Let $F,G\ge 0$, $1<p<\infty$, and $G\in L^p$. Assume the displayed tail inequality and finite positive-threshold superlevel measures. First split $G$ at $a/2$ to obtain $a\mu \{F>2a\}\le {\int }_{\{G>a\}}G$. Layer cake representation and Tonelli theorem imply $F\in L^p$. Integrating the original inequality against $p{a}^{p-2}$ then gives $\Vert F{\Vert }_p^p\le q\int G{F}^{p-1}\le q\Vert G{\Vert }_p\Vert F{\Vert }_p^{p-1}$, with $q=p/(p-1)$. This is the analytic estimate behind the Doob Lp maximal inequality.

Write the Weierstrass elliptic function as $\wp (z)=z^{-2}+{\sum }_{n\ge 1}{c}_n{z}^{2n}$, with $c_1=3G_4$, $c_2=5G_6$ and $c_n=(2n+1)G_{2n+2}$. Differentiate the Weierstrass elliptic differential equation to obtain ${\wp }^{\prime \prime }=6{\wp }^2-g_2/2$. Comparing coefficients of $z^{2n-2}$ gives the displayed recurrence for $n\ge 3$. Every denominator is positive, so induction expresses every higher lattice Eisenstein series as a polynomial in $G_4,G_6$ with positive rational coefficients on its occurring monomials. In particular $G_8=3G_4^2/7$ and $G_{10}=5G_4{G}_6/11$.

A positive bump of the real line is an order automorphism whose nonempty support of a permutation is one open interval and which moves every point of that interval to the right. It fixes the complement. Iterating any interior point tends to the right endpoint in positive time and to the left endpoint in negative time.

An ordered permutation group is order n-transitive if it can send any increasing $n$-tuple to any other increasing $n$-tuple. This compares tuples with the same relative order, unlike unrestricted n-transitive group actions. In the usual transitive chain setting, order two-transitivity implies every finite order-transitivity for a lattice-ordered permutation group.

Every lattice-ordered group embeds in a lattice-ordered permutation group on one totally ordered set. Choose a value in a lattice-ordered group for each nonidentity element, act on its ordered right cosets, and combine these chains in an order sum. The original element moves its own value's identity coset, making the action faithful. The coset lattice formulas make the embedding preserve both join and meet.

A transitive lattice-ordered permutation group is order-primitive when its only invariant equivalence relations with convex classes are equality and the universal relation. This concerns convex blocks in a chain, rather than arbitrary blocks of a primitive group action.

Let $Z_n$ clip the minimum of $k$ independent exponential sample means into $[a,b]$, with $0<a<1/\lambda <b$. The rate function under scalar clipping is the minimum cost on this interval. Apply Varadhan's lemma to $\log z$. On the lower side its objective is increasing; on the upper side its derivative is $(k+1)/z-k\lambda$. Thus the optimum is $z_{\ast }=(k+1)/(k\lambda )$ once $k\ge (\lambda b-1{)}^{-1}$, giving the displayed limit. The expectation is of the high power, rather than a power of the mean; the latter has limit $-\log \lambda$.

In the self-coupling-dominated approximation $d{\alpha }_H/d\ln {\mu }^2=b{\alpha }_H^2$, $b>0$, the running coupling reaches a perturbative Landau pole at the displayed scale. With the conventional Higgs potential, $m_H^2=2{\lambda }_H{v}^2$ and one can normalize ${\alpha }_H={\lambda }_H/(4\pi )$. A larger Higgs boson mass therefore decreases the logarithmic interval before the self-coupling becomes strong. Requiring perturbative validity up to a fixed high cutoff gives an upper Higgs-mass bound in this approximation. The full Standard Model beta function also contains gauge and Yukawa couplings; its sign is not universally positive. A perturbative pole is not a theorem excluding every nonperturbative ultraviolet completion.

The covariant Landau gauge propagator is the transverse limit of the inversion of the gauge-fixed Maxwell kinetic operator. Contracting with $k^{\mu }$ gives zero away from the pole. The Lorenz gauge condition still has residual transformations satisfying $\square \chi =0$, so boundary conditions and the pole prescription complete the choice.

Let $A$ code application and $C$ code numeral quotation. For a lambda term $F$, choose a fresh $x$ and put $H=\lambda x.F(Ax(Cx))$ and $Q=Hd(H)$. Then $Ad(H)(Cd(H)){\equiv }_{\beta }d(Q)$, so $Q{\equiv }_{\beta }Fd(Q)$. This diagonal argument uses the syntax of the actual term $Q$, including any free variables, not an equivalence-class code.

A leaf of a Lagrangian foliation is an immersed Lagrangian submanifold tangent to its integrable distribution. A prequantum line bundle restricts to a flat connection along such a leaf, because its curvature is proportional to the symplectic form. Nonzero global parallel sections require trivial holonomy along every leaf loop.

At distinct interpolation nodes, differentiating Lagrange interpolation gives $p^{(k)}(x)={\sum }_{i}p(t_i){\ell }_i^{(k)}(x)$. The triangle inequality gives the displayed upper bound, attained by prescribing each nodal value to be the sign of the corresponding derivative weight. For increasingly ordered real nodes, the Markov interlacing lemma and monotonicity of polynomial critical points in their roots imply that $(-1{)}^i{\ell }_i^{(k)}(x)$ has at most one sign change, ignoring zero weights. Thus an extremizer has alternating nodal signs with at most one reversal of that alternating pattern. This reduces a choice among $2^{n+1}$ sign patterns to $n+1$ candidate polynomials.

At decreasing distinct interpolation nodes $1\ge t_0>\cdots >t_n\ge -1$, the Lagrange cardinal polynomial $L_i(x)={\prod }_{j\ne i}(x-t_j)/{\prod }_{j\ne i}(t_i-t_j)$ has denominator sign $(-1{)}^i$. For $1\le k\le n$, its numerator's $k$th derivative at one is $k!$ times a sum of products of $n-k$ nonnegative factors $1-t_j$. At most one factor vanishes, and at least one product avoids it, so that sum is strictly positive. The same alternating nodal signs therefore maximize every endpoint derivative simultaneously.

For a cusp form $f=\sum c_n{q}^n$ and coprime integers $a,N$, the additive twist inserts the additive character $e^{2\pi ian/N}$. It is the Dirichlet series associated with the rational translate $f(a/N+\tau )$. Its completed Mellin transform uses the scale $N^s$ and the Gamma function.

In the Known-variance Gaussian mixture EM example, initialize both means at the same value $m$. The mixture responsibilities are then $1/2$ for every observation. Each M-step sets both means to the sample mean, zero here; the next E-step again gives $1/2$. Thus $(0,0)$ is a fixed point with log-likelihood $-14.593631199$, despite the better fit found from $(-1,1)$.

One can check the failure without trusting an iteration plot. Separate the means as $(-a,a)$ and use the hyperbolic cosine identity:The Taylor expansion uses ${\sum }_i{y}_i^2=18.16$. Small nonzero separation therefore improves the log-likelihood, whereas translating both means to $m$ changes it by $-3m^2$. This stationary point has both improving and worsening directions. Identical initialization cannot discover the separation because every observation keeps identical label probabilities. Trying several distinct initial fits and comparing their final log-likelihoods is useful, but does not prove global optimality.

Keeping the variance fixed also removes a different pathology. With known $\sigma =1$, every component probability density function is at most $(2\pi {)}^{-1/2}$, so $\ell \le -3\log (2\pi )$. If separate component variances are allowed to shrink without restriction, one mixture component can concentrate at an observation while another retains positive density at the other observations, making the likelihood function unbounded. The fixed positive variance in this example prevents that collapse; it does not remove initialization traps.

For $n\ge 2$, in coalescent units, independent waiting times satisfy $T_j\sim \mathrm{Exp}(j(j-1)/2)$, and total branch length is $L={\sum }_{j=2}^{n}j{T}_j$. Therefore $EL=2{\sum }_{i=1}^{n-1}1/i$. The variables $j{T}_j/2$ are independent exponentials with rates $j-1$, so $W=L/2$ has the same distribution as the maximum of $n-1$ independent unit exponentials, with density $(n-1)e^{-w}(1-e^{-w}{)}^{n-2}$. The last identity follows either from its Laplace transform or from the independent spacings of exponential order statistics.

In Kingman's coalescent, the epoch with $j$ ancestral lineages lasts $T_j\sim \mathrm{Exp}((\genfrac{}{}{0ex}{}{j}{2}))$, independently across epochs. If each lineage carries an independent Poisson process of mutations with rate $\theta /2$, its epoch mutation count satisfies $Y_j\mid T_j\sim \mathrm{Poisson}(j\theta T_j/2)$. The total mutation count is $S={\sum }_{j=2}^n{Y}_j$. Under the infinite sites mutation model, this also counts segregating sites; without that assumption it still counts mutation events, which need not correspond one-to-one to observed polymorphisms.

For normalized Legendre polynomials, this kernel for density estimation reproduces evaluation at zero on every polynomial of degree below $\ell$: expand $p$ in the orthonormal basis to obtain $\int K_{\ell }p=p(0)$. Taking monomials proves integral one and zero moments of orders $1,\dots ,\ell -1$; compact support gives finite absolute moments. If exact order requires a nonzero moment of order $\ell$, add $({\varphi }_{\ell }(0)+1){\varphi }_{\ell }{1}_{[-1,1]}$. Lower moments remain unchanged. Writing $a_{\ell }=⟨u^{\ell },{\varphi }_{\ell }⟩\ne 0$ and evaluating the full expansion of $u^{\ell }$ at zero shows that the new moment is $a_{\ell }$. A useful three-derivative example is $K_3=(9-15u^2)1_{[-1,1]}/8$, whose first two moments vanish and whose squared L2 norm is $9/8$.

Suppose a probability density function and its third derivative are bounded. A square-integrable unit-integral kernel for density estimation with first two moments zero and ${\mu }_3=\int |u{|}^3|K(u)|du<\infty$ has bias of a kernel density estimator at most $\Vert f^{(3)}{\Vert }_{\infty }{\mu }_3{h}^3/6$, by Taylor theorem with Lagrange remainder and moment cancellation. independence gives variance at most $\Vert f{\Vert }_{\infty }\Vert K{\Vert }_2^2/(nh)$. The Cauchy-Schwarz inequality bounds the expected centred absolute deviation by the square root of this variance. Adding the bias proves the inequality uniformly in $x$. The choice $h=n^{-1/7}$ gives expected absolute error $O(n^{-3/7})$.

Let the kernel for density estimation have integral one, be bounded and have compact support. For a probability density function with bounded derivative, assume $h\to 0$, $nh\to \infty$ and $n{h}^3\to 0$. The bias of a kernel density estimator is $O(h)$, negligible on the $\sqrt{nh}$ scale. For $Y_{n,i}=K((x-X_i)/h)$, $h^{-1}\mathrm{Var}{Y}_{n,i}\to f(x)\Vert K{\Vert }_2^2$. The centered summands divided by $\sqrt{nh}$ are uniformly small, so the Lindeberg-Feller central limit theorem applies. The condition $n{h}^3\to 0$ alone is insufficient: with $h=n^{-2}$ and a compactly supported kernel, with probability tending to one no observation enters the kernel window, producing a degenerate limit at a point with positive density.