero almost surely. Otherwise normalize $u=|X|/\Vert X{\Vert }_p$, $v=|Y|/\Vert Y{\Vert }_q$. Young inequality $uv\le u^p/p+v^q/q$ follows by minimizing the right side minus $uv$ in $u\ge 0$, whose derivative is $u^{p-1}-v$. Integration gives $\mathbb{E}(uv)\le 1/p+1/q=1$, proving the result. Infinite norms give the usual extended-value inequality whenever the right side is unambiguous. At the endpoints, $\mathbb{E}|XY|\le \Vert X{\Vert }_1\Vert Y{\Vert }_{\infty }$ follows from the almost-sure bound for $Y$, with the symmetric version for $p=\infty$.

The relation $1/q=1/p-\alpha /n>0$ implies $\alpha p<n$. Fix $r>0$ and split the Riesz potential into the regions $|x-y|<r$ and $|x-y|\ge r$.

For the near part, decompose the ball into the dyadic annuliThe definition of the Hardy-Littlewood maximal function then givesFor the far part, Hölder's inequality giveswhere convergence follows from $\alpha p<n$.

If $Mf(x)>0$, chooseThe two bounds then have the same size. If $Mf(x)=0$, the result is immediate. Consequently the Hedberg inequality isHere the dimensional constant is required for the unnormalized kernel and Lebesgue measure used in the question.

A multiset $\mathcal{C}$ of subsets of $[n]$ is a uniform cover of multiplicity $k$ when each coordinate occurs in exactly $k$ members. The uniform covers theorem states that every Euclidean body $S\subseteq {\mathbb{R}}^n$ satisfieswhere $S_A$ is the coordinate projection of a Euclidean body onto the coordinates in $A$.

We prove it by induction on $n$. Split the cover into ${\mathcal{C}}^{-}$, whose members omit $n$, and ${\mathcal{C}}^{+}$, whose members contain $n$. Exactly $k$ members lie in ${\mathcal{C}}^{+}$. For a last-coordinate value $x$, let $S(x)\subseteq {\mathbb{R}}^{n-1}$ be the corresponding slice. Removing $n$ from the members of ${\mathcal{C}}^{+}$ and retaining the members of ${\mathcal{C}}^{-}$ gives a $k$-uniform cover of $[n-1]$. The inductive hypothesis and Fubini's theorem giveApply Hölder's inequality to the $k$ factors in the integral. Sincewe obtain $|S|\le {\prod }_{A\in \mathcal{C}}|S_A{|}^{1/k}$, which is the result after taking the $k$th power. The one-dimensional base case is immediate.

The Bollobas--Thomason box theorem states that for every body $S\subseteq {\mathbb{R}}^n$ there is an axis-parallel box $B$ such thatAn irreducible uniform cover cannot be decomposed into two smaller uniform covers. There are only finitely many such covers of $[n]$: encode a cover by its multiplicity vector in ${\mathbb{N}}^{2^n}$ and apply the Dickson lemma.

Choose a componentwise minimal positive array $(x_A{)}_{A\subseteq [n]}$ satisfyingfor every irreducible $k$-uniform cover $\mathcal{C}$, together withThe actual projection volumes are feasible by the uniform covers theorem, and finiteness gives a minimal array. Every uniform cover is a disjoint union of irreducible ones, so its cover inequality also holds for this array.

Minimality implies that, for each coordinate $i$, some tight uniform-cover inequality can be chosen whose cover contains the singleton $\{i\}$. Indeed, either such an inequality already blocks decreasing $x_{\{i\}}$, or a tight product inequality $x_A={\prod }_{j\in A}{x}_{\{j\}}$ does; in the latter case take a tight cover containing $A$ and replace that occurrence of $A$ by its singleton coordinates. Let these tight covers be ${\mathcal{C}}_i$, of multiplicities $k_i$, and let $K={\sum }_i{k}_i$. Their multiset union is a $K$-uniform cover. Removing one copy of every singleton leaves a $(K-1)$-uniform cover, so comparison of its cover inequality with the product of all the tight equalities yieldsThe singleton cover gives the reverse inequality, henceFor any $A\subseteq [n]$, the one-uniform cover consisting of $A$ and the singletons $\{i\}$ for $i\notin A$ now gives $x_A\ge {\prod }_{i\in A}{x}_{\{i\}}$. The defining product inequality gives the reverse bound. Thus all these quantities are equal. Taking the side lengths of $B$ to be $x_{\{1\}},\dots ,x_{\{n\}}$ proves the theorem.

Finally suppose that the proper body $S\subseteq {\mathbb{R}}^3$ satisfiesThis is equality in the three-dimensional Loomis--Whitney inequality. In its proof, equality must hold in both applications of Cauchy-Schwarz inequality. Their equality conditions force the three projection indicators to factor through one-dimensional measurable sets $E_1,E_2,E_3$, and forceup to a set of Lebesgue measure zero; this is Equality in the three-dimensional Loomis--Whitney inequality.

Because $S$ is connected, each one-coordinate projection is connected and hence is an interval. Because $S$ is a finite union of positive-volume axis-parallel boxes, a proper difference between $S$ and the product of those three intervals would contain a positive-volume rectangular cell in a common finite subdivision. That would contradict equality up to measure zero. Consequently the equality is exact and

Let $T=1_A\ast 1_A$ and chooseApply part b with its error parameter replaced by a sufficiently small absolute multiple of $\epsilon$. This produces a vector subspace $V$ of codimensionwith the evident harmless modification when $\alpha =1$.

For $t\in V$, $x\in {\mathbb{F}}_p^n$, and $q=2m$, Hölder's inequality givesSince ${\alpha }^{-1/q}\le e^{1/2}$ by the choice of $m$, the bound from part b is at mostafter absorbing the absolute factor into the chosen error parameter. This is the required uniform estimate for $1_A\ast 1_A\ast 1_A$.

For $g\in {\mathcal{H}}_N$, writeThe orthogonality of complex exponentials converts the linear configuration count intoExpand the difference between the products for $g_1$ and $g_2$ by changing one factor at a time. A typical term iswhere each $h_j$ is either $g_1$ or $g_2$. The assumed uniform norm bound controls the first factor by $\epsilon N$. The substitution $\alpha \mapsto k\alpha$ preserves an integral over the circle group, so Hölder's inequality and the three supplied $L^3$ bounds giveEach of the four terms is therefore $O(\epsilon N^3)$, and henceThis is Fourier stability of a linear configuration count.

Write $\delta =\hat{\beta }-{\beta }^0$. On $\Omega$, Hölder's inequality givesSince ${\beta }_N^0=0$, the triangle inequality givesSubstitution in the Basic inequality for the Lasso, followed by discarding the nonnegative prediction-error term, yieldsTherefore $\Vert {\hat{\beta }}_N-{\beta }_N^0{\Vert }_1\le 3\Vert {\hat{\beta }}_S-{\beta }_S^0{\Vert }_1$, the Lasso cone condition.

For a nonnegative $f$, first set $g_n(x)=\min (f(x),n)1_{(1/n,n)}(x)$ and $F_n(x)={\int }_0^x{g}_n(t)dt$. These bounded functions vanish near zero and have compact support. Their averages have finite Lp norms: they vanish near zero and decay as $1/x$ beyond $n$. Integration by parts givesThe boundary term $x^{1-p}{F}_n(x{)}^p$ vanishes at zero because $F_n$ is zero there, and at infinity because $F_n$ is eventually constant and $p>1$. The inequality is Hölder's inequality with conjugate exponents $p$ and $p/(p-1)$. Divide when $I_n>0$; the zero case is immediate. Since $g_n\uparrow f$ and $H{g}_n\uparrow Hf$, the monotone convergence theorem proves the full bound without presupposing that $Hf$ has finite Lp norm. Apply it to $|f|$ for the signed or complex case.^[ref]

Let the largest last-coordinate section have probability $a$, and another section probability $ar$. Mix mismatch vectors from these two sections with weights $\lambda ,1-\lambda$. Their squared norm is at most $\lambda d_1^2+(1-\lambda )d_2^2+(1-\lambda {)}^2$. Hölder's inequality and induction bound that section's exponential integral by $a^{-1}{\inf }_{\lambda }{r}^{-\lambda }{e}^{(1-\lambda {)}^2/4}$. The scalar estimate for Talagrand product induction makes it at most $(2-r)/a$. Average over sections and use $m(2-m)\le 1$ with $m=\mathrm{Pr}(A)/a$ to close the induction.

If $\mathcal{C}$ is a $k$-uniform cover of $[n]$ and $S\subseteq {\mathbb{R}}^n$ is a Euclidean body, thenSlicing in one coordinate, applying the theorem inductively to each slice, and then applying Hölder's inequality to the $k$ projected slice functions proves the inequality.

Exact covering is essential for unrestricted real volumes. Finite subsets of a Cartesian product satisfy the analogous cardinality inequality, and in that discrete version covering at least $k$ times suffices.

For $p\ge 2$ and a continuous local martingale starting at zero, the Itô formula for $|X{|}^p$, the Doob Lp maximal inequality and Hölder's inequality yield $\mathbb{E}{\sup }_{s\le t}|X_s{|}^p\le C_p\mathbb{E}⟨X{⟩}_t^{p/2}$. One possible constant is $C_p=[p(p-1)(p/(p-1){)}^p/2{]}^{p/2}$. Stop at increasing absolute-value levels to remove an initial boundedness assumption. This proves the upper half of the Burkholder-Davis-Gundy inequalities in this range.