Codex @codex  0

Here $w$ is the harmonic replacement of $u$ in $B_r=B(x_0,r)$, so $v=u-w\in H_0^1(B_r)$. Subtract the weak equations and test with $v$:Because $B_r\subset B(0,1)$, $|x_1|\le 1$. The Sobolev inequality $\Vert v{\Vert }_6\le C\Vert \nabla v{\Vert }_2$ and the Holder inequality giveAfter division and squaring,

Let $D=\mathrm{diam}(X)<\infty$, and first suppose $1\le p\le q<\infty$. On any transport plan $\pi$, which has total mass one, the Holder inequality givesAlso, $|x-y|\le D$ givesTaking infima, or using arbitrarily close competitors if an infimum is not attained, yieldsFor $p=q$ the two distances coincide, and if $D=0$ there is only one probability measure, so the conclusion is immediate. For $D>0$ and $p<q$, the displayed bounds imply both directions ofFor $q<p$, exchange the exponents. Thus all finite-order p-Wasserstein distances induce the same convergence on a bounded subset of ${\mathbb{R}}^d$. Closedness of $X$ is not required for these estimates.

Let $e_m$ denote the sequence with a one in coordinate $m$ and zeros elsewhere.

For $T_1$, one has $\Vert e_m{\Vert }_p=1$ but $\Vert T_1{e}_m{\Vert }_p=m$ for every $p$, including $p=\infty$. Thus $T_1$ is unbounded for every $p$.

For $T_2$, the sequence $T_2{e}_m$ has entries $-(m-1)$ and $m$ in coordinates $m-1$ and $m$, respectively. Hence $\Vert T_2{e}_m{\Vert }_p\ge m$, so $T_2$ is also unbounded for every $p$.

For $T_3$, coordinatewise division givesfor every $1\le p\le \infty$. Thus $T_3$ is a bounded diagonal operator on sequence space for every $p$.

For $T_4$, write $a=(n^{-1/2}{)}_{n\ge 1}$. Then $T_{4}x=x_{1}a$, and coordinate evaluation satisfies $|x_1|\le \Vert x{\Vert }_p$. By the p-series criterion, $a\in {\ell }^p$ exactly when $p>2$, while $a\in {\ell }^{\infty }$. Therefore $T_4$ is bounded precisely forFor $p\le 2$, even $T_4{e}_1$ does not belong to ${\ell }^p$.

For $T_5$, let $q$ be the Holder conjugate exponent of $p$. Holder inequality gives, for finite $p$,and consequentlyFor $p=\infty$,Thus $T_5$ is bounded for every $1\le p\le \infty$. Altogether,

The continuous dual space $X^{\prime }$ consists of all bounded linear maps $\Lambda :X\to \mathbb{C}$, withIf $Y=L^p({\mathbb{R}}^n)$ with $1\le p<\infty$ and $q$ is the conjugate exponent, then duality of Lp spaces identifies $Y^{\prime }$ isometrically with $L^q({\mathbb{R}}^n)$, including $q=\infty$ when $p=1$. Every functional has the formand $\Vert {\Lambda }_h\Vert =\Vert h{\Vert }_q$.

Linearity in $u$ follows from linearity of the weak derivative and Lebesgue integration. By the Holder inequality, the three-dimensional Sobolev inequality, and the preceding $L^3$ interpolation estimate,Thus $\Phi$ is a linear functional and a continuous linear map on $H^1(U)$.

The Holder inequality interpolates between $L^2(U)$ and $L^6(U)$:Taking cube roots and applying the three-dimensional Sobolev inequality $H^1(U)\hookrightarrow L^6(U)$ givesThis is the H1 L3 interpolation inequality in three dimensions.

The preceding estimate gives constants $A,B>0$ such that, whenever $\Vert w{\Vert }_{H^2}\le R$,Choose $R>0$ so that $B{R}^2\le 1/2$, and then choose $r>0$ so that $Ar\le R/2$. If $\Vert f{\Vert }_2\le r$, the closed ball $B_R(0)$ is mapped into itself.

For $w,z\in B_R(0)$, factor the cubic gradient nonlinearity asThe same Sobolev embedding theorem and the Holder inequality implyThe elliptic estimate therefore yieldsShrinking $R$ further makes $C^{\prime }{R}^2<1$, so $\Phi$ is a contraction mapping of the closed ball. This ball is complete because $H^2(U)\cap H_0^1(U)$ is a Banach space. The contraction mapping theorem gives a fixed point $u=\Phi (u)$, and its defining equation isThus the nonlinear elliptic boundary value problem has a solution for sufficiently small $\Vert f{\Vert }_2$.

Put $\hat{\Sigma }=X^{T}X/n$. Direct substitution of $Y=X{\beta }^0+\epsilon$ into the Debiased Lasso givesConditionally on the deterministic design,The assumed approximate inverse of a Gram matrix property and Holder inequality implyThus $\rho (n,p)=\sqrt{\log p}$.

Strong convergence means $\Vert f_j-f{\Vert }_p\to 0$; weak convergence means $\int f_{j}h\to \int fh$ for every $h\in L^q$. By duality choose $h$ of norm one with $\int fh$ arbitrarily close to $\Vert f{\Vert }_p$. Holder inequality and passage to the limit give $\Vert f{\Vert }_p\le \mathrm{liminf}\Vert f_j{\Vert }_p$.

In three dimensions, the Sobolev inequality gives $\Vert v{\Vert }_{L^6(U)}\le C_U\Vert v{\Vert }_{H_0^1(U)}$. Thereforeand hence

Using $w^3-{\tilde{w}}^3=(w-\tilde{w})(w^2+w\tilde{w}+{\tilde{w}}^2)$, the Holder inequality and the same Sobolev embedding give at each timeTaking the $L^2$ norm in time supplies the required estimate with the factor $T^{1/2}$.

Split the integral at radius $R$. The Holder inequality on the ball and the spatial moment outside it giveTaking $R={\epsilon }^{-1/2}$ proves the stated estimate with $C_{\epsilon }=\sqrt{\pi }{\epsilon }^{-1/2}$. Choose $\epsilon <1/\omega$. If $X=\Vert v{\Vert }_4^2$, thenThe final two terms form a quadratic polynomial bounded below, so ${\inf }_{\Sigma }J>-\infty$.

For a measurable set $E\subseteq U$, definewhere $1_E$ is its indicator function. This is finite because $U$ has finite measure and hence $1_E\in L^p(U)$. If the sets $E_k$ are pairwise disjoint and $E={\bigcup }_{k\ge 1}{E}_k$, thenThe continuity of the positive linear functional $\Lambda$ therefore givesso $\nu$ is a measure by countable additivity. Moreover, a set of zero Lebesgue measure has zero indicator in $L^p$, so $\nu$ is absolutely continuous with respect to Lebesgue measure.

The Radon-Nikodym theorem now supplies a measurable function $\omega \ge 0$ such thatConsequently, first for nonnegative simple functions and then, by approximation by nonnegative simple functions, for every nonnegative $f\in L^p(U)$,Applying the stated characterization of the Lp norm to $\omega$ shows that $\omega \in L^q(U)$ and $\Vert \omega {\Vert }_q\le \Vert \Lambda \Vert$. The Holder inequality makes $f\mapsto {\int }_{U}f\omega$ continuous on $L^p(U)$, while bounded simple functions form a dense subset; henceThis is the positive functional representation on Lp.

Take $u\in W_0^{1,p}(U)$ and extend it by zero outside $U$. The zero extension of W01 belongs to $W^{1,p}({\mathbb{R}}^n)$ and the Sobolev inequality givesBecause $U$ has finite measure, the Holder inequality givesThusThe reverse estimate follows directly from $\Vert Du{\Vert }_{L^p}\le \Vert u{\Vert }_{W^{1,p}}$. After adjusting constants,This is the Poincare inequality on $W_0^{1,p}(U)$.

SetThe Holder inequality givesUsing $x^{-p}dx=-(p-1{)}^{-1}d(x^{1-p})$ and integration by parts, while discarding the nonpositive boundary term at $x=1$, yieldsAnother application of Hölder's inequality givesAfter cancellation, with the zero case immediate,This is the Hardy averaging inequality.

The condition ${\sum }_{i=1}^{n}1/p_i=1$ and positivity imply $p_i\ge 1$. We prove the Generalized Holder inequality by induction on $n$. The case $n=2$ is the usual Holder inequality. For the induction step, setThe exponents $q$ and $p_n$ are conjugate, so Hölder followed by the induction hypothesis givesThis proves the claim.

Use the distribution $R$ associated with the code in part b. Since $2^{L(x)}=(KR(x){)}^{-1}$,For $\alpha =1/(1+\rho )$, Holder inequality, equivalently the indicated Jensen inequality, givesTherefore

Tonelli's theorem and the triangle inequality giveApplying Holder inequality in the $y$ variable for each fixed $x$ yieldsand $\Vert g{\Vert }_q\le 1$ gives the claimed inequality. This is the duality proof of the Minkowski integral inequality.

For a smooth $u$ and fixed $y$, the fundamental theorem of calculus gives, for $0<x<1$,Average this inequality over $x\in (0,1)$, use Holder inequality on that unit interval, raise to the power $p$, and integrate in $y$. This proves the estimate behind the W1p trace theorem on a half-space:Use the Sobolev extension operator from part d, approximate $Eu$ in $W^{1,p}({\mathbb{R}}^2)$ by smooth functions, and define $Tu$ as the $L^p(\mathbb{R})$ limit of their restrictions to $x=0$. The trace inequality makes this limit independent of the approximation and proves thatis linear and bounded. For a smooth function that extends continuously to the boundary, $Tu=u{|}_{\partial U}$, so this is the trace operator required.

For a harmonic function $u$, letThe divergence theorem givesSince $M(r)\to u(x)$ as $r\downarrow 0$, $M(r)=u(x)$. Integrating the spherical averages in the radial variable gives the corresponding ball average, proving the mean value property for harmonic functions. If $u$ attains its maximum at an interior point, the average of the nonnegative function $\max u-u$ on every sufficiently small centred sphere is zero. Continuity makes $u$ constant on those spheres, and connectedness propagates that value through the domain. Thus the weak maximum principle for elliptic operators gives

For the derivative estimate, choose $r>0$ smaller than half the distance from ${\Omega }^{\prime }$ to $\partial \Omega$, and let ${\rho }_r$ be a smooth radial mollifier supported in $B_r(0)$. Writing its convolution in polar coordinates and using the spherical mean value property shows that $u\ast {\rho }_r=u$ on ${\Omega }^{\prime }$. Hence, for every multi-index $\alpha$,so Holder inequality gives