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The tail integral identity gives ${\int }_0^{\infty }\mathbb{P}(S>v)dv=\mathbb{E}S$. Therefore the limiting and stationary occupancy distribution isIt is finite under the stated finite-mean assumption.

An arrival at time $u\in [0,t]$ is still present at $t$ exactly when its independent service time exceeds $t-u$. Thinning the marked Poisson process with this retention probability gives a Poisson count with parameterThusThis reasoning allows atoms in the service-time distribution; the strict survival inequality correctly treats a customer whose service has ended.

The model is a general-service infinite-server queue, usually denoted $M/G/\infty$: arrivals form a Poisson process, each wandering time is a service time with distribution that of $S$, and trapping ends service. Infinitely many servers mean that mosquitoes wander concurrently without waiting or delaying one another. General service times are not necessarily exponential, so the count alone need not be a Markov chain.

Choose a step function $s$ with $\Vert f-s{\Vert }_1<\epsilon$, by part (i). The Fourier transform satisfiesBy part (ii), ${\mathrm{limsup}}_{|t|\to \infty }|\hat{f}(t)|\le \epsilon$. Since $\epsilon$ is arbitrary,This proves the Riemann-Lebesgue lemma with the transform convention in the question.

For $t\ne 0$, direct integration of each interval in the Fourier transform givesHence $|\hat{s}(t)|\le 2{\sum }_k|c_k|/|t|$, andThe finite value at zero is ${\sum }_k{c}_k(b_k-a_k)$ and does not affect this limit.

Work in the real $L^1$ space, consistent with the specified real step coefficients. First approximate an integrable function by a finite-valued measurable function with bounded support. Truncate its domain to $[-M,M]$ and its values to $[-M,M]$; the $L^1$ error tends to zero by integrability. Quantize those remaining values in intervals of mesh $\delta$, producing a simple function $u={\sum }_{j=1}^m{c}_j{1}_{B_j}$ with finite-measure Borel sets $B_j$ and error at most $2M\delta$ on the truncated domain. Thus $\Vert f-u{\Vert }_1$ can be made arbitrarily small.

For each $B_j$, use the allowed approximation by a finite union of intervals $A_j$, choosing the symmetric-difference errors soThese unions can be written as finite disjoint bounded intervals, and changing endpoints to the required half-open convention changes only a null set. The function $s={\sum }_j{c}_j{1}_{A_j}$ is therefore a step function of the specified form, andCombining the two approximations proves the step functions are dense in $L^1(\mathbb{R})$. In the complex version of $L^1$, the same argument applies to real and imaginary parts when complex step coefficients are allowed.

The assertion is false without further global hypotheses. For a concrete non-spherical surface of constant Gaussian curvature one, fix $a=1/2$ and setHere $f>0$, $g^{\prime }>0$ and $f^{\prime 2}+g^{\prime 2}=1$, so this is a connected regular embedded surface of revolution with unit-speed meridian. Its metric is $d{u}^2+f(u{)}^{2}d{v}^2$, whose Gaussian curvature is $-f^{\prime \prime }/f=1$. This formula also follows by multiplying its principal curvatures $-f^{\prime \prime }/g^{\prime }$ and $g^{\prime }/f$.

At $u=0$ these principal curvatures are $a=1/2$ and $1/a=2$. Every point of a radius-one sphere is umbilical with both principal curvature magnitudes one. The displayed surface therefore cannot lie in any such sphere, despite having $K=1$ everywhere.

A simple closed curve on the sphere bounds two topological disks. Equal areas give area $2\pi$ for each, since the unit sphere has area $4\pi$. Apply Gauss-Bonnet theorem to either disk $D$, with $K=1$ and $\chi (D)=1$:A continuous geodesic curvature that were strictly positive everywhere, or strictly negative everywhere, would have a correspondingly strict sign for this integral. Therefore neither everywhere strictly positive nor everywhere strictly negative geodesic curvature is possible. Orientation reversal changes its sign but not this conclusion.

Parametrize a regular curve by arclength $s$, write its unit tangent as $T$, and choose an oriented surface unit normal $N$. Its signed geodesic curvature isIt vanishes exactly when the acceleration of the unit-speed curve is normal to the surface, equivalently when the curve is a geodesic with arclength parametrization. For an arbitrary regular parametrization, this means the curve becomes a geodesic after reparametrization.

The global Gauss-Bonnet theorem for a compact oriented surface with smooth boundary, positively oriented with the surface on the left, isFor piecewise smooth boundary, add the signed exterior turning angles at its corners to the left side.

For a compact triangulated surface, the Euler characteristic is $\chi =V-E+F$, independent of the triangulation. For a connected closed orientable surface, the genus $g$ is its number of handles, characterized byA connected compact orientable surface with $b$ boundary components instead has $\chi =2-2g-b$. For multiple connected components the Euler characteristics add, with genus specified for each component. A compact boundaryless surface embedded in ${\mathbb{R}}^3$ is orientable, so the usual closed-surface formula applies to each connected component.

Apply the proved projection result to $V=\overline{X}$ and write $f=v+w$ with $w\perp V$. By part (iii), $A_{n}v\to 0$, while part (iv) gives $U^{k}w=w$ and hence $A_{n}w=w$. ThusThe limit is the orthogonal projection onto the fixed vectors of $U$. This is norm convergence, not just weak convergence.

If $w\perp X$, then $⟨w,(I-U)g⟩=0$ for all $g$. The adjoint identity gives $(I-U^{\ast })w=0$, so $U^{\ast }w=w$. Multiplying by $U$ givesSince $X\subseteq V=\overline{X}$, $w\perp V$ implies $w\perp X$ and yields the same result. Conversely a fixed vector is perpendicular to $X$ by this calculation, identifying $V^{\perp }$ as the fixed subspace.

If $v\in \overline{X}$, choose $x\in X$ with $\Vert v-x\Vert <\epsilon$. By part (i),The last term tends to zero by part (ii). Taking the limit superior and then $\epsilon \downarrow 0$ proves $A_{n}v\to 0$ for every $v\in \overline{X}$.

The set $X$ is the range of the linear operator $I-U$, hence is a linear subspace. If $f=g-Ug$, the sum telescopes:Since $U$ is an isometry, $\Vert A_{n}f\Vert \le 2\Vert g\Vert /n\to 0$. Hence the averages tend to zero on $X$.

For the unitary operator,Thus it is an isometry, and every power has norm one. The triangle inequality gives

Take a homotopy equivalence $f:X\to Y$ with inverse $g$, and universal covering maps $p_X,p_Y$. Since the covering spaces are simply connected, the path lifting theorem constructs lifts $\tilde{f}$ of $f{p}_X$ and $\tilde{g}$ of $g{p}_Y$, after specifying one value in the appropriate fiber. Explicitly, define the lift at a point by lifting the image of a path from the chosen base point; independence from the path follows from simple connectedness and homotopy lifting.

Lift a homotopy $gf\simeq 1_X$ starting at $\tilde{g}\tilde{f}$. The homotopy lifting theorem for a covering map makes its endpoint a map $d_X:\tilde{X}\to \tilde{X}$ with $p_X{d}_X=p_X$. Such a lift of the identity is a deck transformation and is a homeomorphism: lift the identity again with the base point image reversed, and uniqueness of path lifts makes the two compositions identities. Likewise $\tilde{f}\tilde{g}\simeq d_Y$ for a deck transformation of $\tilde{Y}$.

Thus both compositions are homotopy equivalences, even though the lifted homotopies need not end at the particular identity lift. Part (i) then provesKeeping the possible deck transformations is essential when the original homotopies do not fix base points.

If $fg\simeq 1_Y$ and $hf\simeq 1_X$, composing homotopies givesThus $gf\simeq hf\simeq 1_X$, while $fg\simeq 1_Y$ already holds, so $g$ is a homotopy inverse of $f$.

More generally let $a:Y\to Y$ be a homotopy inverse of $fg$ and $b:X\to X$ one of $hf$. Then $g^{\prime }=ga$ satisfies $f{g}^{\prime }\simeq 1_Y$, and $h^{\prime }=bh$ satisfies $h^{\prime }f\simeq 1_X$. The first argument applies. Hence the separate one-sided hypotheses imply that $f$ is a homotopy equivalence, as expressed by separate homotopy inverses combine into a homotopy equivalence.

The polynomial $T^4-2$ is irreducible by Eisenstein criterion at two, so $1,\alpha ,{\alpha }^2,{\alpha }^3$ is a rational basis and each of these elements is integral. The nontrivial automorphism of $K/k$ sends $\alpha$ to $-\alpha$. For an algebraic integer $\theta$, its relative trace and the trace of the algebraic integer $\alpha \theta$ areBoth are in ${\mathcal{O}}_k$, proving $2a,2b,2c,4d\in \mathbb{Z}$. Multiplying $\theta$ by its conjugate gives the field normAgain this is in ${\mathcal{O}}_k$, so both displayed coefficients are integers.

To finish the denominator argument, put $A=2a$, $B=2b$, $C=2c$, $D=4d$, all integers. The two norm coefficients implyThe second congruence modulo two first gives $D$ even. Modulo four it then gives $B$ even. Therefore $2BD$ vanishes modulo four, and the first congruence reads $A^2+2C^2\equiv 0(\mathrm{mod}4)$. The possible squares modulo four force $A$ even and then $C$ even. Returning to the second congruence now gives $D^2\equiv 0(\mathrm{mod}8)$, so $D$ is divisible by four. Thus $a,b,c,d$ are all integers.

Conversely, every integer linear combination of these powers is integral. We have proved the integral basis of the fourth-root-of-two field:

An integral basis of a degree-$d$ number field is a $\mathbb{Z}$-basis of its ring of integers: each algebraic integer has a unique integer linear combination of those $d$ elements. For $k=\mathbb{Q}(\sqrt{2})$, the standard quadratic-integer criterion gives ${\mathcal{O}}_k=\mathbb{Z}[\sqrt{2}]$, since $2$ is squarefree and is not congruent to one modulo four. Thus

Induct on the order of the p-group. Every complex irreducible representation of an abelian group is one-dimensional: commuting group operators preserve a common eigenspace, and irreducibility makes that eigenspace the whole space. This handles the abelian cases.

If $V$ is not faithful, it is an irreducible representation of a smaller quotient $G/K$, where $K$ is its kernel. Induction expresses it as induced from a one-dimensional representation of a subgroup of $G/K$. Pulling back the subgroup and inflating the character gives the same induction statement for $G$, because the coset spaces of the quotient subgroup and its inverse image agree.

Now suppose $V$ is faithful and $G$ is nonabelian. Use the allowed abelian normal subgroup $A$ not contained in the centre. If restriction to $A$ were isotypic, its irreducible type would be one-dimensional, so every $a\in A$ would act as a scalar. It would therefore commute with every representing matrix of $G$. Faithfulness would imply $ga=ag$ in $G$ for every $g$, contradicting the choice of $A$. By part (a), $V={\mathrm{Ind}}_H^{G}W$ for an irreducible representation of a proper subgroup $H$. Induction on $H$ gives $W={\mathrm{Ind}}_K^H\chi$ with $\chi$ one-dimensional. Transitivity of induction yieldsThis proves the monomiality of irreducible representations of finite p-groups.