The James–Stein estimator in the specified unit-variance model iswith an arbitrary value at $X=0$, a probability-zero event. The risk of $X$ is $p$. Gaussian Stein identity states $E_{\theta }[(X_i-{\theta }_i)g_i(X)]=E_{\theta }[{\partial }_i{g}_i(X)]$ for suitably integrable differentiable functions; it extends to the present singular function by truncation, since $p>2$ makes $E\Vert X{\Vert }^{-2}$ finite.

Put $c=p-2$ and $g(x)=-cx/\Vert x{\Vert }^2$. Then $\Vert g{\Vert }^2=c^2/\Vert x{\Vert }^2$ and $\mathrm{div}g=-c(p-2)/\Vert x{\Vert }^2$. Expanding the quadratic risk and applying Stein's lemma yieldsThus domination is strict for every finite $\theta$.

To identify the worst-case risk, let $r=\Vert \theta \Vert \to \infty$. On $\Vert X\Vert >r/2$, the reciprocal square is at most $4/r^2$. On the complementary ball, the Gaussian density is at most $(2\pi {)}^{-p/2}{e}^{-r^2/8}$, so its reciprocal-square integral is bounded by a constant times $e^{-r^2/8}{r}^{p-2}$, tending to zero. Therefore $E_{\theta }\Vert X{\Vert }^{-2}\to 0$, and $R(\theta ,\hat{\theta })\to p$ from below. HenceThe second supremum is approached at infinity and need not be attained at a finite parameter.

The planar planar isoperimetric inequality iswith equality precisely for a disc. Let the positively oriented boundary be parametrized by arclength as $r(s)=(x(s),y(s))$, $0\le s\le L$. Translate its mean to zero. Then $|r^{\prime }|=1$, and Green's theorem gives $2A={\int }_0^L(x{y}^{\prime }-y{x}^{\prime })ds$.

The periodic Wirtinger inequality says that for a real absolutely continuous $L$-periodic function $h$ of mean zero and square-integrable derivative,with equality exactly for $h(s)=a\cos (2\pi s/L)+b\sin (2\pi s/L)$. Apply it to both coordinates and use Cauchy-Schwarz inequality:If equality holds, the Wirtinger equalities give $r=p\cos (2\pi s/L)+q\sin (2\pi s/L)$. The arclength condition forces $|p|=|q|=L/(2\pi )$ and $p\cdot q=0$, so the boundary is a circle. A circle attains equality, making the constant optimal.

The inequality and equality characterization persist for finitely many corners. A rectifiable piecewise smooth simple closed curve has a Lipschitz arclength parametrization, with $|r^{\prime }|=1$ almost everywhere. Green's theorem and the absolutely continuous version of Wirtinger's inequality used above still apply. More generally one may approximate a rectifiable Jordan curve by smooth curves while preserving the limiting length and area. If the allowed isolated singularities give infinite length, the inequality is trivially satisfied in the extended sense.

At a noncritical point a holomorphic map has a local holomorphic inverse. Fibres of a nonconstant map between compact connected Riemann surfaces are finite: zeros are isolated and compactness precludes infinitely many. Around a nonbranch value $w$, choose disjoint inverse neighbourhoods of every point in its fibre. Compactness of the remaining source ensures a sufficiently small disc about $w$ has no additional preimages. Its preimage is therefore a disjoint union of open sets each mapped biholomorphically onto that disc. This proves a finite-sheeted unbranched covering map after removing the branch fibres.

Here the word “regular” must mean locally unbranched or evenly covered, rather than the stronger modern normal-cover condition. A general holomorphic map does not give a normal covering, and the example below itself shows the distinction.

Lift a loop based at $w$ uniquely from each point of $f^{-1}(w)$. Its endpoint defines a permutation; reversing the loop gives the inverse permutation, and concatenation gives composition. Homotopic loops yield the same endpoints. The monodromy group of a covering consists of these permutations. Removing finitely many points from a connected surface leaves it path-connected. Any two points in the same fibre can therefore be joined upstairs; the projection of that path is a loop whose lift joins them. Thus the monodromy action is transitive.

For the rational map of degree three, differentiation givesAt $z=0$, the local degree is three, so a small loop around its value $0$ acts as a 3-cycle. At $z=\pm \sqrt{3}$, the local degree is two; their distinct branch values are $\mp 3\sqrt{3}/2$, and a loop around either acts as a transposition. The poles $\pm 1$ and the point at infinity are simple and unramified. A 3-cycle and a transposition generate the full symmetric group, soThe corresponding degree-three connected cover is not normal: a normal cover would have a regular monodromy action, whereas $S_3$ on three letters has nontrivial point stabilizers. Thus the stronger interpretation of the printed terminology cannot hold.

For an odd prime $p$, the Legendre symbol $(a/p)$ is $0$ if $p\mid a$, $1$ if $a$ is a nonzero square modulo $p$, and $-1$ otherwise. Euler's criterion statesIf $p\mid a$ both sides vanish. Otherwise write $a\equiv g^j$ for a primitive root $g$. The squares are precisely the even powers of $g$, while $g^{(p-1)/2}$ has order two and therefore equals $-1$. Hence $a^{(p-1)/2}=(-1{)}^j$, proving the criterion and the useful special case $(-1/p)=(-1{)}^{(p-1)/2}$.

For odd $n$, $n^2+4\equiv 5(\mathrm{mod}8)$. Every prime divisor $p$ of this number is odd, and $p\nmid n$, so $(n/2{)}^2\equiv -1(\mathrm{mod}p)$. Euler's criterion forces $p\equiv 1(\mathrm{mod}4)$, hence $p\equiv 1$ or $5(\mathrm{mod}8)$. A product of such primes can be $5(\mathrm{mod}8)$ only if a prime $5(\mathrm{mod}8)$ occurs to an odd power.

Suppose the primes $5(\mathrm{mod}8)$ were $p_1,\dots ,p_r$. Choose the odd integer $n=p_1\cdots p_r$ (or $n=1$ for an empty list). None of these primes divides $n^2+4$, since the latter is $4$ modulo each $p_i$. Yet the preceding argument gives a prime divisor $5(\mathrm{mod}8)$, outside the list. This contradiction proves infinitely many primes congruent to $5$ modulo $8$.