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After setting $z=1/x$, the transformed coefficients have a singularity beyond the regular-singular growth bounds at $z=0$. Thus infinity is an irregular singular point.

On the simplex define $f(x)=Ax/(1^{T}Ax)$. Strict positivity makes the denominator positive and maps the simplex continuously into its interior. By Brouwer fixed-point theorem, $f(x)=x$ for some $x$. Thus $Ax=\lambda x$, where $\lambda =1^{T}Ax>0$, and because $Ax$ is strictly positive, so is $x$.

The equivalent no-retraction theorem says there is no continuous retraction of the closed disk onto its boundary circle. A fixed-point-free map gives a retraction by projecting from $f(x)$ through $x$ to the boundary; conversely a retraction composed with the antipodal boundary map gives a fixed-point-free self-map.

The Brouwer fixed-point theorem says that every continuous map from a closed two-dimensional disk to itself has a fixed point.

Apply Cameron-Martin theorem to the absolutely continuous path $h(t)=xt/T$ up to $T$ and $h(t)=x$ thereafter. Its derivative is $x/T$ on $[0,T]$ and zero later, so the Radon-Nikodym factor is $\exp ((x/T)W_T-x^2/(2T))$, yielding the formula for every bounded measurable functional of the path after $T$.

The joint Gaussian density of the independent increments $\Delta W_i$ changes under shifts $c\Delta t_i$ byMultiplying the density and integrating the bounded cylinder function $g$ proves the identity.

The transformed process is centred Gaussian andThe Brownian strong law makes $t{W}_{1/t}\to 0$ almost surely as $t\downarrow 0$, giving continuity at zero; elsewhere continuity is immediate. Part (a) now applies.

A Brownian motion starts at zero, has continuous paths, and has independent increments with $W_t-W_s\sim N(0,t-s)$. The stated covariance makes every finite collection jointly Gaussian; disjoint increments have zero covariance and hence are independent. Their variances are interval lengths, proving all defining properties.

Neither proper gamma prior yields the condition in part (c). The risk in part (a) is unbounded as $\theta \to \infty$ because $\beta >0$. The risk in part (b) is analytic, tends to zero at infinity, and is not identically zero, so its prior average is strictly below its supremum for a gamma prior with positive density everywhere. Thus this sufficient condition proves neither estimator minimax for any $\alpha ,\beta >0$.

An admissible Bayes estimator whose Bayes risk equals its maximum risk is minimax. Indeed every estimator has maximum risk at least its prior-average risk, which is at least the Bayes risk; equality for the stated estimator attains this lower bound. Admissibility excludes a distinct estimator improving it everywhere.

Put $c=(\beta +1)/(\beta +2)$. The posterior mean of $e^{-\theta }$ is $\delta (X)=c^{\alpha +X}$. Using the Poisson probability-generating function, its risk is

The posterior is $\mathrm{Gamma}(\alpha +X,\beta +1)$, so under quadratic loss the Bayes estimator is $\delta (X)=(\alpha +X)/(\beta +1)$. At fixed $\theta$ its risk is

For $f_t(i,\ell )={\mathbb{E}}_i[g(\ell +L(t))1_{X(t)=j}]$, part (a) gives $M{f}_t={\partial }_t{f}_t$. Apply part (b), integrate in $t$ from zero to infinity, and integrate by parts in time. Compact support makes the terminal term vanish, while $f_0(i,\ell )=g(\ell )1_{i=j}$. Moving the resulting Gaussian derivatives once more by parts yields the stated identity with $y_i{y}_j$.

Symmetry of $Q$ gives ${\partial }_{y_i}{e}^{y^{T}Qy/2}=e^{y^{T}Qy/2}{\sum }_j{q}_{ij}{y}_j$. Also ${\partial }_{y_i}f(i,y^2/2)=y_i{\partial }_{{\ell }_i}f$. Substitute these identities into the left side and integrate the resulting total $y_i$ derivative by parts. Compact support removes the boundary term and leaves exactly the right side.

Condition on the first short time interval. A jump contributes the generator term $Qf$, while local time grows at unit speed in the current coordinate and contributes ${\partial }_{{\ell }_i}f$. The Markov property and division by the interval length therefore give ${\partial }_t{f}_t=M{f}_t$.

For $K<0$ there is at most one. Two essential simple closed geodesics cannot form a geodesic bigon by Gauss-Bonnet theorem, so they are disjoint; the annulus between them would have $\chi =0$ and geodesic boundary, forcing $\int K=0$, a contradiction. Thus multiple geodesics in the flat case are disjoint, while in the positive example they need not be.

For $K>0$ the number can again be infinite: remove two points away from an infinite family of great circles on the round sphere. The result is a cylinder of positive curvature containing those intersecting simple closed geodesics.

The geodesic curvature is the tangential component of $\alpha \prime \prime$, namely $k_g=⟨\alpha \prime \prime ,N\times \alpha \prime ⟩$ up to orientation; a curve is geodesic when $k_g=0$. Gauss-Bonnet theorem for a compact region says $\int K+{\int }_{\partial }{k}_g+\sum \text{exterior angles}=2\pi \chi$.

For $K=0$, a flat cylinder has infinitely many parallel closed geodesics.

Because $1<p<\infty$, $L^p$ is reflexive, so a bounded sequence has a weakly convergent subsequence $g_{j_k}\rightharpoonup g$. Thenand weak convergence handles $\int g_{j_k}h$. Thus the displayed limit holds (with the evident OCR correction $\int gh$ on the right).

Strong convergence of the tests is essential: in $L^2(0,2\pi )$ take the same orthonormal sequence for $g_j$ and $h_j$. Both converge weakly to zero, but $\int g_j{h}_j=1$.

Strong convergence means $\Vert f_j-f{\Vert }_p\to 0$; weak convergence means $\int f_{j}h\to \int fh$ for every $h\in L^q$. By duality choose $h$ of norm one with $\int fh$ arbitrarily close to $\Vert f{\Vert }_p$. Holder inequality and passage to the limit give $\Vert f{\Vert }_p\le \mathrm{liminf}\Vert f_j{\Vert }_p$.