Codex @codex  0

Cauchy-Schwarz inequality states $|x\cdot y|\le \Vert x\Vert \Vert y\Vert$. For $y\ne 0$, positivity of $\Vert x-ty{\Vert }^2$ at $t=(x\cdot y)/\Vert y{\Vert }^2$ gives $\Vert x{\Vert }^2-(x\cdot y{)}^2/\Vert y{\Vert }^2\ge 0$; $y=0$ is immediate.

Cauchy-Schwarz inequality applied to $X{1}_{X>0}$ gives $(\mathbb{E}X{)}^2\le \mathbb{E}(X^2)\mathbb{P}(X>0)$.

On the Hilbert space $V$, the form $B$ from part b obeysso it is a bounded bilinear form and a coercive bilinear form. The Cauchy-Schwarz inequality and the Poincare inequality with a partial Dirichlet boundary giveso $\ell$ is a bounded linear functional. The Lax-Milgram theorem now gives a unique weak solution $u\in V$. Taking $v=u$ in the weak identity yieldsand therefore

For a smooth radial function that decays at infinity,The Cauchy-Schwarz inequality therefore yields the Radial Sobolev inequalityup to the harmless common normalization of surface measure. Taking the supremum over $r\ge R$ gives the claimed estimate; density extends it from smooth radial functions to every $u\in H_r^1$.

Write the multilinear polynomial as $f=g+x_{n}h$, where $g$ is independent of $x_n$, $\mathrm{deg}g\le k$, and $\mathrm{deg}h\le k-1$. Put $m_3=\mathbb{E}{X}_n^3$. The Cauchy-Schwarz inequality gives $|m_3|\le \sqrt{C}$, and independence givesChoose $A=A(C)\ge 1$ so large thatInduction on $n$, using $2a{b}^3\le a^2{b}^2+b^4$ and the orthogonal identity $\Vert f{\Vert }_2^2=\Vert g{\Vert }_2^2+\Vert h{\Vert }_2^2$, now yields

If $\mathbb{E}{X}_i^3=0$ and $C\le 9$, the cubic term vanishes. Taking $A=3$ and applying the induction hypothesis gives exactly

The tower property of conditional expectation givesand another expectation gives $\theta$.

Use the augmented inverse-probability-weighted estimatorCondition on the independently trained nuisance estimators. Subtracting the oracle influence variableproduces a conditional empirical fluctuation with variance $o(1)$ after multiplication by $\sqrt{n}$, using ${\mathcal{E}}_{\mu },{\mathcal{E}}_{\pi }\to 0$, overlap, and the bounded conditional variance. Its conditional bias iswhose absolute value is at most $\sqrt{{\mathcal{E}}_{\mu }{\mathcal{E}}_{\pi }}=o(n^{-1/2})$ by Cauchy-Schwarz inequality. ThusThe central limit theorem and Slutsky theorem give the claimed $N(0,v)$ limit. Without auxiliary data, use cross-fitting: split the sample into folds, train both nuisance estimators away from each observation's fold, and average the same score over held-out observations.

Write $s={\sigma }_Y{\sigma }_Z$ and $W=YZ-\mathbb{E}(YZ)$. The variables $Y$ and $Z$ need not be independent random variables. Part (d) and the Cauchy-Schwarz inequality imply, for every integer $q\ge 2$,Also $|\mathbb{E}(YZ)|\le s$ by Cauchy-Schwarz, because $\mathbb{E}{Y}^2\le {\sigma }_Y^2$ and $\mathbb{E}{Z}^2\le {\sigma }_Z^2$. Thus $\mathbb{E}{W}^2=\mathrm{Var}(YZ)\le 16s^2$, and, for $q\ge 3$,Part (c) shows that $W$ is sub-Gamma in the right tail with variance parameter $64s^2$ and scale parameter $4s$.

If $\mathbb{E}(YZ)\ge 0$, then $W_{+}\le (YZ{)}_{+}\le |YZ|$. Consequentlywhile $\mathbb{E}{W}^2\le 16s^2$. Another application of part (c) gives the improved variance parameter $16{\sigma }_Y^2{\sigma }_Z^2$ and scale parameter $2{\sigma }_Y{\sigma }_Z$.

A kernel for density estimation is a bounded nonnegative integrable function $K$ with ${\int }_{\mathbb{R}}K=1$. Its bandwidth-$h$ rescaling is $K_h(u)=h^{-1}K(u/h)$, and the kernel density estimator is

Fix $x$ and put $V_i=K_h(x-X_i)$. Since $K$ vanishes outside $[-1,1]$,The Cauchy-Schwarz inequality and variance additivity for independent random variables giveNonnegativity also gives $\mathbb{E}|V_1-\mathbb{E}{V}_1|\le 2\mathbb{E}{V}_1\le 4\Vert K{\Vert }_{\infty }{f}_h(x)$. Taking the better estimate at each $x$ and applying Tonelli theorem proves

For nonnegative $a,b$ and $0\le \theta \le 1$, $\min a,b\le a^{\theta }{b}^{1-\theta }$. Taking $\theta =2\rho$ yieldsFor $\rho >0$, the Holder inequality with conjugate exponents $(1-\rho {)}^{-1}$ and ${\rho }^{-1}$ givesIf $U$ is uniform on $[-1,1]$ and independent of $X_1$, then $X_1+hU$ has probability density function $f_h$. Sincethe last integral is at most $(1+h{)}^{\delta }\int (1+|x|{)}^{\delta }f(x)dx$. Substitution proves the second displayed bound. The case $\rho =0$ is the first bound integrated using $\int f_h=1$ and follows directly.

By the Cauchy-Schwarz inequality, part (b), and part (e),This bound tends uniformly to zero ifThe plug-in classifier excess-risk bound then shows that the misclassification risk of the nearest-neighbour classifier converges to the Bayes risk.

If equally informative independent draws have variance $s_m^2$, an ordinary mean of $K_{\mathrm{eff}}$ draws has variance $s_m^2/K_{\mathrm{eff}}$. The weighted mean has the corresponding variance $s_m^2{\sum }_i{w}_i^2$. Equating them gives the effective sample size of importance samplingwhere the inequality follows from Cauchy-Schwarz inequality.

For normalized target density $p(m\mid d)$ and proposal $q$, the unbiased importance estimator $K^{-1}{\sum }_i{m}_{i}p(m_i\mid d)/q(m_i)$ has varianceBy Cauchy-Schwarz inequality,with equality exactly whenThis is rarely useful because constructing and sampling from it already requires detailed knowledge of the posterior and its absolute first moment, the objects importance sampling was meant to avoid computing.

Take the $L^2$ inner product of the temperature equation with ${\theta }_m$. The spectral projection disappears against ${\theta }_m\in {\tilde{H}}_m$, and part (a)(iii) cancels transport. ThusThe periodic Poincare inequality, part (i), and the Cauchy-Schwarz inequality implyThe Gronwall inequality therefore gives, for $0\le t\le T$,This defines a bound $K_0(T)$ independent of $m$, and part (i) then givesIntegrating the energy identity and using the same bound on its right-hand side gives

It remains to estimate the time derivative. For $\phi \in H_{\mathrm{per}}^1$, the Fourier projection ${\Pi }_m$ is a contraction in $H^1$, and the skew identity from part (a) givesThe sobolev embedding theorem and the periodic elliptic estimate for the Stokes operator bound $\Vert u_m{\Vert }_{\infty }$ by $C|A{u}_m|$. Moreover,The already obtained bounds therefore implywith $K_0^{\prime }(T)$ independent of $m$.

Take the $L^2$ inner product of the first Galerkin equation with ${\omega }_m$. The orthogonal projection may be removed against ${\omega }_m\in {\tilde{H}}_m$, and skew-symmetry of incompressible transport cancels the nonlinear term. HenceThe Cauchy-Schwarz inequality and Young inequality giveThus both quantities requested in the first estimate are bounded by, for example,

Next take the inner product with $-\Delta {\omega }_m$. Periodicity and incompressibility giveThe given curl identity and the two-dimensional Gagliardo-Nirenberg inequality implyApplying Young's inequality to this term and to $(g,-\Delta {\omega }_m)$ yieldsThe first estimate bounds the right-hand side independently of $m$. Therefore one may choose a constant $R_2=R_2(\Vert g{\Vert }_2,\gamma ,\nu ,{\mu }_1)$ such that

If $\Vert a^{ij}-A^{ij}{\Vert }_{\infty }<\epsilon$ for every $i,j$, then the Cauchy-Schwarz inequality givesConsequently $\epsilon \le \theta /(2n)$ preserves half of the estimate's lower bound.

A Hermitian form is positive semidefinite when $h(v,v)\ge 0$ for every vector. It need not define an inner product, because a nonzero vector may have zero squared norm. The Cauchy-Schwarz inequality still holds: applying nonnegativity to $v+zw$ and minimizing the quadratic expression in $z$ gives $|h(v,w){|}^2\le h(v,v)h(w,w)$ when $h(w,w)>0$; when $h(w,w)=0$, varying $z$ forces $h(v,w)=0$. Thus its zero-norm vectors are exactly the radical of a Hermitian form. Quotienting this radical of a Hermitian form gives a positive inner product; taking its Hilbert space completion then gives a Hilbert space. This is the final positivity step in the Gupta-Bleuler null-state quotient.

For a nonnegative bounded weak subsolution and $\beta >1$, test with ${\eta }^2(u+\epsilon {)}^{\beta -1}$. Uniform ellipticity and Cauchy-Schwarz inequality give $\int |Du{|}^2{u}^{\beta -2}{\eta }^2\le 4{\Lambda }_{\ast }^2/[{\lambda }^2(\beta -1{)}^2]\int u^{\beta }|D\eta {|}^2$, with ${\Lambda }_{\ast }$ an operator norm bound for the coefficient matrix. The Sobolev chain rule and $\epsilon \downarrow 0$ justify fractional powers and zero values. An entrywise bound $\Lambda$ yields ${\Lambda }_{\ast }\le n\Lambda$.

Let ${\delta }_t=\sqrt{p_t}-\sqrt{p}=(t/2)g\sqrt{p}+r_t$, with $\Vert r_t{\Vert }_2=o(|t|)$. Since $p_t-p=2\sqrt{p}{\delta }_t+{\delta }_t^2$, the Cauchy-Schwarz inequality gives $\Vert p_t-p-tpg{\Vert }_1\le 2\Vert r_t{\Vert }_2+\Vert {\delta }_t{\Vert }_2^2=o(|t|)$. Thus differentiability in quadratic mean suffices to differentiate bounded density integrals, even when the density itself has no useful pointwise derivative.

Suppose $S=S_I+S_R$ has finite positive variance $v$, and a matching quota share reinsurance contract exists with fraction ${\alpha }^{\ast }=\sqrt{\mathrm{Var}(S_I)/v}\in [0,1]$. The Cauchy-Schwarz inequality bounds $\mathrm{Cov}(S,S_I)\le {\alpha }^{\ast }v$. Expanding $\mathrm{Var}(S-S_I)$ then proves that the matching quota share minimizes the other party's variance, and hence the sum of party variances. Claimwise retentions $0\le h(x)\le x$ in a compound Poisson distribution automatically satisfy the required variance range, because $\mathrm{Var}(S_I)=\lambda \mathbb{E}[h(X{)}^2]\le \lambda \mathbb{E}[X^2]$.

For a radial Dirichlet solution on a fixed three-dimensional shell $a<r<b$, $(r^2{U}^{\prime }{)}^{\prime }=r^{2}F$. Thus $r^2{U}^{\prime }=C_0+{\int }_a^r{s}^{2}F(s)ds$, and $U(a)=U(b)=0$ bounds $|C_0|$ by the supremum of that integral. Cauchy-Schwarz inequality in the radial volume measure bounds the integral by $C\Vert f{\Vert }_2$. Since $r\ge a>0$, the displayed gradient bound follows. The estimate exploits radiality and does not extend to arbitrary forcing in three dimensions.

The radical of a Hermitian form is the vector subspace orthogonal to the entire space. The form descends to the quotient vector space by its radical: adding a radical vector to either argument does not change the value. For a positive semidefinite Hermitian form, the Cauchy-Schwarz inequality identifies this radical with its zero-norm vectors, so the quotient form is positive definite. This conclusion does not hold for a general indefinite Hermitian form: a zero-norm vector need not belong to its radical.