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For conjugate exponents $p,q\in [1,\infty ]$, $1/p+1/q=1$, Hölder's inequality states that $\int |fg|\le \Vert f{\Vert }_p\Vert g{\Vert }_q$. If $r=\infty$ in the requested product estimate, every $p_i=\infty$ and multiplication of essential bounds proves it directly. Suppose $r<\infty$. Factors with $p_i=\infty$ can again be bounded pointwise by their essential suprema.

Here is an explicit finite-exponent proof of the remaining Generalized Holder inequality. Put $q_i=p_i/r$, so ${\sum }_{i}1/q_i=1$, and normalize $h_i=|f_i{|}^r/\Vert f_i{\Vert }_{p_i}^r$. If one norm is zero the product vanishes almost everywhere. Otherwise $\int h_i^{q_i}=1$. The weighted arithmetic-geometric mean inequality, applied to $h_i^{q_i}$ with weights $1/q_i$, givesIntegrating gives $\int {\prod }_i{h}_i\le 1$. If only one finite exponent remains, its weight is one and the same argument is an equality. Restoring the normalization and taking the $r$th root provesIn particular the product belongs to $L^r$.

For the projection estimate, use the three-dimensional functional Loomis-Whitney inequalityThis permitted three-dimensional result also follows directly by Cauchy-Schwarz inequality: integrate the square root of $g_1{g}_2$ in $x_3$, obtaining the square root of the product of their marginal integrals, and then apply Cauchy-Schwarz inequality in $(x_1,x_2)$ against $\sqrt{g_3}$.

Let $f_j$ be any nonnegative integrable function of the three coordinates excluding $x_j$. For $j=1,2,3$, define $G_j$ by integrating $f_j$ in $x_4$. First Hölder's inequality with three equal exponents yieldsThe fourth factor $f_4(x_1,x_2,x_3)$ is independent of $x_4$. A further Hölder's inequality with exponents $3$ and $3/2$, followed by the three-dimensional estimate, therefore givesThe marginal-integral identifications use Tonelli's theorem. Now take $f_j=1_{K_j}$. The projections are compact and hence measurable with finite measure. Every $x\in K$ has all its projected coordinates in the corresponding $K_j$, so $1_K(x)\le {\prod }_j{1}_{K_j}({\hat{x}}_j{)}^{1/3}$. ConsequentlyThis includes zero-measure projections and is sharp for axis-parallel rectangular boxes.

The interpolated exponent is strictly between $p_0$ and $p_1$. Apply Hölder's inequality to $|f{|}^p=|f{|}^{p(1-\theta )}|f{|}^{p\theta }$ with conjugate exponents $p_0/[p(1-\theta )]$ and $p_1/(p\theta )$. Their reciprocals sum to one, soThis gives $L^{p_0}\cap L^{p_1}\subseteq L^p$.

For the other inclusion take any $a>0$ and use the threshold decomposition between Lp spacesBecause $p_0-p<0<p_1-p$,Thus $g\in L^{p_0}$, $h\in L^{p_1}$ and $f=g+h$, proving $L^p\subseteq L^{p_0}+L^{p_1}$. Raising the two norm bounds to powers $1-\theta$ and $\theta$ gives a power of $a$ equal toand a power of $\Vert f{\Vert }_p$ equal to one. HenceFor later use, choose $a=\Vert f{\Vert }_p$ when $f\ne 0$. Each endpoint norm is then at most $\Vert f{\Vert }_p$, so $\Vert f{\Vert }_{L^{p_0}+L^{p_1}}\le 2\Vert f{\Vert }_p$ for the sum of Lp spaces norm. If $f=0$, take $g=h=0$.

The continuous dual space $X^{\ast }$ consists of bounded complex-linear functionals $\varphi :X\to \mathbb{C}$, with $\Vert \varphi \Vert ={\sup }_{\Vert x\Vert \le 1}|\varphi (x)|$. For $1\le s<\infty$, ${\ell }^s$ consists of sequences with ${\sum }_j|x_j{|}^s<\infty$, norm $({\sum }_j|x_j{|}^s{)}^{1/s}$; ${\ell }^{\infty }$ consists of bounded sequences with supremum norm.

For $a\in {\ell }^p$, define ${\varphi }_a(x)={\sum }_j{a}_j{x}_j$ on ${\ell }^q$. Hölder's inequality gives absolute convergence and $\Vert {\varphi }_a\Vert \le \Vert a{\Vert }_p$. For a finite truncation, choose $x_j=\overline{a_j}|a_j{|}^{p-2}/({\sum }_{j\le N}|a_j{|}^p{)}^{1/q}$ for $j\le N$, zero elsewhere, interpreting the numerator as zero when $a_j=0$. Its $q$-norm is one and its pairing equals $({\sum }_{j\le N}|a_j{|}^p{)}^{1/p}$. Letting $N$ grow proves equality of the norms.

Conversely, for a bounded $\varphi$ let $a_j=\varphi (e_j)$. The same finite test vector gives ${\sum }_{j\le N}|a_j{|}^p\le \Vert \varphi {\Vert }^p$, so $a\in {\ell }^p$. Finite truncations converge in ${\ell }^q$, and continuity gives $\varphi (x)={\sum }_j{a}_j{x}_j$. Thus $\boxed{({\ell }^q{)}^{\ast }\cong {\ell }^p\text{ isometrically}.}$

For ${\ell }^1$, boundedness gives $|a_j|\le \Vert \varphi \Vert$, while any bounded sequence defines an absolutely convergent pairing bounded by $\Vert a{\Vert }_{\infty }\Vert x{\Vert }_1$. Density of finite sequences again proves surjectivity, and single-coordinate tests give norm equality. Hence $\boxed{({\ell }^1{)}^{\ast }\cong {\ell }^{\infty }\text{ isometrically}.}$ The reversed argument fails for ${\ell }^{\infty }$: finite sequences are not dense in its supremum norm; their closure is $c_0$. For example the limit functional on convergent sequences extends by the Hahn-Banach theorem to ${\ell }^{\infty }$, vanishes on every $e_j$, yet takes value one on the constant-one sequence. It cannot be represented by pairing with an ${\ell }^1$ coefficient sequence.

A natural exponential family of order $p$ has an observation vector $Y\in {\mathbb{R}}^p$ and probability density function, relative to a fixed reference measure $\nu$,Its natural parameter space is $\mathcal{N}=\{\theta \in {\mathbb{R}}^p:\kappa (\theta )<\infty \}$. A full natural exponential family allows every $\theta \in \mathcal{N}$. A regular natural exponential family has an open parameter space; fullness and regularity together mean that $\mathcal{N}$ itself is open. A minimal exponential family additionally requires that no nonzero linear combination of the components of $Y$ be constant almost surely. This is the condition that the genuinely needed number of canonical sufficient statistics is $p$.

Hölder's inequality shows that $\mathcal{N}$ is convex: for $0<t<1$, its normalizing integral at $t\theta +(1-t)\eta$ is at most the product of the respective integrals to powers $t$ and $1-t$. Taking logarithms also proves convexity of the cumulant function $\kappa$. In the minimal case its Hessian matrix is positive definite on the interior, as the covariance calculation below establishes. Keeping fullness, regularity and minimality separate matters when discussing existence or uniqueness of a maximum-likelihood estimator.

The space $W_0^{1,2}(\Omega )$ is the closure of $C_c^{\infty }(\Omega )$ in the Sobolev norm; membership imposes the homogeneous Dirichlet condition in the weak sense, without assuming that the boundary is smooth. A weak solution is an element of that space satisfyingRepeated indices are summed. The minus sign comes from integration by parts in the divergence equation. Boundedness of the coefficients, Hölder's inequality and density extend this identity to every $\varphi \in W_0^{1,2}(\Omega )$. Symmetry permits interchanging the derivative indices in the left side.

For finite $p>n$, chooseSince $n\ge 3$ and $p>n$, we have $2<q<2^{\ast }=2n/(n-2)$. The zero-boundary versions of the Sobolev embedding theorem and Rellich-Kondrachov compactness theorem say that $W_0^{1,2}(\Omega )$ embeds continuously into $L^{2^{\ast }}$ and compactly into $L^q$ for $q<2^{\ast }$ on a bounded open set. These versions hold on arbitrary bounded open sets by zero extension; a boundary extension operator is not needed.

If $(v_j)$ is bounded in $L^2$, then $(T{v}_j)$ is bounded in $W_0^{1,2}$ by the preceding part. A subsequence consequently converges in $L^q$ to some $z$. Hölder's inequality givesMultiplication by $w$ is bounded from $L^q$ to $L^2$, and the same estimate proves that $K$ is bounded and well defined. Every bounded input sequence has a subsequence whose images converge in $L^2$, so $\boxed{K\text{ is compact}}$. If $p=\infty$ is included, take $q=2$ and use the same argument. This is the compact L2 potential perturbation of the Dirichlet Laplacian mechanism.

Concavity of $\log$ gives $\log (a/p+b/q)\ge p^{-1}\log a+q^{-1}\log b$ because $1/p+1/q=1$. Exponentiating proves Young inequalitywith equality precisely when $a=b$. Equivalently it follows from convexity of the exponential.

Define ${\ell }^p=\{x=(x_j):{\sum }_j|x_j{|}^p<\infty \}$ with $\Vert x{\Vert }_p=({\sum }_j|x_j{|}^p{)}^{1/p}$. Positivity, definiteness and homogeneity are immediate. Summing Young inequality for normalized sequences gives Hölder's inequality $\sum |x_j{y}_j|\le \Vert x{\Vert }_p\Vert y{\Vert }_q$. For a finite truncation $z=x+y$, it yieldsDividing and passing to larger truncations proves Minkowski inequality and shows $x+y\in {\ell }^p$, so this is a normed vector space.

Let $x^{(m)}$ be a norm-Cauchy sequence. Each coordinate is Cauchy because $|x_j^{(m)}-x_j^{(l)}|\le \Vert x^{(m)}-x^{(l)}{\Vert }_p$, and let its limit be $x_j$. The sequence norms are bounded by some $C$. For every finite $N$, passage to coordinate limits gives ${\sum }_{j\le N}|x_j{|}^p\le C^p$, so $x\in {\ell }^p$. Given $\epsilon >0$, choose $M$ with $\Vert x^{(m)}-x^{(l)}{\Vert }_p<\epsilon$ for $m,l\ge M$. Letting $l\to \infty$ in finite truncations and then $N\to \infty$ gives $\Vert x^{(m)}-x{\Vert }_p\le \epsilon$ for $m\ge M$. Thus ${\ell }^p$ is complete, hence a Banach space.

Let $q=p/(p-1)$. The scalar Young inequality $ab\le a^p/p+b^q/q$ follows by maximizing $ab-a^p/p$ over $a\ge 0$, obtaining $b^q/q$. If both norms are nonzero, apply it to $|x_j|/\Vert x{\Vert }_p$ and $|y_j|/\Vert y{\Vert }_q$, and sum over a finite initial segment. Passing to the increasing limit givesZero norms give the trivial case. This proves Hölder's inequality without assuming convergence of the product series in advance.

Apply Hölder's inequality to the finite sum with the second sequence $|x_j+y_j{|}^{p-1}$:Divide by the last power if the sum is nonzero. The resulting bound is independent of $N$, so passing to the limit proves both membership in ${\ell }^p$ and the Minkowski inequality

Use the bilinear pairing $T_g(f)={\int }_{{\mathbb{R}}^n}fg$, so the map $g\mapsto T_g$ is complex linear. With the sesquilinear convention $\int f\overline{g}$ the representing map would instead be conjugate linear. For conjugate exponents $p,q$, Hölder's inequality gives $|T_g(f)|\le \Vert g{\Vert }_q\Vert f{\Vert }_p$. If $g\ne 0$, take $v=|g{|}^{q-2}\overline{g}$, defined as zero where $g=0$. ThenConsequently $\Vert T_g\Vert =\Vert g{\Vert }_q$. This also proves injectivity of the representation.

For surjectivity, let $\ell$ be a nonzero element of the continuous dual space of $L^p$ and choose $f_0$ with $\ell (f_0)=1$. Its kernel $K$ is a closed linear subspace. Use the allowed best-approximation fact to choose $h\in K$ and put $z=f_0-h$. Thus $\ell (z)=1$ andFor each $f$, the vector $f-\ell (f)z$ belongs to $K$. Therefore $\Phi (f)=\ell (f)\Vert z{\Vert }_p^p$, givingMembership follows from $(p-1)q=p$; the zero functional is represented by zero. This proves the isometric isomorphism $(L^p({\mathbb{R}}^n){)}^{\ast }\cong L^q({\mathbb{R}}^n)$ without using a representation theorem as a substitute for the proof. It is the closed-hyperplane proof of Lp duality.

The relevant Banach-Alaoglu theorem says that for $1<p<\infty$ the closed bounded balls of $L^p({\mathbb{R}}^n)$ are compact in the weak topology. In particular, every bounded sequence has a weakly convergent subsequence whose limit has norm no larger than the common bound. Applying the preceding duality of Lp spaces also identifies this as weak-star compactness of $L^p=(L^q{)}^{\ast }$; its weak and weak-star test spaces are both $L^q$.

Here is a direct proof of both compactness and the subsequence assertion. Let $\Vert f_j{\Vert }_p\le M$ and choose a countable dense set $\{g_m\}$ in the separable Lp space $L^q$. Each scalar sequence $\int f_j{g}_m$ is bounded. Successive subsequence selection followed by the diagonal subsequence argument gives $f_{j_k}$ for which these pairings converge for every $m$. For arbitrary $g\in L^q$,First approximate $g$ by $g_m$ and then take $k,l$ large. This proves convergence of every pairing, and the limit $\Lambda (g)={\lim }_k\int f_{j_k}g$ is linear with $|\Lambda (g)|\le M\Vert g{\Vert }_q$. Apply the already proved duality of Lp spaces with $q$ in place of $p$: $\Lambda (g)=\int fg$ for an $f\in L^p$ with $\Vert f{\Vert }_p\le M$. ThusTo obtain compactness rather than only a sequential statement, on the radius-$M$ ball defineDensity and Hölder's inequality show that this is a metric space distance and that its topology is exactly the weak topology on this ball: every remaining test pairing is uniformly approximated there by a dense-set pairing. The subsequence argument proves sequential compactness for this metric. A sequentially compact metric space is compact: otherwise either an infinite separated sequence contradicts total boundedness or a Cauchy sequence without a limit contradicts sequential compactness; completeness and total boundedness give compactness by successive finite coverings. This completes the proof. The same argument is the Sequential Banach-Alaoglu theorem for a separable predual.

One potential form of the Hardy–Littlewood–Sobolev inequality is: if $0<\alpha <n$, $1<p<n/\alpha$, and $1/r=1/p-\alpha /n$, thenA normalization constant for the Riesz potential only changes $C$. Equivalently, for $0<\lambda <n$, $p,s>1$ and $1/p+1/s+\lambda /n=2$,The exponent relation is necessary under dilation. We prove the potential estimate and then derive the bilinear version.

First establish the needed maximal estimate, so it is not an unproved step in the argument. Define the centered Hardy-Littlewood maximal function by $Mf(x)={\sup }_{t>0}|B_t{|}^{-1}{\int }_{B_t(x)}|f|$. For integrable $f$, cover a compact subset of $\{Mf>a\}$ by finitely many balls whose averages exceed $a$. Order these balls by decreasing radius and greedily retain a ball if it is disjoint from those already retained. Every discarded ball intersects a retained ball of at least its radius, so lies in its triple. HenceThe passage from compact subsets uses inner regularity of Lebesgue measure; the superlevel set is open because fixed-radius local averages are continuous. For $f\in L^p$, split $f=f{1}_{\{|f|>a/2\}}+f{1}_{\{|f|\le a/2\}}$. The second summand's maximal function is at most $a/2$, and the first is integrable. ThereforeThe layer cake representation and Tonelli theorem giveThus the Strong Lp bound for the Hardy-Littlewood maximal function is proved for every $p>1$.

For $I_{\alpha }|f|(x)$, split at radius $R$. Dyadic shells with radii $2^{-j}R$ giveFor the outer part, Hölder's inequality givesThe last integral is finite exactly because $p<n/\alpha$. Balance the two bounds by $R^{n/p}=\Vert f{\Vert }_p/Mf(x)$ to obtain the Hedberg inequalityThe zero cases follow directly or by a limit; $Mf$ is finite almost everywhere by the maximal estimate. Since $r(1-\theta )=p$, integration and the proved maximal bound yield $\Vert I_{\alpha }f{\Vert }_r\le C\Vert f{\Vert }_p$. Taking $\alpha =n-\lambda$ and pairing $I_{\alpha }|f|$ with $|g|\in L^{r^{\prime }}$ proves the bilinear form by Hölder's inequality, because $s=r^{\prime }$. All integrals are justified first for nonnegative truncations and then by monotone convergence and the resulting finite bound. The strong estimate does not include $p=1$ or $p=n/\alpha$.

As usual, the bounded set is assumed Lebesgue measurable and an $L^q$ exponent is positive; the conventional Banach-space assertion concerns $1\le q<2^{\ast }=2n/(n-2)$. No boundary regularity of $\Omega$ is needed. Choose a smooth compactly supported cutoff $\eta$ equal to one on a ball containing $\Omega$. Weak convergence and the Uniform boundedness principle give a common $H^1$ bound for $f_j$, hence also for $g_j=\eta f_j$. These functions have a common compact support.

We prove their local $L^2$ compactness directly. Translation and the fundamental theorem along line segments give, by density from smooth functions,For a compactly supported mollifier ${\rho }_{\epsilon }$, integration of this estimate givesAt fixed $\epsilon$, Hölder's inequality bounds $\Vert g_j\ast {\rho }_{\epsilon }{\Vert }_{\infty }$ and $\Vert \nabla (g_j\ast {\rho }_{\epsilon }){\Vert }_{\infty }$ uniformly by the $L^2$ norm of the corresponding smoothing kernels. Their supports lie in one fixed compact set, so the Arzelà-Ascoli theorem makes this smoothed family precompact in $L^2$. The uniform $O(\epsilon )$ approximation makes the original family totally bounded in $L^2$: choose a finite net for the smoothed family and add the approximation error. Completeness gives relative compactness.

Every strongly convergent subsequence of $g_j$ has limit $\eta f$, since multiplication by $\eta$ preserves weak $L^2$ convergence. Consequently the whole sequence converges strongly: otherwise a subsequence with distance at least a fixed positive number would have a further convergent subsequence and contradict that unique limit. In particular $\Vert {\chi }_{\Omega }(f_j-f){\Vert }_2\to 0$.

The critical Sobolev inequality supplies a uniform bound in $L^{2^{\ast }}$. For clarity, this estimate follows from part 2(a), not from compactness at the critical exponent. For a compactly supported smooth $v$, integrating by parts against the vector field $(x-y)/|x-y{|}^n$ givesIts distributional divergence is the sphere-area multiple of a point mass; applying the Hardy–Littlewood–Sobolev inequality with $\alpha =1,p=2$ yields $\Vert v{\Vert }_{2^{\ast }}\le C\Vert \nabla v{\Vert }_2$. Density extends this to $H^1({\mathbb{R}}^n)$.

For $2<q<2^{\ast }$ choose $0<\theta <1$ with $1/q=\theta /2+(1-\theta )/2^{\ast }$. Hölder's inequality givesFor $1\le q<2$, use $\Vert v{\Vert }_{L^q(\Omega )}\le |\Omega {|}^{1/q-1/2}\Vert v{\Vert }_{L^2(\Omega )}$. The same integral estimate also works for $0<q<1$ if $L^q$ is understood with its usual quasi-norm. Thus ${\chi }_{\Omega }{f}_j\to {\chi }_{\Omega }f$ strongly for every positive $q<2^{\ast }$. We have never asserted that multiplying by the possibly nonsmooth indicator preserves $H^1$.

We first construct a genuinely compact Besicovitch set; a sequence of unrelated small-area sets would not suffice. The compact Kakeya construction from small projections below gives the limiting compact set together with its segments.

Let $\mathcal{X}$ consist of nonempty compact sets $K\subset [0,1]\times \mathbb{R}$ whose horizontal projection is exactly $[0,1]$. With the Hausdorff metric this is a complete metric space. Indeed a Hausdorff-Cauchy sequence is eventually contained in one bounded closed rectangle and converges there to a nonempty compact set; horizontal projection is Lipschitz for Hausdorff distance, so its value $[0,1]$ is preserved in the limit.

For $t\in [0,1]$ put $L_t(x,y)=y+tx$ and $m_t(K)=|L_t(K)|$. The function $(K,t)\mapsto m_t(K)$ is upper semicontinuous. To verify this, cover the compact set $L_t(K)$ by finitely many open intervals with total length less than $m_t(K)+\epsilon$. Joint continuity of the compact images in the Hausdorff metric puts every sufficiently nearby image inside the same cover. This proves the required upper bound, uniformly over a compact parameter interval when a strict bound holds throughout it.

Fix $q\in [0,1]$ and $\epsilon >0$. The setis open. It is also dense, by the following explicit finite approximation. Given $K$ and tolerance $h>0$, partition $[0,1]$ into $N$ equal subintervals and choose $(j/N,y_j)\in K$ at each left endpoint. Above that subinterval put the segmentEach segment lies within $\sqrt{1+q^2}/N$ of its chosen point in $K$. Add a finite $h/2$-net of points of $K$ to their union, ensuring the reverse Hausdorff inclusion. For large $N$ the resulting compact set $K^{\prime }$ belongs to $\mathcal{X}$ and has Hausdorff distance less than $h$ from $K$. The image of each segment under $L_t$ has length $|t-q|/N$, and the added points contribute zero length. Henceso $K^{\prime }\in U(q,\epsilon )$.

Take a finite grid of $q$'s whose $\epsilon$-neighborhoods cover $[0,1]$. The finite intersection of the open dense sets $U(q,\epsilon )$ is dense and implies ${\sup }_{0\le t\le 1}{m}_t(K)<2\epsilon$. Thusis open and dense. The Baire category theorem gives a compact $K\in {\bigcap }_{r\ge 1}{V}_{1/r}$, and therefore $m_t(K)=0$ for every $t\in [0,1]$.

DefineThis is the continuous image of the compact set $K\times [0,1]$, hence compact. Its section at horizontal coordinate $t$ is $L_t(K)$, of measure zero. The Fubini theorem gives $|B|={\int }_0^1{m}_t(K)dt=0$. For every slope $x\in [0,1]$, horizontal surjectivity supplies $(x,y)\in K$, and $B$ contains the segment $(t,y+tx)$, $0\le t\le 1$, of length $\sqrt{1+x^2}\ge 1$. Four rotated copies, through angles $0,\pi /4,\pi /2,3\pi /4$, cover all unoriented directions. Their finite union $\mathcal{B}$ is compact, has area zero, and contains a unit subsegment in every direction:The point-line dual construction is the mechanism in Körner's construction; the density calculation above supplies the needed proof, rather than assuming that small-area sets automatically admit a suitable nested limit.

For the multiplier discussion, define $T_D$ by $\widehat{T_{D}g}(\xi )=1_{\{|\xi |\le 1\}}\hat{g}(\xi )$. The Plancherel theorem gives boundedness on $L^2$. We explain why a putative $L^p$ bound, $1<p<2$, would contradict Kakeya packing; duality of Lp spaces then treats $p>2$.

First, such a bound would force the vector-valued inequalityfor arbitrary directions $v_j$, where $P_v$ is the half-plane Fourier projection onto the half-plane $\xi \cdot v<0$. To see the important curvature step, let $T_R$ project onto the radius-$R$ disk and modulate by $e^{iR{v}_j\cdot x}$. Its conjugated symbol is $1_{\{|\xi +R{v}_j|\le R\}}$, tending almost everywhere to $1_{\{\xi \cdot v_j<0\}}$. Dilations give the same putative norm bound for every $R$. Apply this common operator to ${\sum }_j{ϵ}_j{e}^{iR{v}_j\cdot x}{g}_j$, average over independent random signs, and use the Khintchine inequality. Letting $R\to \infty$ gives the displayed estimate by $L^2$ convergence and Fatou lemma, first for finitely many smooth function functions and then by approximation. The disk has a boundary normal in every direction; a fixed polygon does not.

The geometric input for this overview is the standard finite Perron tree packing used in Fefferman's original proof: for arbitrarily small $\eta >0$ one obtains long thin rectangles $R_j$ with $A={\sum }_j|R_j|$ and $|{\bigcup }_j{R}_j|\le \eta A$, while fixed longitudinal translates $R_j^{\prime }$ are pairwise disjoint. This is a quantitative finite arrangement, stronger than merely having a compact null set. The splitting/sliding procedure subdivides an angular triangle into narrow triangles and overlaps their bases; the extensions along their axes separate in reverse order. Its area estimate has the form ${\alpha }^{2m}+2(1-\alpha )$ relative to the original total triangle area: the merged central part shrinks by ${\alpha }^2$ at each generation, and the accumulated caps contribute at most $2(1-\alpha )$. Taking $\alpha$ close to one and then $m$ large makes the ratio arbitrarily small. Inscribed rectangular portions retain a fixed proportion of the total area. The reflected extensions explain why the translates can be disjoint even though the original rectangles overlap strongly.

Here is the analytic contradiction, with the relevant powers explicit. Take $g_j=1_{R_j}$. In coordinates parallel and perpendicular to the long axis $v_j$, the directional Hilbert transform integrates only along that axis. If the longitudinal interval is $[0,\ell ]$, then outside itOn a fixed translate such as $s\in [5\ell ,6\ell ]$, this has magnitude at least ${\pi }^{-1}\log (6/5)$. Since $P_v=(I-i{H}_v)/2$ up to the harmless choice of half-plane sign, $|P_{v_j}{g}_j|\ge c_0>0$ on $R_j^{\prime }$. Their disjointness gives a lower bound $c_0{A}^{1/p}$ for the vector-valued output norm.

Writing $h={\sum }_j{1}_{R_j}$, Hölder's inequality for $p<2$ givesThe putative vector-valued inequality would implywhich is impossible as $\eta \downarrow 0$. The disk multiplier is self-adjoint, so a bound for $p>2$ would give one for its conjugate exponent below two. Thus the disk multiplier is bounded only at $p=2$ among $1<p<\infty$. This explains both the random-sign step and the precise geometric amplification; it is not an assertion that frequency rectangles by themselves produce a logarithmic norm blow-up.

Put $V_t=⟨X{⟩}_t$ and $S_t={\sup }_{s\le t}|X_s|$. Suppose first that $|X|\le K$. The Itô formula for $|x{|}^p$, which is twice continuously differentiable for $p\ge 2$, givesFor $p=2$, the second derivative is interpreted as the constant $2$. The stochastic term has mean zero: its integrand is bounded, and $\mathbb{E}{V}_t<\infty$ makes it a square-integrable martingale. ConsequentlyThis is the first required estimate.

A bounded local martingale is a true martingale. Apply the allowed Doob Lp maximal inequality, and then Hölder's inequality with conjugate exponents $p/(p-2)$ and $p/2$ for $p>2$. With $D_p=(p/(p-1){)}^p$, this yieldsIf the left side is zero there is nothing to prove; otherwise divide by its indicated power and raise to $p/2$. For $p=2$, the same conclusion follows directly from $\mathbb{E}{X}_t^2=\mathbb{E}{V}_t$ and Doob's inequality. A usable constant is thereforeIn particular $C_2=4$. This is the upper maximal moment bound for a continuous local martingale.

For an unbounded continuous local martingale, stop at ${\tau }_n=\inf \{s:|X_s|\ge n\}$. Its stopped path is bounded, and its bracket at $t$ is $V_{t\wedge {\tau }_n}\le V_t$. The proved inequality gives a bound by $C_p\mathbb{E}{V}_t^{p/2}$, independent of $n$. Continuity makes ${\tau }_n\uparrow \infty$, and the stopped maxima increase to $S_t$. Monotone convergence proves the same inequality for the original unbounded process, with the same constant. This localization step also shows that the assumed bracket moments supply all the needed maximal moments.