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Thomson principle states that the effective resistance is the minimum energy of a unit flow from $a$ to $z$:An edge cutset separating $a$ and $z$ is a set of edges whose removal disconnects them. Every unit flow has net flux one across each such cutset ${\Pi }_k$. By Cauchy-Schwarz inequality,For disjoint cutsets, summing these energy lower bounds and applying Thomson's principle gives the Nash-Williams inequalityThe commute time identity is

The limiting conditional error of one-nearest-neighbour classification is $1-{\sum }_k{p}_k(x{)}^2$. Put $M={\max }_k{p}_k(x)$ and $r=1-M$. By the Cauchy-Schwarz inequality, the other $K-1$ probabilities satisfyConsequentlyNow $\mathbb{E}r=R_{\mathrm{Bayes}}$, and Jensen inequality gives $\mathbb{E}{r}^2\ge (\mathbb{E}r{)}^2$. Taking expectations proves

Take the $\tilde{H}$ inner product of the finite-dimensional equation with $u_m$. Part a gives exact cancellation of the nonlinear term:By the Cauchy-Schwarz inequality and the Poincare inequality,Thus every solution satisfiesThe bound is independent of $m$.

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 Poincaré 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

The hypothesis implies $v_1,v_2\ne 0$, since the zero vector is a scalar multiple of every vector. SetSince $v_1$ is not a scalar multiple of $v_2$, $w\ne 0$. Positive definiteness and bilinearity of the real inner product giveMultiplying by $\Vert v_2{\Vert }^2>0$ and taking square roots proves the strict Cauchy-Schwarz inequalitySince every real number $s$ satisfies $s\le |s|$, this proves, in particular,The argument derives the required inequality directly from the inner product axioms.

The centered regressors make the first normal equation read $n{\hat{\beta }}_0={\sum }_i{Y}_i$. DefineThe remaining equations are $A{\hat{\beta }}_1+B{\hat{\beta }}_2=u$ and $B{\hat{\beta }}_1+C{\hat{\beta }}_2=v$. Provided $D=AC-B^2>0$, matrix inversion givesBy the Cauchy-Schwarz inequality, $D\ge 0$. If $D=0$, the two regressor columns are linearly dependent, and their separate slopes are not uniquely identifiable. One still has well-defined linear least squares fitted values, but cannot use the displayed divisions by $D$.

For random variables with finite second moments, their covariance isThe Cauchy-Schwarz inequality ensures the product is integrable. Expand and use linearity of expectation:The same identity holds whenever the quantities involved exist with sufficient integrability.

At $(t,t)$ with $t^2=1/2$, differentiating the gradient yields the Hessian matrix $H=-tI$. At the positive point it is negative definite, and at the negative point it is positive-definite; in both cases $f_{xy}=0$ and $\det H=1/2$. Therefore the positive point is a maximum, the negative point is a minimum, and there are no saddle points.

They are global extrema: with $r=\sqrt{x^2+y^2}$, the Cauchy-Schwarz inequality gives $|f|\le \sqrt{2}r/(1+r^2)\le 1/\sqrt{2}$, with equality at exactly those two points. This is the extremum bound underlying the level sets of a linear numerator over a radial quadratic.