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Let $B$ and $F$ satisfy the hypotheses of the Lax-Milgram theorem. By the Riesz representation theorem, there are a bounded linear operator $T:H\to H$ and $z\in H$ such thatCoercivity and the Cauchy-Schwarz inequality implyso $\Vert Tu{\Vert }_H\ge \alpha \Vert u{\Vert }_H$. Thus $T$ is injective and its range is closed. If $y$ is orthogonal to its range, then $B(x,y)=0$ for every $x$; taking $x=y$ and using coercivity gives $y=0$. The range is therefore dense as well as closed, hence all of $H$. There is a unique $u=T^{-1}z$, and it satisfies $B(u,v)=F(v)$ for all $v$. The lower bound also gives $\Vert u{\Vert }_H\le \Vert F\Vert /\alpha$.

LetApplying the Cauchy-Schwarz inequality in $t$ givesTake absolute values and apply Cauchy-Schwarz to $(r_1,r_2)$. Since ${\sum }_{r_1,r_2}|c_{r_1}{c}_{r_2}{|}^2=\Vert c{\Vert }_2^4$, expanding the remaining square givesTaking fourth roots proves the claimed bilinear quadratic exponential sum fourth-moment bound.

WriteApply the Cauchy-Schwarz inequality first in the variable carrying $f_1$ and then in the variable carrying $f_2$. After the invertible linear changes of variables permitted by $2,3\nmid |G|$, apply the assumed three-function $U^2$ estimate to the resulting multiplicative derivatives. The standard calculation givesBy the Derivative identity for the Gowers U3 norm, the last two factors are $\Vert f_3{\Vert }_{U^3}^8$ and $\Vert f_4{\Vert }_{U^3}^8$. Taking eighth roots proves

If $A\subseteq G$ has density $\alpha$, write $1_A=\alpha +f$. Expanding ${\Lambda }_4(1_A,1_A,1_A,1_A)$, the constant term is ${\alpha }^4$, and the displayed inequality bounds every nonconstant term after translation of one factor by a $U^3$ norm of the balanced function $f$. Thus sufficiently small $\Vert 1_A-\alpha {\Vert }_{U^3}$ makes the normalized number of four-term arithmetic progressions close to ${\alpha }^4$.

Choose a test function $\chi$ that equals one on the unit ball and vanishes outside the ball of radius two. Sinceintegration by parts givesThe first term on the right is supported where $1\le |x|\le 2$, where $u^2\le |x{|}^{\alpha }{u}^2$. The Cauchy-Schwarz inequality and Young inequality bound the second term byAfter absorbing the first integral,

This estimate also proves completeness. Indeed, a Cauchy sequence in $\Sigma$ is Cauchy in $H^1$ locally, while its gradients and the functions $|x{|}^{\alpha /2}{u}_n$ converge in $L^2$. The limits agree locally with a function $u$, so $u\in \Sigma$ and $u_n\to u$ in the energy norm. Thus $\Sigma$ is a Hilbert space; it is the confining-potential energy space for $V(x)=|x{|}^{\alpha }$.

For a partition $\pi$ of $[0,t]$, apply the first inequality to each increment of the covariation and then the Cauchy-Schwarz inequality:Taking the supremum over partitions proves the Kunita-Watanabe inequality

Under $H_0$, conditional independence givesMultiplication by the measurable sign $s(z_1)$ and the law of total expectation prove the first identity.

WriteThenConditional orthogonality under $H_0$ and the variance bounds in (ii) giveand similarly for the term containing ${\xi }_i{a}_i$. The Cauchy-Schwarz inequality bounds the last term bywhich is $o_p(n^{-1/2})$ by $n{\mathrm{MSPE}}_f{\mathrm{MSPE}}_g\to 0$.

The leading summands ${ϵ}_i{\xi }_{i}s(z_i)$ are independent, centered, and have variance $\mathrm{Var}({ϵ}_1{\xi }_1)$ because $s^2=1$. The central limit theorem and consistency of ${\tau }_D$ therefore give, by the Slutsky theorem,This is the generalized covariance measure statistic.

Without the null, condition first on $(x_1,z_1)$. Sincethe term involving $\mathbb{E}(x_1\mid z_1)$ vanishes, givingFor the specified alternative, independence and $\mathbb{E}{x}_1=0$ make the right sideThe constant choice $s=1$ gives zero because $\mathbb{E}h(z_1)=0$, so no first-order power is expected. Takinginstead gives $\mathbb{E}(x_1^2)\mathbb{E}|h(z_1)|>0$, producing asymptotic power.

The Group Lasso penalty isBy the Cauchy-Schwarz inequality within each group,

Put $\delta =\hat{\beta }-{\beta }^0$. Comparing the objective at $\hat{\beta }$ and ${\beta }^0$ givesOn $\Omega$, the preceding duality inequality givesThe triangle inequality $P(\delta )\le P({\beta }^0)+P(\hat{\beta })$ then yields

When $m_j=\sqrt{r}$ and $X_{G_j}^T{X}_{G_j}=n{I}_r$,Writing ${\delta }_0=n{\lambda }^2-1\in (0,1)$, the exponential Markov inequality and the supplied chi-square moment-generating-function bound giveThe prescribed equation makes this $q^{-(A+1)}$. A union bound over the $q$ groups therefore gives

The pushforward measure of $\mu$ under $g$ isThe Lebesgue decomposition theorem says that for sigma-finite measures $P,Q$ there are unique measures $P_{\mathrm{ac}}$ and $P_{\mathrm{s}}$ such that

For a convex lower-semicontinuous function $f$ with $f(1)=0$, choose any measure $\lambda$ dominating probability measures $P,Q$, put $p=dP/d\lambda$ and $q=dQ/d\lambda$, and define the f-divergenceusing the lower-semicontinuous perspective value when $q=0$. This definition includes the singular part and is independent of $\lambda$.

To prove the data processing inequality for f-divergences, let $\mathcal{G}=\sigma (g)$ and use $\lambda =P+Q$. The densities of $g_{\ast }P$ and $g_{\ast }Q$, pulled back to $\mathcal{G}$, are ${\mathbb{E}}_{\lambda }(p\mid \mathcal{G})$ and ${\mathbb{E}}_{\lambda }(q\mid \mathcal{G})$. Since the perspectiveis jointly convex, conditional Jensen inequality givesIntegration proves

The chi-squared divergence iswhen $P\ll Q$, and is infinite otherwise.

Fix a probability measure $Q$. Let $J$ be uniform on $\{1,\dots ,M\}$ and, conditionally on $J=j$, draw $X$ from $P_j$. Compare this joint law with the reference law under which $J$ is uniform and independent of $X\sim Q$. Forthe target expectation is $M^{-1}{\sum }_j{P}_j(A_j)$, while its reference expectation isbecause the $A_j$ form a partition. Its reference variance is $M^{-1}(1-M^{-1})$. The Cauchy-Schwarz inequality applied to the likelihood ratio givesThe last divergence separates over $J$:Taking the infimum over all probability measures $Q$ proves the required inequality.

Take $|L⟩=|0⟩$ and $|R⟩=|1⟩$. Initially the two masses are in the product stateThe branch-dependent Newtonian gravitational potential energy is $-G{m}^2/d_{ab}$. Under the stated approximation, only the $|RL⟩$ branch acquires an appreciable relative phase, so after time $t$,The determinant of its two-by-two coefficient matrix is $(1-e^{i\varphi })/4$, which is nonzero unless $\varphi$ is a multiple of $2\pi$. Thus the state generally has Schmidt rank two: the branch-dependent gravitational phase creates gravitationally induced entanglement.

An entanglement witness has a bound obeyed by every separable quantum state and violated by at least one entangled state. For a product state with Bloch vectors $\mathbf{r}$ and $\mathbf{s}$,by the Cauchy-Schwarz inequality. Convexity gives the same bound for every separable mixed state. Consequently $W>1$ certifies entanglement; in conventional operator form, one of $I\mp (X\otimes Z+Y\otimes Y)$ has negative expectation whenever the absolute-value criterion is violated.

For the state above, direct use of the Pauli matrices givesWith the supplied values,which is close to $\pi$. Hence $W\simeq 2.00$, and to the nearest integerThis is the operating principle of the Bose--Marletto--Vedral experiment.

Let $T_{\alpha }=A^{\ast }A+\alpha I$. For every $f\in X$,Thus $T_{\alpha }$ is a coercive operator. If it is invertible and $T_{\alpha }f=g$, the lower bound and the Cauchy-Schwarz inequality giveand therefore $\Vert T_{\alpha }^{-1}g\Vert \le {\alpha }^{-1}\Vert g\Vert$. Hence

The extremal number $\mathrm{ex}(n,H)$ is the maximum number of edges in an $n$-vertex graph containing no subgraph isomorphic to $H$.

Let $G$ be triangle-free with $e$ edges. For every edge $uv$, the neighbourhoods of $u$ and $v$ are disjoint apart from their endpoints, so
$d(u)+d(v)\le n$. Summing over edges and applying Cauchy-Schwarz inequality giveswhileThus $e\le n^2/4$, proving the required Mantel theorem bound.

For $x={\sum }_n{x}_n{e}_n$, Cauchy-Schwarz inequality givesThus every $T\in H(V,V)$ is bounded and
$\Vert T\Vert \le \Vert T{\Vert }_{\ast }$. The truncationshave finite rank and satisfyHence every such Hilbert-Schmidt operator is compact.

DefineThis is an inner product inducing $\Vert \cdot {\Vert }_{\ast }$. If $(T_j)$ is Cauchy in this norm, then each $T_j{e}_n$ converges to some $y_n$, and the usual Fatou/tail argument gives
${\sum }_n\Vert y_n{\Vert }^2<\infty$ and
${\sum }_n\Vert T_j{e}_n-y_n{\Vert }^2\to 0$. Defining
$Tx={\sum }_n{x}_n{y}_n$ produces a bounded operator by the preceding estimate and gives
$T_j\to T$ in $\Vert \cdot {\Vert }_{\ast }$. Thus $(H(V,V),\Vert \cdot {\Vert }_{\ast })$ is a Hilbert space.

The norms are not equivalent in infinite dimension. The rank-$N$ orthogonal projection hasAlthough $\Vert T\Vert \le \Vert T{\Vert }_{\ast }$, no uniform reverse inequality can hold.

Colour every vertex of a graph independently and uniformly with three colours, and retain precisely the edges whose endpoints have different colours. Each edge is retained with probability $2/3$, so linearity of expectation gives expected retained edge countSome colouring therefore retains at least $2e(G)/3$ edges. Its retained subgraph is three-colourable, and deleting edges if necessary leaves exactlyedges without increasing its chromatic number. This is the three-colourable two-thirds subgraph lemma.

To prove sharpness, take $G=K_n$. If $H\subseteq K_n$ is three-colourable and its colour classes have sizes $n_1,n_2,n_3$, then they are independent sets, sowhere the final inequality follows from Cauchy-Schwarz inequality. Since $e(K_n)=n(n-1)/2$,Given $\epsilon >0$, choose $n$ with $2/[3(n-1)]\le \epsilon$. Then every three-colourable subgraph has at most $(2/3+\epsilon )e(G)$ edges.

The periodic Wirtinger inequality says that if $f$ is continuously differentiable, $L$-periodic, and has mean zero, thenEquality holds exactly for a linear combination of $\cos (2\pi s/L)$ and $\sin (2\pi s/L)$.

First suppose $\partial \Omega$ is one positively oriented simple closed curve, parametrized by arc length as $\gamma (s)=(x(s),y(s))$ for $0\le s\le L$. Translating the origin, which changes neither length nor area, makes both $x$ and $y$ have mean zero. Green theorem and the Cauchy-Schwarz inequality giveApplying Wirtinger's inequality to both coordinate functions and using the unit-speed identity $(x^{\prime }{)}^2+(y^{\prime }{)}^2=1$ yieldsand therefore the planar isoperimetric inequalityFor several boundary components, apply the simple-curve result to the relevant enclosed regions and use ${\sum }_j{L}_j^2\le ({\sum }_j{L}_j{)}^2$; holes only decrease the area.

Equality in both inequalities forcesfor constant vectors $A,B$, with the unit-speed and Cauchy-Schwarz equality conditions making $A$ and $B$ perpendicular and equally long. Thus the boundary is a circle. Conversely, a circular domain has area $\pi R^2$ and perimeter $2\pi R$, so equality holds.

PutIndependence and $\mathrm{var}(Y_i)=\theta x_i$ givewhereas ${\sum }_i{Y}_i\sim \mathrm{Poisson}(\theta S_1)$ givesThe Cauchy-Schwarz inequality, applied to $(x_i^{1/2})$ and $(x_i^{3/2})$, yieldsConsequentlyEquality holds exactly when all positive exposures $x_i$ are equal. This is the variance comparison for Poisson exposure estimators.

The functionalis bounded on $H_0^1(\Omega )$ by the Cauchy-Schwarz inequality and the Poincare inequality:The Riesz representation theorem therefore supplies a unique $u\in H_0^1(\Omega )$ satisfyingfor every test function. This is exactly the weak identity from part (a). Uniqueness also follows by testing the homogeneous difference with itself.

Choose a cutoff function $\eta$ supported in $B_{\rho }(x)$, equal to one on $B_{{\rho }^{\prime }}(x)$, and satisfying $|D\eta |\le (\rho -{\rho }^{\prime }{)}^{-1}$. Use ${\eta }^{2}u$ as a test function in the weak formulation:The Cauchy-Schwarz inequality followed by Young inequality givesand henceThis is the Caccioppoli inequality. The numerical constant is convention-dependent and is normally absorbed into the displayed estimate; replacing the radii by fixed intermediate radii gives the stated form with one universal constant.

First replace $u$ by $u+\epsilon$ and later let $\epsilon \downarrow 0$. In the weak subsolution inequality use the admissible truncations approximating ${\eta }^2{u}^{\alpha -1}$. Uniform ellipticity, the coefficient bound, Cauchy-Schwarz inequality, and Young inequality giveand thereforeApply the Sobolev embedding theorem to $w=\eta u^{\alpha /2}$. The preceding estimate yields, for concentric balls $B_r\subset B_R\subset B_1$,Starting with $\alpha =p>1$, taking ${\alpha }_k=p{\sigma }^k$, and choosing radii decreasing to $1/2$, the product of constants converges because ${\sum }_k{\alpha }_k^{-1}<\infty$. Letting $k\to \infty$ provesThis exponent-raising argument is Moser iteration.

Use the four families from the hint. Put $N=2^n$, $u=|{\mathcal{U}}_1|/N$, and $v=|{\mathcal{V}}_1|/N$. The families ${\mathcal{U}}_1,{\mathcal{V}}_1$ are up-sets, while their complements ${\mathcal{U}}_2,{\mathcal{V}}_2$ are down-sets. Applying the Harris-Kleitman inequality to the up-sets, and equivalently to the down-sets after taking complements of the ground set, givesThe cross-incomparability assumption givesConsequentlyThe Cauchy-Schwarz inequality now yieldsThis is the sharp two-family Cross-Sperner inequality.

For $K>0$, the Cauchy-Schwarz inequality and part a giveThus $(Y_n^{-})$ is uniformly integrable. Combining this with the weak convergence of random variables from part b yields convergence of the first moments:By symmetry of the standard normal density,