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For unknown normalization constants, use self-normalized importance sampling with $\tilde{w}=\tilde{p}/\tilde{q}$:Under the same support condition, $E_q\tilde{w}=Z_p/Z_q>0$ and $E_q[\tilde{w}f]=(Z_p/Z_q)\mu$. Moreover $E_q|\tilde{w}f|=(Z_p/Z_q)E_p|f|<\infty$, by Cauchy-Schwarz inequality and the supplied $E_p{f}^2<\infty$. The strong law of large numbers applies to numerator and denominator sample averages. Their ratio tends almost surely, hence in probability, to $\mu$. If a finite sample happens to give zero denominator, it can be assigned an arbitrary value; this does not affect the eventual almost sure limit. The ratio estimator is generally biased at finite $N$.

Differentiability at $a$ means that there is a linear map $L:V\to V_1$ withThis map is the total derivative $D{f}_a$. In the Frobenius norm, $|AB|\le |A||B|$, by the Cauchy-Schwarz inequality applied to matrix entries. Expanding with the order of all factors preserved givesThe remainder after the terms linear in $H$ has norm at most $3|A||H{|}^2+|H{|}^3=o(|H|)$. HenceFor $|C|<1$, the Neumann series $I-C+C^2-\cdots$ converges absolutely in the finite-dimensional space of matrices. Multiplying its partial sums by $I+C$ and passing to the limit shows that it is $(I+C{)}^{-1}$. Its remainder after $I-C$ has norm at most $|C{|}^2/(1-|C|)=o(|C|)$, proving the required expansion.

If $A$ is an invertible matrix and $|H|<1/|A^{-1}|$, then $A+H=A(I+A^{-1}H)$ is an invertible matrix. Thus the general linear group is an open subset of the matrix space. The preceding expansion also givesApply the chain rule to $g(X)=(X^{-1}{)}^3$, using the cubic total derivative at $A^{-1}$ and this total derivative of inversion:The three terms generally cannot be combined by commuting $A$ and $H$.

For a single species both panels of the figure in c contribute, now with $j=1$. They are the $s$-channel annihilation and $t$-channel scalar exchange, where the Mandelstam variables are $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$ and $s+t+u=4m^2$. Fix the external-state order consistently with c. The exchange of identical external fermionic operators gives a relative minus sign:Changing the convention for ordering all external states can change the overall sign, but not this relative fermionic sign.

The exchange $p_1\leftrightarrow -p_4$ must include crossing symmetry the associated external spinors, as well as continuing the four-momenta. It interchanges $s$ and $t$, leaves $u$ fixed, and interchanges the two spinor pairings. Hence the analytically continued invariant scattering amplitude is antisymmetric:The first relation is understood with consistently chosen spinor crossing phases. It is not a substitution of negative-energy arguments into the original positive-energy $u$ spinors without changing external states.

Without repeating the spin sum calculation, c fixes the square of the first diagram. Crossing symmetry fixes the second by $s\leftrightarrow t$. Consequently the strongest immediate inference isThe interference function is not determined by the separate diagram squares. In particular adding the two squares and dropping interference is unjustified. The Cauchy-Schwarz inequality applied to the arrays of spin amplitudes also gives $|C(s,t)|\le 2|(s-4m^2)(t-4m^2)|$ on the physical scattering region.

For completeness, its value can be checked with one contracted trace, without another diagram calculation. The cross product isThe relative minus in the invariant scattering amplitude gives $C=-2K$. Thus the fully evaluated result, beyond the symmetry inference alone, isEvery term respects the inferred $s,t$ symmetry. In the massless limit the interference numerator reduces to $st$, which is nonzero at a generic scattering angle.

For a normalized state and a self-adjoint operator, the expectation value and dispersion areUse the physics convention that the inner product is conjugate-linear in its first argument, and assume the displayed products and second moments are defined. Self-adjointness gives $\overline{⟨\psi |A\psi ⟩}=⟨A\psi |\psi ⟩=⟨\psi |A\psi ⟩$, so the expectation is real. MoreoverIt vanishes exactly when $\psi$ is an eigenstate, with eigenvalue its expectation.

The product rule gives $XPf=-i\hslash x{f}^{\prime }$ and $PXf=-i\hslash (f+x{f}^{\prime })$, hence $\boxed{[X,P]f=i\hslash f}$ on their common differentiable domain. For general operators, put $z=⟨A\psi |B\psi ⟩$. ThenThe Cauchy-Schwarz inequality yieldsApply this to centered position and momentum operators, whose commutator is unchanged. Their squared norms are the variances, givingEquality requires both proportionality in Cauchy-Schwarz and a purely imaginary inner product. Thus for real means $x_0,p_0$,with real $\gamma$. Solving this first-order equation givesNormalizability on the real line forces $\gamma >0$. Writing ${\sigma }^2=\hslash /(2\gamma )$, all normalized equality states are thereforeThey have $\Delta X=\sigma$ and $\Delta P=\hslash /(2\sigma )$. A nonconstant quadratic phase would increase the product, so the full family here has a real positive Gaussian width and only a linear spatial phase.

Set $u=A\psi$, $v=B\psi$ and $z=⟨u,v⟩$. The self-adjointness of $A,B$ impliesThe Cauchy-Schwarz inequality followed by $|z{|}^2\ge (\mathrm{Im}z{)}^2$ givesBoth inequalities must be equalities. The exact conditions are $A\psi$ and $B\psi$ are linearly dependent, and $\mathrm{Re}⟨A\psi ,B\psi ⟩=0$. Equivalently, equality holds if either vector is zero; when both are nonzero it holds precisely whenThe zero-vector cases matter: they saturate the inequality but cannot all be expressed by solving for $B\psi$ as a multiple of $A\psi$. The real-part condition can also be written ${\mathbb{E}}_{\psi }(AB+BA)=0$, using the anticommutator. This is the uncentered second-moment commutator inequality.

Let $m$ be the ice mass for which the stated drag coefficients are defined, $K=k_1+k_2$ and $f=2\Omega \sin \ell$. With stationary water, the steady horizontal ice equations areThe linear system givesFor fixed wind speed $V=10\mathrm{m}{\mathrm{s}}^{-1}$, the Cauchy-Schwarz inequality minimizes this component whenThus the wind must blow towards a direction east of south by ${\tan }^{-1}(mf/K)$ in the northern hemisphere, west of south in the southern hemisphere, and due south at the equator. This compensates the Coriolis force; at the optimum $v_x=0$ and the ice moves due south. If the drag coefficients are per unit mass, replace $K/m$ by the corresponding total drag rate. “Blows towards” avoids the opposite convention used when naming winds by their origin.

For every $m$, $D_m$ is a continuous square-integrable vector martingale. Its norm is a submartingale, so the Doob L2 maximal inequality and the estimate from part (c) giveThe Cauchy-Schwarz inequality now showsBy Tonelli's theorem, the random sum of these suprema is finite almost surely. Therefore the telescoping series $a+{\sum }_m{D}_m(t)$ converges absolutely and uniformly on $[0,T]$ outside one null event. Its limit is continuous and agrees at each fixed time with the $L^2$ limit from part (c). Taking the intersection of the probability-one events for integer $T$ provides a single version on the whole half-line. HenceThis also gives the requested pointwise almost-sure convergence of the full sequence.

The initial value can first be subtracted: it is ${\mathcal{F}}_0$-measurable and has the same integrability under both measures, since the density has conditional expectation one on ${\mathcal{F}}_0$. Put $N=M-⟨M,X⟩$. Since $dZ=ZdX$, the Itô product rule and the quadratic covariation identity $d⟨M,Z⟩=Zd⟨M,X⟩$ giveThis is a local P-martingale. Stop $N$ at increasing bounded levels; the resulting products with $Z$ are true martingales, since they are local martingales dominated at stopping times by a constant times the uniformly integrable density process. Part (c) then shows that the stopped $N$ processes are Q-martingales. ConsequentlyThe unqualified true-martingale assertion printed in this part needs an additional integrability hypothesis. One sufficient version is a square-integrable P-martingale $M$. Indeed $Z_t\in L^2(P)$ by boundedness of $Y$, and the covariation bound givesThe Doob L2 maximal inequality, ${\mathbb{E}}_P⟨M{⟩}_t<\infty$, and Cauchy-Schwarz inequality therefore give ${\mathbb{E}}_Q{\sup }_{s\le t}|N_s|<\infty$. Its local martingale identities pass to the limit, making it a true Q-martingale. This is Girsanov compensation with square-integrable input.

Here is a counterexample to the broader printed assertion. On $[0,1]$, take $X_s=W_s$, so $Y_s\equiv 1$ and $Z_s=e^{W_s-s/2}$. Put $a=1/2$ andUnder P, $W_a$ has density ${\pi }^{-1/2}{e}^{-x^2}$, so ${\mathbb{E}}_{P}H<\infty$. Independence of future Brownian increments proves $M$ is a continuous true P-martingale. Under Q, $W_s^Q=W_s-s$ is Brownian and the density of $W_a$ is multiplied by $e^{x-1/4}$, so ${\mathbb{E}}_{Q}H=\infty$. The compensated process after $a$ isThe future Q-Brownian increment is independent of $H$ and has positive finite expected absolute value. Thus ${\mathbb{E}}_Q|N_s|=\infty$ for every $s>a$, so it cannot be a true martingale. This proves that Bounded Girsanov drift does not preserve true martingales under compensation.

For the requested application, the Brownian driver does satisfy the needed conditions. To distinguish the state from the density-driving integral, call the state $U$. Set $Y_s=b(U_s)/a(U_s)$; the stated coefficient bounds make this integrand bounded. Applying the local result to $M=W$ givesIt is a continuous local Q-martingale with quadratic variation $s$. The Lévy characterization of Brownian motion therefore makes it a Q-Brownian motion. Substituting $d{W}_s=d{\tilde{W}}_s+[b(U_s)/a(U_s)]ds$ into the original state equation givesThe Brownian driver after the change is $\tilde{W}$; writing it as $W$ again in the last printed equation amounts to renaming that driver.

Take an orthonormal basis $\{|j⟩\}$ and insert its completeness relation:No normalization or orthogonality of $u,v$ is required. This is the trace of a rank-one operator; in a Hilbert space the sum converges by the Cauchy-Schwarz inequality.

Interpret the two displayed exchange components as belonging to the common nearest-neighbour sum with coefficient $-J$. Write $H_f=|h|$ and, for $h\ne 0$, $s=\mathrm{sgn}h$. The source explicitly fixes $J>0$ but does not separately constrain $h$ in this question; keeping $H_f$ covers either field direction.

For classical vectors of length $S$, define ${\mathbf{q}}_j=(S_j^x,S_j^y)$. The Cauchy-Schwarz inequality gives ${\mathbf{q}}_j\cdot {\mathbf{q}}_{j+1}\le |{\mathbf{q}}_j||{\mathbf{q}}_{j+1}|\le S^2$, and $h{S}_j^x\le H_{f}S$. Each term is therefore bounded below, soUniform alignment along the field saturates every bound:Equality requires all planar vectors to be parallel with length $S$, and the nonzero field then fixes their direction. At $h=0$, every uniformly oriented vector in the $xy$ plane is a classical ground state:A nonuniform configuration cannot improve these simultaneously saturated bond bounds, including on the periodic ring.

The constraint is a compact set, so the continuous objective attains both extrema by the extreme value theorem. Use Lagrange multipliers with $q(x,y)=9x^2+y^2$:Since the second component makes $\lambda \ne 0$, the two equations give $4/(18x)=1/(2y)$, or $y=9x/4$. Substitution into the constraint yieldswith matching signs. The objective values are thereforeThe constraint gradient never vanishes on the ellipse, so these are all the possible extrema. As a direct check, set $u=3x$, $v=y$; then $u^2+v^2=36$ and the Cauchy-Schwarz inequality gives $|4u/3+v|\le 6\sqrt{16/9+1}=10$, with equality at the same points.

If $\mathbf{b}=0$, the objective is zero at every feasible point, so every point of the ellipsoid is a minimizer. Suppose now $\mathbf{b}\ne 0$. A minimum cannot be interior: a sufficiently small step in direction $-\mathbf{b}$ would remain feasible and decrease the objective. The boundary is active.

Forstationarity gives $\mathbf{b}+2\lambda A\mathbf{x}=0$. A positive-definite symmetric matrix is invertible, and the minimizing sign has $\lambda >0$. Enforcing the boundary gives

For global optimality, use the inner product $⟨u,v{⟩}_A=u^{\mathsf{T}}Av$. The Cauchy-Schwarz inequality givesThe proposed point attains the negative of this bound. Thus the minimum is $-\sqrt{{\mathbf{b}}^{\mathsf{T}}{A}^{-1}\mathbf{b}}$, and equality in the Cauchy-Schwarz inequality, together with the sign and boundary conditions, makes this minimizer unique when $\mathbf{b}\ne 0$.

Put $\mathbf{z}=(x,y{)}^T$, $r^2={\mathbf{z}}^T\mathbf{z}$ and $f=\sqrt{1+r^2}$. The chain rule yieldsFor every nonzero vector $\mathbf{v}$, the Cauchy-Schwarz inequality givesThus the Hessian is a positive-definite matrix everywhere. The Hessian criterion for convexity implies that $f$ is strictly convex on ${\mathbb{R}}^2$. Its unique global minimum is at $\mathbf{z}=0$, where $f=1$.

Multiply the complex Fourier series by $e^{-i\omega rt}$ and integrate over one period. Fourier orthogonality removes every term except the one with index $r$, givingFor general square-integrable functions, this is an orthogonal projection onto the corresponding Fourier mode. It remains valid when the expansion converges only in L2 space: integrate finite Fourier partial sums first, then pass to the limit using the Cauchy-Schwarz inequality. No unjustified pointwise exchange of an infinite series and an integral is needed.

The Pearson correlation coefficient is the covariance divided by the product of the standard deviations. For real random variables with finite, strictly positive variances,It is the expected value of the product of the two centered variables after dividing each by its standard deviation. The Cauchy-Schwarz inequality gives $-1\le {\rho }_{X,Y}\le 1$. This removes measurement units, while retaining the direction and strength of the linear relationship. Constant random variables have zero standard deviation, so their Pearson correlation coefficient is undefined.

For the covariance model $Y=2X+Z$, with $X,Z$ independently uniform on $\{-1,0,1\}$, the calculation givesThe positive correlation is less than one because the independent $Z$ term introduces extra variation. For $Y=-2X+Z$ the correlation is $-2/\sqrt{5}$. Both examples appear in the covariance diagram.

For nonzero real $a,c$, the correlation of $aX+b$ and $cY+d$ is $\mathrm{sgn}(ac){\rho }_{X,Y}$. Positive rescaling and added offsets therefore leave the correlation unchanged; reversing one scale reverses its sign. Equality $|{\rho }_{X,Y}|=1$ means that one random variable is an affine function of the other almost surely, with a nonzero slope of the corresponding sign. This is the equality condition in the Cauchy-Schwarz inequality, and includes perfect positive correlation.

Zero correlation says that the centered variables are orthogonal in their square-integrable inner product. It permits nonlinear statistical dependence, as the example $Y=X^2$ in uncorrelated random variables demonstrates. More generally, a correlation is a statement about a joint probability distribution; identifying effects of interventions is the separate task of causal inference. This distinction keeps an observed association from being mistaken for a demonstrated mechanism.

An early source for standardized association is Francis Galton's 1888 paper “Co-relations and their Measurement, chiefly from Anthropometric Data”. He compared stature and forearm measurements and changed each measurement to units of its own variability so the two regression directions could be compared. His scale used probable error, so it should not be identified word for word with the modern standard deviation formula above. ^[ref].

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.

The Cauchy-Schwarz inequality for the Hilbert-Schmidt inner product gives $0\le J\le 1$ for $n$-dimensional unitary operators. Equality $J=0$ holds exactly when $U=e^{i\varphi }{U}_T$. When absolute gate phase matters, use the phase-sensitive Hilbert-Schmidt distance $\Vert U-U_T{\Vert }^2/(2n)=1-\mathrm{Re}\mathrm{Tr}(U_T^{†}U)/n$ instead.

Photon bunching is enhanced near-coincident detection relative to an appropriate independent-count reference. It need not require nonclassical light. In a classical intensity model with nonnegative instantaneous light intensity $I$, ideal detection with a conditional Poisson distribution givesA fluctuating intensity therefore bunches counts. For a stationary classical intensity with finite second moment, the Cauchy-Schwarz inequality further gives $⟨I(t)I(t+\tau )⟩\le ⟨I^2⟩$. A zero-delay correlation smaller than one at a later delay cannot be explained by this model.

Choose $M\asymp {\delta }^{-1}$ angularly separated unit segments in a bounded Kakeya set, and let $F$ sum the indicators of their Kakeya tubes. Then $\int F\asymp 1$. Two tubes at angular separation $\alpha$ intersect in area at most $C{\delta }^2/(\delta +\alpha )$, so summing each harmonic angular row gives $\int F^2\lesssim \log (2/\delta )$. The Cauchy-Schwarz inequality on their union proves the displayed neighborhood lower bound. Since $|E_{\delta }|\lesssim {\delta }^2{N}_{\delta }(E)$, both planar Minkowski dimensions equal two.

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.