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Every $2\times 2$ principal submatrix of a kernel matrix is positive semidefinite, soThe Cauchy-Schwarz inequality for integrals therefore givesThus every entry of $k(x,y)=\int k_{\tau }(x,y)d\tau$ is well defined. For any finite coefficients $c_r$ and points $x_r$, linearity of the integral givesbecause the integrand is nonnegative. Hence $k$ is a positive-semidefinite kernel. This proves the integral closure of positive-semidefinite kernels.

Conditional on the covariates and all responses used to construct $\hat{g}$, the residual ${\epsilon }_i$ has conditional second moment at most $C$; conditional independence under the null is what permits this conditioning. ConsequentlyThe Markov inequality proves

Next, the Cauchy-Schwarz inequality givesThe first factor converges to zero in probability. The weak law of large numbers and part a make the second $O_p(1)$, so the product converges to zero in probability.

Two applications of the Cauchy-Schwarz inequality giveandHere the empirical residual second moment is again $O_p(1)$ by the weak law of large numbers, while the assumed product error and the conclusion of part b are $o_p(1)$.

WriteExpanding $R_i^2$ gives the leading term ${\epsilon }_i^2{\xi }_i^2$ and terms of the forms treated in parts b and c, together with their versions obtained by interchanging $(\epsilon ,F)$ and $(\xi ,G)$. For example, the pure error terms are ${\epsilon }_i^2{G}_i^2$, ${\xi }_i^2{F}_i^2$, and $F_i^2{G}_i^2$, and each cross term is controlled by Cauchy-Schwarz from these. HenceAssuming this limit is positive, the continuous mapping theorem yields ${\tau }_D\stackrel{p}{\to }\{\mathbb{E}({\epsilon }_1^2{\xi }_1^2){\}}^{1/2}$. Combining this with the assumed convergence in distribution of $\sqrt{n}{\tau }_N$ and applying the Slutsky theorem givesThis is the studentization of the generalized covariance measure statistic.

A centered random variable $X$ is sub-Gaussian with parameter $a>0$ whenfor every $t\in \mathbb{R}$.

The Bernstein concentration inequality for products of sub-Gaussian variables quoted in the course says that if each coordinate of the identically distributed pairs $(U_i,V_i)$ is sub-Gaussian with parameter $\sigma /4$, thenSince $U^{T}V={\sum }_i{U}_i{V}_i$, this is the required bound. It follows by observing that a product of sub-Gaussian variables is sub-exponential and applying Bernstein's inequality to the independent centered products.

For any vector $\delta$ admissible in the definition of ${\varphi }_{\Sigma }^2$, one has $\Vert \delta {\Vert }_1\le 4$. The entrywise maximum norm bound therefore impliesBy the definition of the compatibility constant, ${\delta }^T\Sigma \delta \ge {\varphi }_{\Sigma }^2/s$, and henceTaking the infimum proves ${\varphi }_{\Theta }^2\ge {\varphi }_{\Sigma }^2/2$. This is the stability of a compatibility constant under entrywise perturbation.

Put $C=X^{T}X/n$. Applying the product concentration bound with the stated $t$ and using $t\le {\sigma }^2/3$ gives, for every $j,k$,There are $p(p+1)/2$ distinct entries in the symmetric matrix, so the union bound shows that the eventhas probability at least $1-p/(p+1{)}^2\ge p/(p+1)$.

On $\mathcal{E}$, $|C_{jj}-1|\le t$. Since $|{\Sigma }_{jk}|\le 1$ by Cauchy-Schwarz, normalization of the sample columns givesChoosing one coordinate of $S$ in the infimum shows ${\varphi }_{\Sigma }^2\le s$, so the assumed bound on $t$ is below one and, more precisely,The perturbation result now gives ${\varphi }_{\hat{\Sigma }}^2\ge {\varphi }_{\Sigma }^2/2$ throughout $\mathcal{E}$, and therefore

Put $G={\max }_i\Vert a_i{\Vert }_2$. By the Cauchy-Schwarz inequality,Interchanging $x$ and $y$ proves the Lipschitz bound

The subgradient method chooses $g_k\in \partial f(x_k)$ and a step size $t_k>0$, then setsAssume, as the question's use of $\min f$ requires, that a minimizer $x_{\ast }$ exists, and write $R=\Vert x_0-x_{\ast }{\Vert }_2$. Since every subgradient here has Euclidean norm at most $G$, the standard best-iterate estimate isTaking a suitable constant step when the target accuracy is known, or a standard diminishing sequence, gives error at most $ϵ$ initerations, so the requested exponent is $p=2$.

Assume $p\in L^{\infty }(-1,1)$ with $p(x)\ge p_0>0$, $q\in L^{\infty }(-1,1)$ with $q\ge 0$, and $f\in L^2(-1,1)$. The natural energy space is the Sobolev space $V=H_0^1(-1,1)$. Multiplyingby $v\in V$ and using integration by parts gives the weak formulationwhereThe form is bounded, and the Poincare inequality together with $p\ge p_0$ makes it coercive:The functional $\ell$ is bounded by the Cauchy-Schwarz inequality. The Lax-Milgram theorem therefore gives a unique weak solution.

Define the energyIf $y$ is the weak solution, then for every $w\in V$,Thus $y$ is the unique minimizer of $J$. Conversely, differentiating $J(y+tw)$ at $t=0$ recovers $a(y,w)=\ell (w)$, so the minimization and weak problems are equivalent.

The Ritz method chooses a finite-dimensional conforming space $V_h\subset V$ and minimizes $J$ over $V_h$. Its minimizer $y_h$ satisfiesFor a basis $\{{\varphi }_j\}$ this becomes $Kc=F$, withCoercivity makes the stiffness matrix $K$ symmetric positive definite, so the discrete minimizer exists and is unique.

Subtracting the continuous and discrete equations gives Galerkin orthogonality,In the energy norm $\Vert v{\Vert }_a=\sqrt{a(v,v)}$, for any $v_h\in V_h$,After cancellation and minimization over $v_h$, this proves the sharp symmetric form of the Céa lemma:

For a mesh $-1=x_0<x_1<\cdots <x_N=1$, take $V_h$ to be the continuous functions that are affine on each interval and vanish at the endpoints. The interior piecewise-linear hat functions form a basis, and each overlaps only its two neighbours, so $K$ is symmetric tridiagonal. In the illustrative case $p=1$ and $q=0$ on a uniform mesh, its diagonal and neighbouring entries areIf $y\in H^2(-1,1)$, its nodal interpolant $I_{h}y\in V_h$ satisfies $\Vert y-I_{h}y{\Vert }_{H^1}=O(h)$. The best-approximation estimate therefore gives $\Vert y-y_h{\Vert }_{H^1}=O(h)$, and a standard duality argument improves the $L^2$ error to $O(h^2)$ under the corresponding elliptic regularity. This completes the route from the boundary-value problem through its variational principle to a sparse, stable finite-element approximation.

Suppose first that the stated unitary and environment states exist. Preservation of the inner product givesThe environment states are normalized, so the Cauchy-Schwarz inequality gives $|⟨e_0|e_1⟩|\le 1$. Therefore

Conversely, writeand assume $|a|\le |b|$. If $b\ne 0$, put $c=a/b$, so $|c|\le 1$, and chooseThen $⟨e_0|e_1⟩=c$ and the desired output vectors have inner product $bc=a$, exactly matching the input vectors. If $b=0$, the inequality forces $a=0$, and one may take $|e_0⟩=|e_1⟩=|0⟩$.

In either case the input pair and output pair have the same Gram matrix. The isometry taking one pair to the other extends to a unitary matrix, so the required $U$ exists. Hencewhich is the environment-assisted two-state pure-state transformation criterion.

For $s>\frac{1}{2}$, the Sobolev trace theorem states that the restriction map initially defined on smooth functions extends uniquely to a bounded linear operator

Take first $u$ in the Schwartz space. Up to the harmless constant determined by the Fourier transform convention,The Cauchy-Schwarz inequality, with weights $1+|{\xi }^{\prime }{|}^2+{\xi }_n^2$, givesAfter the substitution ${\xi }_n=(1+|{\xi }^{\prime }{|}^2{)}^{1/2}t$, the second factor iswhere finiteness is exactly the condition $s>1/2$. Multiplying by $(1+|{\xi }^{\prime }{|}^2{)}^{s-1/2}$ and integrating in ${\xi }^{\prime }$ yieldsDensity of the Schwartz space in the Sobolev space completes the unique bounded extension.

Compact support and the fundamental theorem of calculus give, for every $(x,y)$,whereApplying the same argument in the other coordinate gives

Multiplication and Fubini's theorem now yieldFinally, the Cauchy-Schwarz inequality givesThus the required estimate holds with $C=4$; it is the Ladyzhenskaya inequality in two dimensions.

The set $C$ is the second-order cone. PutWhen $\rho \ge |t|$ and $\rho >0$, one has $0\le \alpha \le 1$, and the proposed pointlies on the boundary of $C$ because $\Vert \alpha u{\Vert }_2=\alpha \rho$.

Set $\delta =(\rho -t)/2\ge 0$. Sincefor any $(v,s)\in C$ we obtainwhere the first inequality is the Cauchy-Schwarz inequality and the second uses $\Vert v{\Vert }_2\le s$. Part (d) therefore giveswhenever $\Vert u{\Vert }_2\ge |t|$ and $u\ne 0$. At $(u,t)=(0,0)$ the projection is plainly $(0,0)$, consistent with the limiting formula.

If $\rho \le -t$, then for every $(v,s)\in C$,Applying part (d) with $\pi =0$ givesTogether with the trivial case $\rho \le t$, this is the full projection onto the second-order cone.

A weak solution is a function $u\in H_0^1(U)$ satisfyingThe zero boundary condition is encoded by membership in $H_0^1(U)$.

On the Hilbert space $H_0^1(U)$ defineThe form is bounded and coercive:The functional ${\ell }_f$ is bounded by the Cauchy-Schwarz inequality. The Lax-Milgram theorem therefore gives a unique weak solution and a boundThus $T:L^2(U)\to H_0^1(U)$ is bounded. Composing it with the compact Rellich-Kondrashov compactness theorem for H01 embeddingshows that $T:L^2(U)\to L^2(U)$ is the compact massive-Laplacian resolvent.

For $\mathbf{a},\mathbf{b}\in {\mathbb{R}}^{n+1}$, the triangle inequality and the Cauchy-Schwarz inequality giveThus $S$ is a continuous function. The unit sphere in the finite-dimensional Euclidean normed vector space ${\mathbb{R}}^{n+1}$ is compact. Moreover, $S(\mathbf{a})=0$ implies that the polynomial $T\mathbf{a}$ vanishes identically, so every coefficient $a_r$ is zero. Hence $S$ is strictly positive on the unit sphere. By the extreme value theorem, it has a positive minimum

The supremum norm is homogeneous. For $\mathbf{a}\ne 0$,It follows directly that $\Vert T\mathbf{a}{\Vert }_{\infty }\to \infty$ whenever $\Vert \mathbf{a}{\Vert }_2\to \infty$.

We prove the estimate first for $u$ in the Schwartz space. By the Fourier inversion theorem and the Cauchy-Schwarz inequality,The integral is finite because $n>n/2$.

For $0<\alpha \le 1$, the elementary boundgivesThe last integral is finite near zero for every $\alpha >0$ and at infinity exactly when $2\alpha <n$. Choose, for example, $\alpha =1/4$, which works for every $n\ge 1$. We obtainDensity of the Schwartz space in $H^n$ extends the estimate and supplies a unique Hölder-continuous representative. Thereforecontinuously. This is the Fourier proof of Hölder regularity from a Sobolev norm.

Differentiate the eigenvalue equationat $t=0$, where $A^{\prime }(0)=E$. Suppressing the argument $0$ givesLeft-multiplication by the transpose of the unit left eigenvector $u$ and use of $u^{T}A=\lambda u^T$ cancel the terms involving $v^{\prime }$:Because the eigenvalue is simple, $u^{T}v\ne 0$, and hence the first-order perturbation of a simple eigenvalue isThe definition of the operator norm, followed by the Cauchy-Schwarz inequality, givesTherefore

For every $N$, the Cauchy-Schwarz inequality givesThe right-hand side is bounded independently of $N$. Since the summands are positive, the partial sums increase to a finite limit, proving the geometric means of two summable positive sequences result.

Another application of the Cauchy-Schwarz inequality givesWhen $p>1/2$, the second factor is bounded because $2p>1$, so the series converges.

At the endpoint, take $a_1=1$ and, for $n\ge 2$,Then $\sum a_n$ converges, whilewhose series diverges. This proves both parts of weighted square roots of a summable sequence.

Write the Laurent expansion on the punctured disc asFor every integer $k\ge 1$, the Laurent coefficient formula and the Cauchy-Schwarz inequality giveLetting $r\downarrow 0$ shows that $a_{-k}=0$. Every coefficient in the principal part vanishes, so the uniform L2 circle bound for a removable singularity proves that

Define on $H$ the bilinear formand the linear functionalThe Hardy inequality on an interval, Cauchy-Schwarz inequality, and the one-sided Poincare inequality show that $a$ is a bounded bilinear form and that

For $v\in H$, $v(x)={\int }_0^x{v}^{\prime }(t)dt$, so Cauchy--Schwarz and Fubini's theorem giveHenceThus $a$ is a coercive bilinear form. The Lax-Milgram theorem supplies a unique $u\in H$ satisfying $a(u,v)=\ell (v)$ for every $v\in H$. This is precisely the unique weak solution described by the weak boundary value problem with an inverse-square potential.