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Replace the sum by the integral against the density of states and set $x=ϵ/(kT)$, $y=\mu /T$:ThereforeFor the given $a>0$, the integral converges at zero and infinity for every finite $y$.

Dominated convergence makes $\hat{f}$ continuous, and $\Vert \hat{f}{\Vert }_{\infty }\le \Vert f{\Vert }_1$. Smooth compactly supported functions are dense in $L^1$. Their Fourier transforms decay at infinity by integration by parts, so uniform approximation proves the Riemann-Lebesgue lemma: $\hat{f}\in C_0$.

For density, smooth compactly supported functions are also uniformly dense in $C_0$ by cutoff and mollification. Each such target $g$ is a Schwartz function; its inverse Fourier transform is another Schwartz function, hence belongs to $L^1$, and Fourier inversion recovers $g$. Therefore the image of a function of $L^1$ is dense in $C_0$.

The $L^1$ Fourier inversion theorem says that if $f,\hat{f}\in L^1$, then $f$ agrees almost everywhere with the inverse integral of $\hat{f}$. If $\hat{f}=0$, this applies and gives $f=0$ almost everywhere. Hence the Fourier transform is injective as a map of $L^1$ equivalence classes, with dense image in $C_0$.

A family $(X_n)$ of integrable random variables is uniformly integrable whenEquivalently the integrals of their tails vanish uniformly. This includes uniform $L^1$ boundedness, since $\mathbb{E}|X_n|\le M+\mathbb{E}[|X_n|1_{|X_n|>M}]$.

The score function isWith common parameter-independent support and justified differentiation under the integral, ${\mathbb{E}}_{\theta }[{\partial }_{\theta }\log f(X_1,\theta )]=\int {\partial }_{\theta }f(x,\theta )dx={\partial }_{\theta }1=0$. Linearity then gives $\boxed{{\mathbb{E}}_{{\theta }_0}{S}_n({\theta }_0)=0}$. These regularity conditions exclude, for example, moving-support uniform models.

The per-observation Fisher information is $I(\theta )={\mathbb{E}}_{\theta }[({\partial }_{\theta }\log f(X_1,\theta ){)}^2]$. Sinceand the integral of $f_{\theta \theta }$ is the second derivative of $1$, the regularity assumptions giveFor independent observations the total Fisher information is $nI(\theta )$, because centered individual score functions have zero cross terms.

The exponent $\Phi (t)=\frac{i}{2}(t-t^{-1})$ has saddles $t=\pm i$, with $\Phi (-i)=1$ and $\Phi (i)=-1$. Since the integrand is analytic off the origin, deform the contour to the unit circle and rotate $t=-i{e}^{i\theta }$. This gives the exact representationThe dominant saddle at $\theta =0$ is a steepest-descent direction for the method of steepest descent: $\cos \theta =1-{\theta }^2/2+{\theta }^4/24-\cdots$. Set $\theta =s/\sqrt{x}$ and expand the analytic amplitude against $e^{-s^2/2}$. Integrating its Gaussian moments gives the leading factor $e^x/\sqrt{2\pi x}$; the rest of the contour is exponentially smaller than this dominant contribution outside any fixed neighbourhood of the maximum.

To obtain every coefficient efficiently, the integral satisfies $x^2{I}_0^{\prime \prime }+x{I}_0^{\prime }-x^2{I}_0=0$, by differentiating and integrating the derivative of $\sin \theta e^{x\cos \theta }$. Substitute $I_0=e^x{x}^{-1/2}\sum a_n{x}^{-n}/\sqrt{2\pi }$ to get $a_0=1$ and $a_n=(2n-1{)}^2{a}_{n-1}/(8n)$. Therefore the full asymptotic expansion isHere $(-1)!!=1$. For each fixed truncation at $n=N$, the relative remainder is $O(x^{-N-1})$ as $x\to +\infty$. The infinite series is generally divergent, and this all-orders expansion does not purport to resolve exponentially small saddle contributions or their Stokes phenomenon.

With the sign convention of the displayed operator, the Lax equation is $L_t=[L,A]$. Differentiating $L\phi =k^2\phi$ shows that ${\phi }_t+A\phi$ is another solution at the same spectral parameter. The normalization at $-\infty$ fixessince $A{e}^{-ikx}=4i{k}^3{e}^{-ikx}$ there. At $+\infty$, comparison of the two plane-wave coefficients givesThus the transmission denominator $a$ is time independent. This is the Time evolution of KdV scattering data for the stated normalization.

The logarithmic derivative $S={\phi }_x/\phi +ik$ obeys the Riccati equationIts formal large-$k$ expansion givesThere is an important qualification to the integral identity in the question. For real $k$ with nonzero reflection, $e^{ikx}\phi \sim a+b{e}^{2ikx}$ has no limit at $+\infty$, so the literal improper integral of the exact $S$ need not exist. Continue to $\mathrm{Im}k>0$, where the reflected exponential decays relative to $e^{-ikx}$, and take sufficiently large $k$ away from zeros and with a consistent logarithm. Then the normalization givesThe second equality is an asymptotic expansion, not an asserted convergent series. Time independence of $a$ fixes every coefficient, so each $\int S_{n}dx$ is a conserved quantity of the Korteweg-De Vries equation. The first two nontrivial examples are $-\int udx$ and $\int u^{2}dx$.

For the usual real-valued KdV potential and real formal $k$, write $S=X+iY$ with real coefficient functions. Its imaginary part of the Riccati equation is $Y^{\prime }-2kX+2XY=0$, henceIn the formal expansion, $X$ contains precisely the even-indexed $S_n$ and $Y$ the odd-indexed ones. Rapid decay of $u$ and its derivatives makes every coefficient of the logarithm vanish at both spatial ends. Therefore $\boxed{{\int }_{\mathbb{R}}{S}_{2j}dx=0(j\ge 1)}$. This is a coefficientwise formal identity; it does not claim that the exact real-axis reflected solution has a convergent integral of $X$. The analytically continued scattering identity and the formal expansion are the conventions needed to make the requested argument valid.

The L1 norm properties of absolute homogeneity and the triangle inequality follow from the corresponding properties of the absolute value and the integral. If $\Vert f{\Vert }_1=0$, the nonnegative continuous function $|f|$ has zero integral. Were $|f(x_0)|>0$, continuity would make it bounded below by a positive number on a nontrivial interval, contradicting the zero integral. Thus $f=0$, proving positive definiteness.

Put $M=\Vert f{\Vert }_{\infty }$ and choose the continuous cutoff functionThen $g_n(0)=g_n(1)=0$, $\Vert g_n{\Vert }_{\infty }\le M$, and $f-g_n$ is supported in the two endpoint intervals of total length $2/n$. Therefore

Now suppose ${\int }_0^{1}fg=0$ for every $g\in \mathcal{S}$. Apply this to the above $g_n$. By the uniform norm bound,Hence ${\int }_0^1{f}^2=0$, and continuity again gives $f=0$.

For independent and identically distributed random variables with distribution function $F$, let $M_n={\max }_{1\le i\le n}{X}_i$ be the sample maximum. In extreme value theory, $F$ belongs to the maximum domain of attraction $D(G)$ of a nondegenerate distribution function $G$ when there exist $a_n>0$ and $b_n\in \mathbb{R}$ such thatat every continuity point of $G$. Thus the normalized sample maxima have convergence in distribution to $G$.

A positive measurable function $q$ has regular variation at infinity with index $\rho$, written $q\in {\mathrm{RV}}_{\rho }$, ifThe case $\rho =0$ defines a slowly varying function; equivalently a regularly varying function is $q(t)=t^{\rho }\ell (t)$ with $\ell$ a slowly varying function.

Write $\overline{F}=1-F$ for the survival function and $x_F=\sup \{x:F(x)<1\}$ for the right endpoint of a distribution. For $\alpha >0$, use the following standard extreme value theory normalization conventions:They are respectively the Fréchet distribution, the negative Weibull distribution, and the Gumbel distribution for maxima. The negative Weibull distribution is supported to the left of its finite endpoint; it is not the usual positive Weibull distribution.

The necessary and sufficient maximum domain of attraction criteria are:

For the corresponding sufficient hazard function conditions, assume $F$ is absolutely continuous near $x_F$, its survival function tends to zero there, and its probability density function $f$ is positive there. Write $\eta (t)=f(t)/\overline{F}(t)$. The Von Mises conditions for extreme values giveThe last condition additionally assumes that the reciprocal hazard function is continuously differentiable near the endpoint. It concerns the derivative of the reciprocal hazard function, not the derivative of the hazard function itself. Equivalently, when $f$ is differentiable, it is $\overline{F}(t)f^{\prime }(t)/f(t{)}^2\to -1$.

To see the connection, taking an integral of the hazard function givesThe first two limits make the logarithmic tail ratios over multiplicative or endpoint-distance intervals tend to $-\alpha \log u$ or $\alpha \log u$, giving the required regular variation. For the Gumbel distribution, set $a(t)=1/\eta (t)$. The derivative condition makes $a(t+a(t)v)/a(t)\to 1$ uniformly on bounded $v$-intervals, soThe endpoint assumptions ensure these shifts are admissible: at an infinite endpoint $a(t)/t\to 0$, and at a finite endpoint $a(t)/(x_F-t)\to 0$. These are sufficient smooth-tail conditions; the tail-ratio characterizations above are necessary and sufficient without differentiability assumptions. For a source using the same reciprocal-hazard convention, see Smith and Weissman's university lecture notes, equations (2.18)–(2.24).

For the Gamma distribution with integer shape $m$ and rate one, let ${\overline{F}}_m(t)={\int }_t^{\infty }{u}^{m-1}{e}^{-u}du/(m-1)!$. The base case is ${\overline{F}}_1(t)=e^{-t}$, and integration by parts givesInduction provesThis survival function has the asymptotic form ${\overline{F}}_m(t)\sim e^{-t}{t}^{m-1}/(m-1)!$; for $m=1$ the relation is exact. Put $t_n={\beta }_n+x$. For every fixed real $x$, $t_n\to \infty$ and $t_n/\log n\to 1$. The proposed centering satisfiessoOnly the highest-degree term contributes to the limit, with the same reasoning covering $m=1$. The Taylor expansion $\log (1-u)=-u+O(u^2)$ now givesbecause $n{\overline{F}}_m({\beta }_n+x{)}^2\to 0$. Therefore the Gumbel limit for gamma maxima isEquivalently, $M_n-{\beta }_n$ has convergence in distribution to the standard Gumbel distribution, with scale $a_n=1$.

Assume $m^2>0$ and take $\lambda \to 0^{+}$. The free Gaussian integral gives $Z_0=\sqrt{2\pi }/|m|$, and the normalized free Gaussian measure has covariance $C=m^{-2}$. Expanding only the interaction exponential gives the asymptotic expansionIn this zero-dimensional scalar field theory, the path integral has become an ordinary integral, but its combinatorics are precisely those of sextic scalar field theory.

The free generating functional is $⟨e^{J\varphi }{⟩}_0=e^{C{J}^2/2}$. Differentiating it proves Wick theorem: odd moments vanish, while each even moment is the sum over pairings of its factors. Each Wick contraction contributes $C$, soThus the Feynman rules are a quantum field theory propagator $C$ for each internal line and a factor $-\lambda$ for each six-valent interaction vertex. The factorials $V!$ and $(6!{)}^V$ remove vertex and half-edge labels; the remaining weight is $(-\lambda {)}^V{C}^I/S$, where $S$ is the Feynman-diagram symmetry factor, the order of the automorphism group of the resulting diagram, including its half-edge symmetries. A tadpole diagram has an additional interchange symmetry of the two ends of its loop.

There is one empty Vacuum Feynman diagram, contributing $1$. At one interaction vertex, the six half-edges form three loops. Its Feynman-diagram symmetry factor is $2^{3}3!=48$, so its contribution is $-\lambda C^3/48$.

At two interaction vertices let $r$ count the lines joining them. Each interaction vertex has $(6-r)/2$ self-loops, so $r$ can only be $0,2,4,6$. These four possibilities exhaust all Vacuum Feynman diagrams at this order. Their Feynman-diagram symmetry factors areHere the leading $2$ exchanges the two interaction vertices, $r!$ permutes the joining lines, and the remaining factors permute and reverse the self-loops. The $r=0$ graph is disconnected and must be included in $Z/Z_0$; only a connected generating functional would discard it.

Consequently,Indeed ${\sum }_r{S}_r^{-1}=77/7680$, independently agreeing with $11!!/[2(6!{)}^2]$ from Wick theorem.

This is an asymptotic expansion, rather than a convergent perturbation series: the coefficient of ${\lambda }^V$ contains $(6V-1)!!/[V!(6!{)}^V]$ and grows too rapidly for a nonzero radius of convergence. For any fixed truncation order, the Taylor expansion remainder bound for $e^{-x}$ on $x\ge 0$, followed by the finite Gaussian moment integral, proves the stated remainder estimate. For negative real $\lambda$ the original integral diverges.

There is an exact rational-map description of sigma-model lumps and a restricted variational rational-map description of Skyrmions. The first follows from a Bogomolny equation; the second separates angular and radial dependence in a field that generally does not saturate a Bogomolny bound.

For the exact example, take the two-dimensional O3 nonlinear sigma model with a unit field $\mathbf{n}$, energy $E=\frac{1}{2}\int |\nabla \mathbf{n}{|}^2{d}^{2}x$, and a fixed limit at infinity. Regular fields then have a one-point compactification to maps $S^2\to S^2$. Choose the orientation so that the stereographic fieldhas positive charge when it is holomorphic. Equivalently, this is the CP1 nonlinear sigma model since the target sphere is the complex projective line. Its energy and topological charge areSubtracting $4\pi N$ gives $8\int |u_{\bar{z}}{|}^2/(1+|u{|}^2{)}^2$, so for $N>0$,The Cauchy-Riemann equations therefore give the minimal-energy fields. A holomorphic map from the compactified domain Riemann sphere to the target Riemann sphere is a rational map $u=p(z)/q(z)$, with common polynomial factors cancelled. Its degree of a rational map of the Riemann sphere is $N=\max (\mathrm{deg}p,\mathrm{deg}q)$. Poles of $u$ are coordinate singularities, not singularities of $\mathbf{n}$. For instance, $u=(z-a)/\lambda$ with $\lambda \ne 0$ is an exact unit lump with energy $4\pi$, arbitrary centre $a$, and scale $|\lambda |$. Its density is $4|\lambda {|}^2/(|z-a{|}^2+|\lambda {|}^2{)}^2$, whose plane integral is $4\pi$. For negative charge use antiholomorphic maps. Scale invariance allows arbitrarily small lumps and does not by itself prevent a singular concentration limit in the time-dependent theory.

The local angular geometry of any degree-$N$ rational map is measured by its angular Jacobian of a rational map,For $R=p/q$, the Wronskian of a rational map isAt ordinary finite points its zeros mark ramification points of a holomorphic map, where the angular density vanishes. At a pole use $q/p$ as the target coordinate, and at infinity use $1/z$ as the domain coordinate. A pole of order $m$ is ramified by $m-1$. The Riemann-Hurwitz formula counts total ramification $2N-2$, including infinity, even if the affine Wronskian has smaller degree. For $R=z^N$, $W=N{z}^{N-1}$ accounts for $N-1$ zeros at zero; the reciprocal coordinate shows the other $N-1$ at infinity.

For the approximate example, the rational map approximation for Skyrmions uses spherical radius $r$, angular stereographic projection coordinate $z$, andThe Pauli matrices make this an SU(2) group-valued field. The boundary conditions give a continuous centre and the vacuum value at infinity; the profile must also make the energy finite. Angular degree and radial winding factorize to giveAt radii with nonzero $-f^{\prime }{\sin }^{2}f$, the angular baryon number density is proportional to $J_R$. Thus the Wronskian of a rational map locates zero-density directions and reveals the holes or face directions in shell-like Skyrmion configurations.

In dimensionless Skyrme model units, insert this ansatz into the quadratic and quartic derivative energies. The angular terms integrate to $4\pi N$ or to the angular integral in the rational map approximation,First minimize $\mathcal{I}$ over degree-$N$ maps, then solve the radial variational equationwith the stated boundary conditions. The Cauchy-Schwarz inequality gives $\mathcal{I}\ge N^2$ because $J_R$ has sphere average $N$. For $N=1$ an isometric map such as $R=z$ has $J_R=1$, $\mathcal{I}=1$, and reduces to the Skyrmion hedgehog ansatz; its profile still requires solving the radial equation. For general $N$, angular and radial separation restrict the allowed fields. The minimum within this class is an upper bound on the full sector minimum, not a claim that every exact Skyrmion has this separated form. The original construction is Houghton, Manton and Sutcliffe's arxiv.org/abs/hep-th/9705151.

A rotational symmetry of a rational map must satisfy $R(gz)=hR(z)$, where $g,h$ are the domain and target rotations written as Möbius transformations from SU(2) matrices. Since a target rotation is an isometry, this identity implies $J_R(gz)=J_R(z)$. For $R=z^N$, $R(e^{i\alpha }z)=e^{iN\alpha }R(z)$, giving axial spatial symmetry accompanied by an internal rotation. This explains the axial symmetry of the degree-two toroidal ansatz.

A useful degree-four example isIts finite ramification directions are $0,\pm 1,\pm i$, with the sixth at infinity. They are the six coordinate-axis directions under stereographic projection, so they are the face normals of a cube. The full map has octahedral rotational equivariance: for example $R(iz)=1/R(z)$, and the cyclic-axis generator $g(z)=(iz+1)/(1-iz)$ obeys $R(gz)=e^{2\pi i/3}R(z)$. These generate the cube's rotational group and give a cubic angular density. The associated profile then produces the familiar cubic charge-four approximation. Symmetry of the Wronskian is a useful necessary diagnostic but is not sufficient: replacing this map by $cR$, $c>0$, leaves the same ramification directions, while the identity under $z\mapsto iz$ becomes $cR(iz)=c^2/(cR(z))$, whose target transformation is a sphere rotation only when $c=1$. Equivariance of the entire rational map, rather than symmetry of its critical-point set alone, determines the physical rotational symmetry.

Since the natural logarithm is increasing, comparison with an integral givesThe lower bound is $n\log n-n+1$. Dividing all three expressions by $n\log n$ and applying the squeeze theorem gives

Bilinearity and symmetry follow from the corresponding properties of the integral. Since $w(x)>0$,If this integral is zero, the nonnegative continuous integrand must vanish everywhere, so $f=0$. Therefore the displayed formula is a real inner product on $C[a,b]$.

Direct evaluation of the integral givesThe continuous value at $k=0$ is zero, in agreement with the vanishing integral of the odd function $f$. Thus

At time $t$ the state isDirect evaluation of the radial integral gives the matrix elementsandIt follows thatThus $R(t)$ oscillates sinusoidally about $39a/8$. Its angular frequency is

Fix $x\in {\mathbb{R}}^n$ and a coordinate direction $e_i$. The fundamental theorem of calculus givesFor sufficiently small $h$, the points $(t,x+sh{e}_i)$ lie in a fixed compact set. The continuity of $D_{i}g$ makes it uniformly continuous there, so the right-hand side converges to $D_{i}g(t,x)$ uniformly in $t$. We may consequently pass the limit through the finite integral:

To prove continuity, if $x_k\to x$, joint continuity makes $D_{i}g(t,x_k)\to D_{i}g(t,x)$ uniformly for $t\in [0,1]$ once the $x_k$ lie in a compact neighbourhood of $x$. Hence $D_{i}G(x_k)\to D_{i}G(x)$. This proves the stated differentiation under the integral sign result and the continuity of every partial derivative.