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A harmonic function on $\Omega$ is a twice continuously differentiable function satisfying $\Delta u=0$ there. For complex-valued functions this means that both its real part and its imaginary part are harmonic functions. The mean value property for harmonic functions characterizes harmonicity: a continuous function is harmonic exactly when its value at each point equals its average over every sphere, or equivalently every ball, whose closed ball lies in $\Omega$. In the forward direction, the derivative of the spherical average vanishes by the divergence theorem. Conversely, radial smoothing leaves a continuous function with the mean value property unchanged locally; it is therefore smooth. Expanding its small-ball average gives $u(x)+r^2\Delta u(x)/(2(d+3))+o(r^2)$ in dimension $d+1$, and hence $\Delta u=0$.

Write $M=\Vert u{\Vert }_{h^1}$ and $v_{d+1}$ for the volume of the unit ball in ${\mathbb{R}}^{d+1}$. We first obtain a pointwise estimate directly from the mean value property for harmonic functions, before using any boundary representation. For the ball of radius $r=t/2$ centered at $(x,t)$,Thus every upward shift of $u$ is a bounded harmonic function, with a continuous bounded boundary value.

The required map is the Poisson integralThe Poisson kernel for the upper half-space is positive, has integral one, and is harmonic as a function of $(x,t)$. Its normalization follows by polar coordinates and $r=\tan \theta$, which reduces the radial integral to ${\int }_0^{\pi /2}{\sin }^{d-1}\theta d\theta$; direct differentiation gives ${\Delta }_{x,t}{P}_t=0$. Differentiation under the integral on compact subsets proves that $P\mu$ is a harmonic function. Tonelli theorem givesHere the total variation norm of a measure is the norm on finite signed or complex Borel measures.

The Poisson kernel for the upper half-space is an approximate identity: for $\psi \in C_0({\mathbb{R}}^d)$, $P_t\ast \psi \to \psi$ uniformly. Indeed, uniform continuity controls translations smaller than a fixed radius, and the kernel mass outside that radius tends to zero. ConsequentlyThe dual characterization of the total variation norm of a measure, as the supremum over $\psi \in C_0$ of supremum norm at most one, now gives the reverse inequality $\Vert \mu {\Vert }_{\mathrm{TV}}\le \Vert P\mu {\Vert }_{h^1}$. Thus $P$ is a linear isometry and is injective.

For surjectivity, fix $s>0$ and put $u_s(x)=u(x,s)$. Bounded harmonic uniqueness in the upper half-space givesTo justify this uniqueness, subtract the Poisson integral of the bounded continuous boundary value from the upward-shifted harmonic function. Their difference is bounded and harmonic and has zero continuous boundary values. Its odd reflection across the boundary is an entire bounded harmonic function; the reflection is harmonic across the flat boundary by harmonic odd reflection across a hyperplane. The Liouville theorem for harmonic functions makes it constant, and oddness makes it zero. The initial pointwise estimate supplies the boundedness needed here. To see the reflection locally, take a ball centered on the hyperplane and extend the boundary data on its upper hemisphere oddly to its lower hemisphere. The harmonic solution on this ball is odd and zero on its equatorial disk. The maximum principle for harmonic functions identifies it with the given difference on the upper half-ball, proving that the reflected function is harmonic across the boundary.

Choose $s_j\downarrow 0$. The measures $u_{s_j}dx$ have total variation norm of a measure at most $M$. By the Banach-Alaoglu theorem, and separability of $C_0({\mathbb{R}}^d)$, a subsequence converges in the weak-star topology to a finite measure $\mu$ with $\Vert \mu {\Vert }_{\mathrm{TV}}\le M$. For fixed $(x,t)$ and $s_j<t$,The kernels in this display belong to $C_0$ and converge uniformly to $P_t(x-\cdot )$. The uniform measure norm bound therefore allows passage to the limit, giving $u(x,t)=P\mu (x,t)$. This proves the Poisson representation of harmonic h1 by finite measures, including onto-ness.

Finally $P_t(x)\le c_d{t}^{-d}$, so the Poisson representation of harmonic h1 by finite measures improves the preliminary pointwise constant:The last occurrence of $h^1({\mathbb{R}}^{d+1})$ in the printed question is understood as the harmonic Hardy space of the upper half-space defined earlier in that question.

We first construct a terminal integrable function without using the martingale convergence theorem. Put $G={\sup }_n|g_n|$ and $d\eta =Gdx$ on $(0,1{]}^d$. This is a finite positive measure. If $G=0$ almost everywhere the conclusion is immediate. Otherwise, for a bounded dyadic step function $h$ measurable with respect to ${\mathcal{F}}_n$ in the dyadic filtration, defineThe martingale property makes the definition independent of the choice of a sufficiently large $n$. Moreover $|\Lambda (h)|\le \int |h|d\eta$. Dyadic step functions are dense in $L^1(\eta )$: continuous functions on the closed unit cube are dense for this finite Borel measure (extend the measure by zero onto the omitted faces), and uniform continuity approximates each continuous function by step functions on fine dyadic partitions. Thus $\Lambda$ extends to a bounded linear functional on $L^1(\eta )$ of norm at most one. The duality of Lp spaces representation of this functional gives $H\in L^{\infty }(\eta )$ with $|H|\le 1$ and $\Lambda (h)=\int hHd\eta$. Set $g=HG$, taking $g=0$ where $G=0$. Then $g\in L^1(dx)$ and testing indicators of all dyadic atoms provesThis argument also works for complex-valued martingales, with a complex linear functional.

To prove convergence, choose a dyadic step function $h$ with $\Vert g-h{\Vert }_1$ arbitrarily small. For all $n$ beyond its dyadic level, $\mathbb{E}(h\mid {\mathcal{F}}_n)=h$. The L1 contraction of conditional expectation yieldsThis proves convergence in $L^1$. For convergence almost everywhere, Doob maximal inequality applied to the conditional expectations of $g-h$ givesLet the approximation error tend to zero, and then take positive rational $a$. Hence $g_n\to g$ both in $L^1$ and almost everywhere.

For the second assertion, write $M={\sup }_n\mathbb{E}|f_n|<\infty$ and take the given first-exit stopping time $\tau$. Each $g_n=f_{\tau \wedge n}$ is integrable, being a finite sum of restrictions of integrable variables. Its increments satisfyThe indicator is ${\mathcal{F}}_n$-measurable, so taking its conditional expectation proves that $(g_n)$ is a martingale.

The overshoot at $\tau$ need not be bounded by $N$. Instead, for a fixed $m$, on $\{\tau =j\}$ with $j\le m$ we have $|f_j|\le \mathbb{E}(|f_m|\mid {\mathcal{F}}_j)$. Integrating over this ${\mathcal{F}}_j$-measurable event, summing the disjoint events, and including $\{\tau >m\}$ givesThe variables $Z_m=|f_{\tau }|1_{\{\tau \le m\}}$ increase to $Z$, and $Z_m\le |g_m|$. The monotone convergence theorem gives $\mathbb{E}Z\le M$. Before stopping, the process has magnitude less than $N$; after stopping it remains at $f_{\tau }$. ConsequentlyThis is the integrable overshoot of a stopped L1-bounded martingale. The first part now proves that every such stopped process converges almost everywhere.

Finally, Doob maximal inequality gives $|\{{\sup }_n|f_n|\ge N\}|\le M/N$, by applying the finite-horizon estimate and then increasing the horizon. Thus ${\sup }_n|f_n|$ is finite almost everywhere. Outside the union of the null sets for all positive integer stopping thresholds, choose an integer $N$ larger than this supremum. No stopping occurs at that point, so $f_n=g_n$ for all $n$, and $f_n$ converges to a finite limit almost everywhere. We have not assumed, or concluded, $L^1$ convergence for this general $L^1$-bounded martingale.

We derive the required Calderón–Zygmund decomposition and apply the preceding two parts. Use the Fourier transform convention $\hat{f}(\xi )=\int e^{-ix\cdot \xi }f(x)dx$, and put $A=\Vert \hat{K}{\Vert }_{\infty }$. The Plancherel theorem givesFor completeness this starts with smooth rapidly decreasing inputs, where $\widehat{K\ast f}=\hat{K}\hat{f}$, and extends by $L^2$ density. It agrees with the pointwise convolution in the question: if $f_j\to f$ in $L^2$, then $\Vert K\ast (f_j-f){\Vert }_{\infty }\le \Vert K{\Vert }_2\Vert f_j-f{\Vert }_2$. Thus there is no need to assume that $K$ itself is integrable.

Fix $f\in L^1({\mathbb{R}}^d)$, let $F=\Vert f{\Vert }_1$, and choose $\lambda >0$. Select the maximal dyadic cubes $Q$ for which $|Q{|}^{-1}{\int }_Q|f|>\lambda$. They exist above every qualifying cube because averages over increasingly large ancestors tend to zero. They are disjoint, their total volume is at most $F/\lambda$, and maximality givesOutside their union, dyadic differentiation gives $|f|\le \lambda$ almost everywhere. One can obtain this differentiation fact from the allowed martingale convergence theorem: on each fixed dyadic cube the averages are conditional expectations of the integrable restriction, their almost-everywhere limit exists, and approximation by dyadic step functions shows in $L^1$ that this limit is $f$.

Let $f_Q=|Q{|}^{-1}{\int }_{Q}f$, define $g=f$ outside the selected cubes and $g=f_Q$ on $Q$, and set $b_Q=(f-f_Q)1_Q$. This constructswith the estimatesThese follow by integrating separately on each selected cube; $|f_Q||Q|\le {\int }_Q|f|$ gives both the good-part norm bound and the last cancellation-error bound.

Let $\Omega ={\bigcup }_Q\hat{Q}$, with the dilation $L=4\sqrt{d}$ from part (ii). Then $|\Omega |\le L^{d}F/\lambda$. On its complement, part (i), followed by part (ii), givesFinally, $|Tf|>\lambda$ implies $|Tg|>\lambda /2$ or $|Tb|>\lambda /2$. The first alternative is controlled by the $L^2$ operator bound, and the second by the last integral outside $\Omega$. HenceThus $T$ has weak type $(1,1)$, with a constant depending only on the dimension, the bounded Fourier transform, and the Hörmander integral kernel condition.

Put $M={\sup }_{y>0}{\int }_{\mathbb{R}}|f(x+iy)|dx$. The goal is to obtain an integrable non-tangential maximal function without assuming in advance that $f$ has an integrable boundary density. We use a square root to place the boundary data in $L^2$, where the Hardy-Littlewood maximal function is strongly bounded.

First recall this strong bound directly from Q2. For a nonnegative $h\in L^2(\mathbb{R})$ and $\lambda >0$, split $h=h{1}_{\{h>\lambda /2\}}+h{1}_{\{h\le \lambda /2\}}$. The second summand has centered maximal function at most $\lambda /2$; the first is in $L^1$. Q2's weak estimate, in dimension one, therefore givesThe layer cake representation and Tonelli theorem then implyThis is the needed Strong Lp bound for the Hardy-Littlewood maximal function, with an explicit adequate constant.

For $\epsilon >0$, let $f_{\epsilon }(z)=f(z+i\epsilon )$ and $h_{\epsilon }(x)=|f(x+i\epsilon ){|}^{1/2}$. Q1's elementary harmonic pointwise estimate makes $f_{\epsilon }$ bounded on the closed upper half-plane. Its boundary value is continuous, and $\Vert h_{\epsilon }{\Vert }_2^2\le M$. The function $|f_{\epsilon }{|}^{1/2}$ is subharmonic: away from zeros, for any $q>0$,and regularization, or the submean inequality at zeros, extends this to the whole domain. Bounded subharmonic comparison with the Poisson integral givesHere $h_{\epsilon }$ need not be in $L^1$. The Poisson integral is nevertheless defined, since $h_{\epsilon }\in L^2\cap L^{\infty }$ and $P_y\in L^1\cap L^2$, and it tends to its continuous boundary values. To justify comparison on this unbounded domain, the difference between the two sides is bounded above, subharmonic, and zero on the real boundary. On an upper half-disk of radius $R$, use the positive harmonic barrierOn its semicircular arc, $b_R(R{e}^{i\theta })=2/\sin \theta \ge 2$, while $b_R(z)\to 0$ for every fixed interior $z$ as $R\to \infty$. The maximum principle for subharmonic functions, with a fixed bound for the difference, proves the comparison. The barrier's positive blow-up at the arc endpoints causes no difficulty.

For $|x-x^{\prime }|\le y$, the elementary inequality $y^2+|x-s{|}^2\le 3(y^2+|x^{\prime }-s{|}^2)$ givesQ2's radial decreasing kernel domination by the maximal function extends to nonnegative $L^2$ inputs by applying it to increasing bounded compactly supported truncations. Hence, taking the supremum over the cone,For any fixed cone point, $f_{\epsilon }(z)\to f(z)$ as $\epsilon \downarrow 0$. Therefore $f^{\ast }(x)\le {\mathrm{liminf}}_{\epsilon \downarrow 0}{f}_{\epsilon }^{\ast }(x)$; use a sequence of positive shifts and the Fatou lemma to concludeThe precise numerical constant is unimportant. Suprema over countable dense sets of cone points give the same values, so the maximal functions used here are measurable.

This yields boundary absolute continuity of measures rather than assuming it. The holomorphic function $f$ is a complex-valued harmonic function, so Q1 represents it as $f(x+iy)=P_y\ast \mu (x)$ for a finite complex measure $\mu$, with $f(\cdot +iy)dx\to \mu$ weak-star. Since $|f(x+iy)|\le f^{\ast }(x)$, every compactly supported continuous $\psi$ satisfiesThe dual characterization of the total variation measure implies $|\mu |\le f^{\ast }dx$, so the Radon-Nikodym theorem gives $\mu =Fdx$ with $F\in L^1$. The approximate identity now gives $f(\cdot +iy)\to F$ in the L1 norm and almost everywhere. Thus the analytic Hardy space has an integrable boundary function, unlike the unrestricted harmonic $h^1$ space.

In the real-line form of the F. and M. Riesz theorem, start with a finite complex measure whose Fourier transform of a finite measure vanishes for $\xi <0$. Its Poisson integral is holomorphic: the transformed operators ${\partial }_y$ and $i{\partial }_x$ act by $-|\xi |$ and $-\xi$, which agree on that spectrum. These identities can first be read as distributional identities and then classically, since the Poisson integral is smooth. Its harmonic $h^1$ norm is finite by Q1, so the preceding argument makes the measure absolutely continuous. A one-sided Fourier spectrum forces a finite boundary measure to have a Lebesgue density.

For the customary circle form, normalize arc length by $dm=d\theta /(2\pi )$. If $\hat{\mu }(n)=\int e^{-in\theta }d\mu (\theta )=0$ for $n<0$, thenis analytic and $F(r{e}^{i\theta })$ is the Poisson integral on the unit disk of $\mu$. Positivity and unit mass of the Poisson kernel on the circle imply ${\sup }_r\int |F(r{e}^{i\theta })|dm\le \Vert \mu {\Vert }_{\mathrm{TV}}$. The same square-root argument on the disk gives an integrable radial maximum, as follows. For $0<\rho <1$, put $h_{\rho }(\theta )=|F(\rho e^{i\theta }){|}^{1/2}$. The maximum principle for subharmonic functions on the disk givesThe supremum of these Poisson integrals is bounded by a constant times the circular Hardy-Littlewood maximal function. Indeed, for $r\ge 1/2$ the circular kernel is bounded by a constant times $(1-r)/((1-r{)}^2+{\theta }^2)$ for $|\theta |\le \pi$; splitting into arcs of radii $2^j(1-r)$ gives a summable geometric series of arc averages. For $r<1/2$ the kernel is uniformly bounded, and the whole-circle average suffices. The covering proof and the truncation argument already used on the line give the strong $L^2$ maximal bound for arcs on the circle as well. ThusLet $\rho \uparrow 1$ and apply Fatou lemma to obtain an integrable radial maximum $R(\theta )={\sup }_{0<r<1}|F(r{e}^{i\theta })|$. The measures $F(r{e}^{i\theta })dm$ tend weak-star to $\mu$ and are dominated in magnitude by $Rdm$. The same continuous-test-function argument gives $|\mu |\le Rdm$. This proves the F. and M. Riesz theorem: a finite complex measure on the circle with all negative Fourier coefficients zero is absolutely continuous with respect to arc length. Reversing the Fourier convention interchanges positive and negative indices.

On a suitable dense domain of smooth rapidly decreasing wave functions, the canonical commutator is $[x,p]=i\hslash$. Expanding the two ladder operators givesTheir sum and difference giveDirect multiplication, retaining the operator order, givesUsing ${\beta }^2=m\omega /\hslash$ identifies the Hamiltonian operator:For any admissible wave function, adjointness givesFor the conventional normalized state $\Vert \Psi \Vert =1$, this is the stated lower bound $E\ge \hslash \omega /2$. For an unnormalized function the norm factor must be retained; rescaling a wave function cannot leave its unnormalized energy integral bounded below by a fixed positive number.

Equality holds exactly when $a{\Psi }_0=0$. In position representation this equation is $(\beta x+{\beta }^{-1}{\partial }_x){\Psi }_0=0$, whose normalized square-integrable solution isThis construction establishes existence of a state attaining the bound, and hence proves that it is the ground state.

The commutator $[a,a^{†}]=1$ implies $[H,a^{†}]=\hslash \omega a^{†}$. Repeatedly commuting through the product givesThese vectors are nonzero: the identity $a(a^{†}{)}^n{\Psi }_0=n(a^{†}{)}^{n-1}{\Psi }_0$ gives their norm squared $n!$ by induction. ThereforeFor time-dependent stationary states multiply each spatial eigenfunction by $e^{-i{E}_{n}t/\hslash }$.

Use first-kind Chebyshev polynomials $T_j$, characterized by $T_j(\cos \theta )=\cos (j\theta )$. Substituting $x=\cos \theta$ converts the weighted inner product into ${\int }_0^{\pi }g(\cos \theta )h(\cos \theta )d\theta$. Hence these polynomials are orthogonal, with squared norms $\pi$ for $j=0$ and $\pi /2$ for $j\ge 1$.

For the specified function, $f(\cos \theta )=\sin \theta$. Its constant projection coefficient is ${\pi }^{-1}{\int }_0^{\pi }\sin \theta d\theta =2/\pi$. For $j\ge 1$ the coefficient is $(2/\pi ){\int }_0^{\pi }\sin \theta \cos (j\theta )d\theta$. Symmetry about $\pi /2$ makes it zero for odd $j$. For $j=2m$, the product-to-sum identity givesTherefore the Chebyshev projection of a semicircle isIn particular the best degree-at-most-$n$ polynomial is even, so for odd $n$ its actual degree is at most $n-1$.

Apply the Gram-Schmidt process explicitly. Start with $f_1=e_1/\Vert e_1\Vert$, and for $i\ge 2$ setAssume inductively that the previous $f_j$ form an orthonormal basis for the span of $e_1,\dots ,e_{i-1}$. Inner products give $⟨v_i,f_j⟩=0$ for every $j<i$. Moreover $v_i\ne 0$: otherwise $e_i$ would lie in the span of its predecessors, contradicting that the $e_j$ form a basis. Thus $f_i$ is well defined and has unit norm. The definition expresses $f_i$ in the span of $e_1,\dots ,e_i$, and also expresses $e_i$ in the span of $f_1,\dots ,f_i$. The two spans therefore agree at each stage. Induction proves the required orthonormal, span-preserving basis.

Let $\pi$ be a prime element of the Gaussian integers. Its norm $\pi \overline{\pi }$ is a positive integer greater than one, so $\pi$ divides some ordinary positive prime number in its integer prime factorization. It cannot divide two distinct such primes $p,q$: an integer Bézout identity $up+vq=1$ would imply $\pi \mid 1$, contrary to its being a nonunit. This proves existence and uniqueness of the rational prime lying below $\pi$.

Up to associates, the complete list of Gaussian prime elements is: $1+i$; the positive rational primes $p\equiv 3(\mathrm{mod}4)$; and, for every rational prime $p\equiv 1(\mathrm{mod}4)$, the two nonassociate elements $a+bi$ and $a-bi$, where $a,b>0$ and $p=a^2+b^2$. Representatives related by multiplication by one of $\pm 1,\pm i$ are associates. The last pair accounts for both factors of a split prime.

Every residue class modulo $p$ has a unique representative $a+bi$ with $0\le a,b<p$, since $p\mid (a-a^{\prime })+(b-b^{\prime })i$ means both coordinate differences are divisible by $p$. Hence $\boxed{|R/pR|=p^2}$ andFor $p=2$, the class of $1+i$ is nonzero but its square is $2i=0$; thus the quotient ring is not a field. For $p\equiv 3(\mathrm{mod}4)$, a root $r^2=-1$ would imply $r^{p-1}=(-1{)}^{(p-1)/2}=-1$, contradicting Fermat's little theorem. The quadratic $X^2+1$ is therefore irreducible and its quotient is a field with $p^2$ elements.

For $p\equiv 1(\mathrm{mod}4)$, choose a generator $g$ of the cyclic group ${\mathbb{F}}_p^{\times }$. Then $r=g^{(p-1)/4}$ satisfies $r^2=-1$. The two nonzero classes $i-r$ and $i+r$ have product zero in $R/pR$. ThusFor $p\equiv 1(\mathrm{mod}4)$ it is instead isomorphic to ${\mathbb{F}}_p\times {\mathbb{F}}_p$, by the distinct linear factors and the Chinese remainder theorem.

Use the standard complex inner product $⟨u,v⟩=u^{\ast }v$, conjugate-linear in its first argument. A unitary matrix satisfies $U^{\ast }U=I$, and a Hermitian matrix satisfies $H^{\ast }=H$, where $\ast$ is conjugate transpose.

If $Hv=\lambda v$ with $v\ne 0$, thenPositivity of $⟨v,v⟩$ proves $\boxed{\lambda \in \mathbb{R}}$. If $Uv=\lambda v$, preservation of the norm gives $\Vert v{\Vert }^2=\Vert Uv{\Vert }^2=|\lambda {|}^2\Vert v{\Vert }^2$, so $\boxed{|\lambda |=1}$.

For eigenvectors $Hv=\lambda v$ and $Hw=\mu w$, the Hermitian matrix identity gives $⟨v,Hw⟩=⟨Hv,w⟩$. The already established reality of the eigenvalues therefore implies $(\mu -\lambda )⟨v,w⟩=0$. Thus distinct eigenvalues have orthogonal eigenvectors.

The integral is finite and nonnegative because $|f|$ is continuous on a compact interval. If it vanishes and $f(x_{\ast })\ne 0$, continuity gives a relative interval of positive length on which $|f|\ge |f(x_{\ast })|/2$, a contradiction. Thus the integral is zero only for $f=0$. For a real scalar $a$, $\Vert af\Vert =|a|\Vert f\Vert$, and the pointwise triangle inequality $|f+g|\le |f|+|g|$ integrates to $\Vert f+g\Vert \le \Vert f\Vert +\Vert g\Vert$. These are all the norm axioms.

The given sequence satisfiesHence it converges to the zero function in this integral norm. Although $f_n(0)=1$ for all $n$, an isolated endpoint contributes no integral. The pointwise limit is discontinuous, and the sequence does not converge in the uniform norm; neither fact obstructs convergence in the stated norm.

Write the hyperbolic toral automorphism as $T_A([v])=[Av]$ on ${\mathbb{T}}^2={\mathbb{R}}^2/{\mathbb{Z}}^2$, where $A\in \mathrm{GL}(2,\mathbb{Z})$. Its determinant is $\pm 1$. The eigenvalues cannot be a nonreal conjugate pair: their product would have absolute value one, forcing each to have absolute value one, contrary to hyperbolicity. The same product condition and hyperbolicity imply two distinct real eigenvalues, labelled so thatWe will proveThis is the entropy of a hyperbolic toral automorphism; the absolute value is important when the expanding eigenvalue is negative.

Choose real eigenvectors $v_s,v_u$ and the adapted norm $\Vert a{v}_s+b{v}_u{\Vert }_{\ast }=\max (|a|,|b|)$. Its quotient metric on the torus isIt is a compatible metric, so part (a) allows its use. Let $\rho ={\min }_{0\ne \ell \in {\mathbb{Z}}^2}\Vert \ell {\Vert }_{\ast }>0$ and $D=\Vert A{\Vert }_{\ast }=|{\lambda }_u|$. Fix $ϵ>0$ small enough that $2ϵ<\rho$ and $(D+1)ϵ<\rho$.

If $[v]$ lies in the Bowen ball $B_n(0,ϵ)$, each $T_A^k[v]$ has a unique lift $z_k$ with $\Vert z_k{\Vert }_{\ast }<ϵ$. Uniqueness follows because two such lifts differ by a lattice vector of norm less than $2ϵ<\rho$. Moreover, $z_{k+1}-A{z}_k$ is a lattice vector, andIt must be zero. Thus $z_k=A^k{z}_0$: there is no hidden lattice jump between successive lifted iterates.

Write $z_0=a{v}_s+b{v}_u$. The simultaneous bounds $\Vert A^k{z}_0{\Vert }_{\ast }<ϵ$ for $0\le k<n$ are exactlyThe stable coordinate is largest at time zero and the unstable coordinate at time $n-1$. Conversely, every vector in this rectangle satisfies all the Bowen ball bounds. The quotient is injective on this rectangle by $2ϵ<\rho$. If $J=|\det (v_s,v_u)|$, its normalized Haar measure is thereforeThe quotient metric is translation invariant and $T_A$ is linear, so $d_n(x,y)=d_n(0,y-x)$; every Bowen ball has this same measure, independently of its centre.

Take an $ϵ$-separated set of largest cardinality $s_n(d,ϵ)$. It is maximal under inclusion, so its radius-$ϵ$ Bowen balls cover the torus: a point outside all these balls could be added. The covering bound givesIts radius-$ϵ/2$ Bowen balls are disjoint, by the triangle inequality for $d_n$. Their measures sum to at most one, givingTaking logarithms, dividing by $n$, and letting $n\to \infty$ yields $h_d(T_A,ϵ)=\log |{\lambda }_u|$ for every sufficiently small $ϵ$. The limit as $ϵ\downarrow 0$ proves the answer. Contraction in the stable direction contributes no exponential growth; only expansion in the unstable direction does.

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 printed symbol $E$ is not defined. Take $E=L^1(\mathbb{R})$, or more generally an $L^1$ subspace with its induced $L^1$ norm; density must be in that norm. Given $g\in E$, choose a sequence $f_n\in F$ withBy the assumed convergence on $F$, each $T_r{f}_n\to T_0{f}_n$ holds outside a null set $N_n$. Their countable union $N={\bigcup }_n{N}_n$ is still null. Thus outside $N$ the convergence statements for every member of this single approximation sequence hold simultaneously. No assertion of simultaneous convergence for the entire uncountable dense class is needed.