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Since the common radius of convergence is infinite, the termwise differentiation of a power series theorem applies on the entire real line. For the even series, differentiation and the index change $m=n-1$ giveThe odd series similarly gives $g^{\prime }(x)={\sum }_{n=0}^{\infty }(-1{)}^n{x}^{2n}/(2n)!=f(x)$. Thus $\boxed{f^{\prime }=-g,g^{\prime }=f}$ everywhere. By the product rule,The mean value theorem implies that a differentiable function with zero derivative is constant on the real line. At zero, the series give $f(0)=1$ and $g(0)=0$. Consequently $\boxed{f(x{)}^2+g(x{)}^2=1\text{ for every }x\in \mathbb{R}}$. This derives the identity from the power series without assuming a trigonometric identity in advance.

Use the Einstein summation convention and the contraction of two Levi-Civita symbols. For the first cross product term,The last term cancels $(F\cdot \nabla )G_i$. Interchanging $F,G$ gives the analogous cancellation with $(G\cdot \nabla )F_i$. The surviving terms are $F_j{\partial }_i{G}_j+G_j{\partial }_i{F}_j={\partial }_i(F_j{G}_j)$ by the product rule. This proves the dot-product gradient identityThe proof only requires continuously differentiable vector fields; neither field needs to be an irrotational vector field.

Under the usual bounded-domain hypotheses, with piecewise smooth boundary and a sufficiently smooth vector field, the divergence theorem stateswhere $\mathbf{n}$ is outward. Apply it to $g\mathbf{F}$ and use the product rule $\nabla \cdot (g\mathbf{F})=\mathbf{F}\cdot \nabla g+g\nabla \cdot \mathbf{F}$ to obtainFor uniqueness, let $w=u_1-u_2$ for two classical solutions with the same Dirichlet boundary data. Then $\Delta w=0$ and $w=0$ on the boundary. Taking $\mathbf{F}=\nabla w$, $g=w$ gives Green's first identity in the formThe continuous nonnegative integrand must vanish, so $w$ is constant on each connected component. The zero boundary data make every such constant zero. Thus the solution of the Dirichlet problem for the Laplace equation is unique, if it exists. This establishes Uniqueness of the Dirichlet problem; existence is assumed. Boundedness, or suitable conditions at infinity, is needed: on an unbounded half-space the harmonic function $w=z$ has zero boundary values but is not identically zero.

Wealth and admissibility. The dollar holding $\theta$ earns the risky return, while $w-\theta$ earns the continuous-time bank account return. Removing consumption therefore givesHere $\theta$ is a dollar amount, not a number of shares; the number of shares is ${\theta }_t/S_t$. Both controls must use available information: take $\theta$ predictable and $c\ge 0$ progressively measurable, withalmost surely on every finite interval. Require a well-defined objective and an admissible trading strategy satisfying $w_t\ge 0$. At zero portfolio wealth this excludes continued risky gambling or positive consumption. The state-price budget constraint rules out doubling strategies. Throughout the diffusion calculations take $\sigma \ne 0$, positive discount $\rho$, positive decay $\lambda$, and finite value; degeneracies are discussed where they affect the conclusions.

Satisfaction and dynamic programming. Write the consumption satisfaction stock asThe product rule gives, almost everywhere in time,This state is a finite-variation process, so it has no quadratic covariation with portfolio wealth. Applying the Itô formula to the discounted value function over a short interval, then using dynamic programming, gives the interior HJB equationAssuming $V_{ww}<0$, completing the square givesThe last supremum is zero if $V_{\xi }-V_w\le 0$ and infinite otherwise. Thus the gradient constraint for unbounded consumption is $V_{\xi }\le V_w$, and $c=0$ wherever the inequality is strict. Since there is no direct penalty for a very large consumption rate, an active boundary can involve singular consumption control. In that relaxed interpretation the HJB equation isWith ordinary rate controls, this describes the supremum and its limiting transfer policy; it does not promise that an instantaneous transfer is attained by a finite rate.

Power reduction. An additive constant in the utility function only adds a control-independent constant divided by $\rho$ to the value, so normalize $U(\xi )={\xi }^{1-R}/(1-R)$. Put $p=1-R$ and $x=w/\xi$ for $\xi >0$. Scaling portfolio wealth, consumption satisfaction and the controls by the same positive number gives the wealth-to-satisfaction reductionConsequently the reduced HJB equation isIn the strict waiting region the first expression vanishes and consumption is zero. On a transfer region the second vanishes; integrating it gives $v(x)=K(1+x{)}^p$. This reflects preservation of $w+\xi$ during an instantaneous wealth-to-satisfaction transfer.

Why a waiting threshold is expected, and its qualification. The gradient constraint for unbounded consumption compares the benefit of increasing consumption satisfaction with the opportunity cost of spending financial portfolio wealth. When consumption satisfaction is already large relative to cash, waiting lets consumption satisfaction decay while financial portfolio wealth earns returns; consuming immediately can be wasteful. Homogeneity makes the comparison depend only on $x$. For fixed $s=w+\xi$, joint concavity of the value function makes $q_s(a)=V(s-a,a)$ concave in financial portfolio wealth $a$. It is nondecreasing because an immediate transfer can reproduce any smaller financial allocation. Consequently its derivative $V_w-V_{\xi }$ is nonnegative and nonincreasing in $a$: a strict waiting region, if present, starts at zero financial portfolio wealth and ends at a single transfer boundary. Homogeneity makes the corresponding boundary a ratio $x_{\ast }$. In the usual finite-boundary regime this gives $c=0$ for $x<x_{\ast }$, and transfers push a larger ratio down towards $x_{\ast }$. The condition is a comparison of marginal values, not the ordinary formula $c=(V_w{)}^{-1/R}$: current utility function here depends on consumption satisfaction, not on current consumption.

A positive threshold is not guaranteed by the printed hypotheses alone. A useful sufficient local test illustrates the intended argument. Let $a=\rho +\lambda p>0$. At zero portfolio wealth,Starting with a small extra portfolio wealth $\epsilon$, holding it in the continuous-time bank account until a fixed time $t$, then transferring it into consumption satisfaction, has right derivative in $\epsilon$ at zero equal toTherefore, if $\rho <r+\lambda R$, the portfolio wealth marginal value is strictly larger than the consumption satisfaction marginal value at zero. With the usual continuity of marginal values, there is a positive interval on which the gradient constraint for unbounded consumption is strict. The fixed-total-resource concavity argument then gives the threshold structure.

For a concrete counterexample to an unconditional positive threshold, take $R=1/2$, $\lambda =1$, $r=\mu =1/10$, $\sigma =1$ and $\rho =2$. There is zero market price of risk. The relaxed value isIndeed $F_{\xi }=F_w$, $F_{ww}<0$, and the optimized waiting residual, after division by ${\xi }^p$, isbecause $(r+\lambda )/a<1$ and the integrand is decreasing. The Itô formula gives an upper bound by $F$, while transferring all portfolio wealth into consumption satisfaction over intervals tending to zero attains that bound in the limit. Hence the transfer boundary is $x_{\ast }=0$ in this example. The positive-threshold explanation needs a parameter regime supporting a genuine waiting region. For $R>1$, even the zero-wealth value is finite only if $\rho >\lambda (R-1)$; otherwise decaying consumption satisfaction gives value $-\infty$.

For a diffeomorphism, the differential and its inverse preserve the natural pairing of a vector with a covector. Consequently the mixed tensor pullback commutes with every tensor contraction $C$:Differentiating at zero gives ${\mathcal{L}}_X(CT)=C({\mathcal{L}}_{X}T)$. The mixed tensor pullback also preserves tensor products, soThe ordinary product rule for differentiation therefore yieldsThis proves the contraction and Leibniz rule properties for all smooth generators, without needing straightening coordinates.

The suggested flow-box theorem applies locally only where $X\ne 0$. For example $X=x{\partial }_x$ vanishes at $x=0$, so it cannot equal a coordinate basis vector there. Its local flow is nevertheless ${\varphi }_s(x)=e^{s}x$ and ${\varphi }_s^{\ast }dx=e^{s}dx$, giving ${\mathcal{L}}_{X}dx=dx$ even at that zero. This Lie derivative at a zero of its generator illustrates why the general flow proof is needed to cover every point.

Write the covector field as $\omega ={\omega }_{\mu }d{x}^{\mu }$. From the contraction and product rule properties,Take $Y={\partial }_{\mu }$. The Lie bracket of vector fields is $[X,{\partial }_{\mu }]=-({\partial }_{\mu }{X}^{\nu }){\partial }_{\nu }$, soEquivalently, the covariant-factor contribution comes from ${\mathcal{L}}_{X}d{x}^{\mu }=d{X}^{\mu }=({\partial }_{\nu }{X}^{\mu })d{x}^{\nu }$. The formula holds in any coordinate basis and is the one-form case of the coordinate tensor Lie derivative.

Apply the product rule and the preceding gradient formula to the vector field $F_i=f(r)x_i$. Summing over the repeated index givesHere ${\delta }_{ii}=n$ for the Kronecker delta. Thus the divergence and curl of a radial vector field give, in any dimension,

Apply the divergence theorem to $\mathbf{F}=\psi \nabla \varphi$, for functions smooth enough on the closure of the bounded piecewise smooth volume. The product rule givesThus Green's first theorem, also called Green's first identity, isThe normal is outward. This identifies the boundary term and its sign directly from the divergence theorem.

For a smooth map $\mathbf{u}$, the Jacobian determinant is the determinant of its Jacobian matrix $A_{ai}={\partial }_i{u}_a$. In Einstein summation convention,Differentiate the supplied expression for $V_i$ by the product rule. The term differentiating ${\partial }_j{u}_a$ vanishes because ${\partial }_i{\partial }_j{u}_a$ is symmetric in $i,j$, while ${ϵ}_{ijk}$ is antisymmetric; the term differentiating ${\partial }_k{u}_b$ vanishes similarly. Only differentiation of $u_c$ remains:The last equality follows by the cyclic relabelling $(i,j,k)\mapsto (j,k,i)$, which has positive sign. This proves the Jacobian determinant as a divergence identity.

For a composition $\mathbf{w}=\mathbf{u}\circ \mathbf{v}$, the chain rule givesTaking determinants yieldsFor the first-order perturbation, use spatial derivative control in perturbations of the identity map: the usual smooth-family interpretation gives $D{\mathbf{u}}_t=I+tD\mathbf{F}+o(t)$ locally. Multilinearity of the determinant shows that its linear term comes from choosing one column of $tD\mathbf{F}$ and all other columns from $I$. This gives the trace:The regularity qualification matters: a pointwise $o(t)$ remainder for maps does not by itself give an $o(t)$ remainder for spatial derivatives. For instance $(x+t^2\sin (x/t^2),y,z)$ has a uniform $o(t)$ displacement and $\mathbf{F}=0$, but its Jacobian determinant at $x=0$ is $2$ for $t\ne 0$. Thus the first expansion needs the remainder in $C^1$ locally, or an equivalent smoothness assumption on the family.

Under the subsequent group law, ${\mathbf{u}}_0$ is the identity and ${\mathbf{u}}_{-t}$ is the inverse of ${\mathbf{u}}_t$. Write $j(t,\mathbf{x})=J[{\mathbf{u}}_t](\mathbf{x})$. Applying the composition formula to ${\mathbf{u}}_{t+h}={\mathbf{u}}_h\circ {\mathbf{u}}_t$ and the short-time expansion gives the Jacobian evolution of a smooth flow:HenceIf only the pointwise generator expansion is assumed initially, the group law still gives ${\partial }_t{\mathbf{u}}_t(\mathbf{x})=\mathbf{F}({\mathbf{u}}_t(\mathbf{x}))$. Uniqueness and smooth dependence for this smooth ordinary differential equation make it a smooth flow map, thereby justifying the differentiated expansion used above. Thus a divergence-free generator hasBy the change of variables formula, these maps preserve ordinary Euclidean volume and orientation. This is the Euclidean instance of a volume-preserving vector field.

The product rule for the divergence givesSubtracting cancels the two gradient products. ThereforeTo obtain the requested order of terms, apply this identity with $u,v$ interchanged and use the divergence theorem on the spherical shell. On the outer sphere the shell's outward unit normal is $\hat{\mathbf{r}}$; on the inner sphere it is $-\hat{\mathbf{r}}$. Write ${\partial }_r=\hat{\mathbf{r}}\cdot \nabla$ on both spheres. Then the resulting Green second identity isThe minus sign records the inward radial direction of the shell's normal on its inner boundary. Thus the normal derivatives in the displayed difference are outward from each individual ball; using shell-outward normals instead would put a plus sign between the two boundary integrals.

Put $u=X_1-1$ and $v=X_2-1$. Evaluation at $(1,1)$ is surjective, andso its kernel is the prime ideal $\boxed{P=(u,v)}$. In particular $P/P^2$ is naturally an $R/P=\mathbb{Z}$-module, generated by the classes of $u,v$.

For integers $a,b$, including negative integers, reduction modulo $P^2$ givesFor negative powers, this follows from $(1+u{)}^{-1}\equiv 1-u$ and $(1+v{)}^{-1}\equiv 1-v$; the mixed product $uv$ vanishes modulo $P^2$. Therefore a Laurent polynomial $f$ satisfiesThe formal derivatives obey the product rule, so $D_i(ab)=\theta (a)D_i(b)+\theta (b)D_i(a)$. If $a,b\in P$, both terms vanish. Hence both $D_i$ vanish on $P^2$ and define a mapIt sends $[u]$ to $(1,0)$ and $[v]$ to $(0,1)$, and the displayed expansion gives its inverse. Thus the conormal module has the explicit basis

Let $A=R_P$ be the localization at a prime ideal, with maximal ideal $\mathfrak{m}=P{R}_P$. Every nonzero integer lies outside $P$ and is therefore inverted. The residue field isIndeed there is an explicit identificationinverting $1+u$ and $1+v$ has no effect in this local ring, and a polynomial denominator lies outside $(u,v)$ precisely when its constant term is nonzero. Conversely, clearing rational denominators turns every such fraction into a fraction from $R_P$.

For the Krull dimension, the prime ideal correspondence for localization preserves the strict chainof prime ideals in $R$. They are prime because the successive quotients are the integral domains $R$, $\mathbb{Z}[X_2,X_2^{-1}]$, and $\mathbb{Z}$. They remain distinct after localization at a prime ideal because they are all contained in $P$. Hence $\dim A\ge 2$.

For the upper bound, $R$ is a Noetherian ring by the Hilbert basis theorem and Localization of a Noetherian ring. Since $P$ is generated by two elements, the Krull height theorem gives $\mathrm{ht}P\le 2$. The prime ideal correspondence for localization identifies $\dim R_P$ with $\mathrm{ht}P$, so

A Noetherian local ring $(B,\mathfrak{n})$ with residue field $\kappa$ is a regular local ring ifBy the Nakayama lemma, the dimension on the left is the minimal number of generators of $\mathfrak{n}$; this is also called the embedding dimension.

The localization of a module is exact and commutes with quotients and products of ideals. Applying it to the already computed conormal module gives the cotangent space of a local ringIts dimension is two, equal to $\dim A$. Therefore $\boxed{R_P\text{ is a regular local ring}}$. Notice that $P/P^2$ has a basis over $\mathbb{Z}$, whereas the localized cotangent space of a local ring has a basis over the residue field $\mathbb{Q}$.

This subpart is also present in the original PDF but missing from the local TeX. Put $p(x)=1-x^2$ and ${\lambda }_n=n(n+1)$. Multiply the Legendre differential equation for $P_n$ by $P_m$, multiply the equation for $P_m$ by $P_n$, and subtract. The product rule givesIntegrating from $-1$ to $1$, the boundary term vanishes: $p(\pm 1)=0$, and the Legendre polynomials and their derivatives are finite there. For distinct nonnegative integers $m,n$, ${\lambda }_m\ne {\lambda }_n$. Thus the Orthogonality of Legendre polynomials isThe vanishing coefficient at the endpoints causes no difficulty here because the solutions are polynomials.

Apply the weak maximum principle for elliptic operators to $q=|Dw{|}^2$. The product rule and the equation for the harmonic function giveThe function $q$ is twice differentiable inside and continuous on the closure, by the prescribed regularity. Hence ${\sup }_{\Omega }q\le {\sup }_{\partial \Omega }q$. Conversely, each boundary value of $q$ is a limit of values at interior points, so the reverse inequality holds for the suprema. Taking square roots proves the gradient maximum principle for harmonic functions:This argument controls the length of the entire gradient, without comparing its separate components.

A wavelet with $p$ vanishing moments satisfies $x^k\psi \in L^1$ and $\int x^k\psi (x)dx=0$ for $0\le k<p$. The moment differentiation of the Fourier transform theorem states that these weighted integrability hypotheses make $\hat{\psi }\in C^{p-1}$, with ${\hat{\psi }}^{(k)}(\xi )=\int (-ix{)}^k\psi (x)e^{-ix\xi }dx$. Hence ${\hat{\psi }}^{(k)}(0)=0$. The MRA projection Fourier identity, together with density of the approximation spaces and continuity of $\hat{\phi }$ at zero, gives $|\hat{\phi }(0)|=1$. These are the two analytic theorems needed in addition to the quadrature mirror filter construction.

For the ordinary derivative conclusion, assume also that the MRA low-pass filter is $C^{p-1}$ near $\pi$ and $\hat{\phi }$ is $C^{p-1}$ near zero. These hypotheses hold, for example, for a compactly supported scaling function with a finite refinement filter. The wavelet identity givesThe denominator is nonzero near zero. Its reciprocal and the exponential are $C^{p-1}$, so the product rule and the zero Taylor polynomial of $\hat{\psi }$ give the smooth-mask vanishing-moment criterion

The printed integrability hypothesis alone does not guarantee ordinary higher derivatives. It gives only a continuous $\hat{\phi }$. Without added smoothness, Taylor theorem for $\hat{\psi }$ still gives $m(\pi +t)=o(|t{|}^{p-1})$ for the representative defined by the quotient: a Peano zero. This establishes the value at $\pi$, and for $p\ge 2$ the first derivative there, but it does not automatically establish iterated derivatives.

Here is a lacunary scaling-phase regularity counterexample for $p\ge 3$. Start with a compactly supported Daubechies wavelet having $p$ vanishing moments, with scaling function ${\phi }_0$ and finite MRA low-pass filter $m_0$. Define the periodic phaseThe Fourier coefficients of $\theta$ are absolutely summable. The Wiener algebra is closed under multiplication and the exponential series, so $a$ and $a^{-1}$ have absolutely summable Fourier coefficients. Thus $\phi$ is an absolutely summable combination of integer function translations of ${\phi }_0$, belongs to $L^1\cap L^2$, and generates the same $V_0$. Unimodularity preserves orthogonality and unit norm of its integer function translations; the inverse phase preserves their complete span. The same approximation spaces therefore give a multiresolution analysis.

The identity $\theta (2t)=\theta (t)+\theta (t+\pi )$ gives $a(2t)=a(t)a(t+\pi )$, so the periodic phase change of a scaling function produces $m(t)=a(t+\pi )m_0(t)$. In the high-pass formula the phases cancel:The associated wavelet is therefore exactly the original one, with unchanged vanishing moments.

At any dyadic point $t_0=2\pi k/2^j$, the difference quotient of the terms with $j\le n\le {\log }_2(1/|h|)$ is $1+O((2^{n}h{)}^2)$ per term. The sum of these errors and the remaining tail divided by $h$ are bounded; the finitely many earlier terms have finite limits. Hence $(\theta (t_0+h)-\theta (t_0))/h={\log }_2(1/|h|)+O(1)$, which has no finite limit. The exponential has the same failure of differentiability. Dyadic points are dense, and $m_0(\pi +t)$ is nonzero for sufficiently small $t\ne 0$. Thus $m$ is not differentiable on any neighborhood of $\pi$, so an ordinary second derivative at $\pi$, understood as the derivative of a locally defined first derivative, need not exist. The Peano zero remains valid. This separates the intended smooth-mask theorem from what the literal assumptions establish.

Under the smoothness hypotheses in the smooth-mask vanishing-moment criterion, write any nonzero integer $j$ as $j=2^r\ell$, where $r\ge 0$ and $\ell$ is odd, possibly negative. Iterating the scaling refinement equation exactly $r+1$ times givesThe last MRA low-pass filter factor is $m(\pi \ell +t/2^{r+1})$. Its derivatives of all orders less than $p$ vanish at $t=0$, because $\ell$ is odd and the MRA low-pass filter is $2\pi$-periodic. If the remaining factors are $C^{p-1}$ at the displayed arguments, the product rule forces every derivative of order less than $p$ of the product to vanish. In particular compact support and a finite MRA low-pass filter supply all these hypotheses. Therefore the integer-frequency zeros of a scaling function are

Under only the printed assumptions, the exact same finite refinement argument gives the Peano zero $\hat{\phi }(2\pi j+t)=o(|t{|}^{p-1})$. Indeed, the last MRA low-pass filter factor has that estimate, the other MRA low-pass filter factors are bounded by one almost everywhere by the quadrature mirror filter identity, and the last Fourier transform factor is bounded because $\phi \in L^1$. The estimate initially holds almost everywhere and extends to every $t$ by continuity of $\hat{\phi }$. It does not imply arbitrary ordinary higher derivatives. In the lacunary scaling-phase regularity counterexample, $\hat{\phi }=a{\hat{\phi }}_0$ is nondifferentiable at dense dyadic points next to the isolated integer-frequency zeros of the smooth base transform. Thus the same regularity qualification is necessary here.

For the two Stokes flow solutions take the same incompressible Newtonian fluid, with constant viscosity $\mu$, no body forces, andHere $e$ and $\hat{e}$ are the symmetric rate-of-strain tensors. The equality of viscosities is part of comparing two solutions in the same fluid; symmetry of an arbitrary constitutive stress alone would not suffice.

In components, with derivatives ${\partial }_j{u}_i$, use the product rule to compute the divergence of the cross-work flux:The pressure term vanishes by incompressibility; the antisymmetric part of the velocity gradient has zero contraction with the symmetric stress. Interchanging the hatted and unhatted fields gives exactly the same scalar. ThusIntegrate over $V$ and apply the divergence theorem. In its usual form the normal is outward from the fluid volume, whereas the printed $n$ points into the fluid. Reversing the normal multiplies both integrals by the same minus sign, leaving the Lorentz reciprocal theorem for Stokes flow:This proof applies to smooth fields or the corresponding finite-energy weak formulation. For an exterior domain one first truncates at a large sphere, applies the same identity, and takes the limit when the outer cross-work terms vanish, as for the decaying fields used below.

Put $\mathbf{t}=\mathbf{\theta }$. The product rule applied to $Q_{ij}=\lambda (t_i{t}_j-{\delta }_{ij}/2)$ givesSince $\mathrm{Tr}{Q}^2={\lambda }^2/2$, substituting this vector into the divergence elastic term proves the requested form of the local density. For the radial amplitude and tangential director, the polar coordinates identities areHence the tangential integer nematic defect core hasThe $1/r$ director-curvature term survives when the amplitude is constant. Replacing the divergence norm by a full tensor-gradient norm would not give this expression; it is the PDF's divergence convention that is being evaluated.

For the survivor functions, $F_A^{\prime }(u)=-f_A(u)$ and $F_B^{\prime }(u)=-f_B(u)$. The product rule givesIntegrating from 0 to $t$, using $F_A(0)=F_B(0)=1$, yieldsThe three terms partition the possible histories: A occurs first by time $t$, B occurs first by time $t$, or neither has occurred by $t$. The first two are the cumulative incidence functions for the two competing risks, while the third is the survival probability of their minimum. The initial value assumes nonnegative event times with no atom at zero.

Differentiate $B_{\alpha \beta }={\nabla }_{\beta }{X}_{\alpha }$ along $X$ and commute the two covariant derivatives:where the sign is the covector form of the Ricci identity. Apply the product rule to the first term:The first term vanishes by the geodesic equation, and the second isFinally the pair symmetries and antisymmetries of the Riemann curvature tensor giveConsequentlyThis is the Riccati equation for a timelike geodesic congruence; taking its spatial trace is the starting point for the timelike Raychaudhuri equation.

Write the density and pressure as $\rho ={\rho }_0+{\rho }^{\prime }$ and $p=p_0+p^{\prime }$, and let $\mathbf{u}$ be the small fluid velocity. The linear homentropic acoustic equations arewhere the adiabatic sound speed isIf the flow is irrotational, write $\mathbf{u}=\nabla \varphi$. The momentum equation integrates spatially toa purely time-dependent integration term having been absorbed into $\varphi$. Substitution into continuity gives the wave equationso the wave speed is $c_0$.

To obtain energy conservation, dot the momentum equation with $\mathbf{u}$ and use the product rule:The continuity and pressure-density relations giveTherefore the acoustic energy conservation law iswith acoustic energy density and intensity