Rotate the polar axis to $\xi$. Since $W_t$ is a finite sphere measure, its Fourier transform iswith value $t$ at $\xi =0$. Fourier transforming the wave equation gives ${\hat{u}}_{tt}+|\xi {|}^2\hat{u}=0$ with initial displacement zero and velocity $\hat{g}$. Its solution is this multiplier times $\hat{g}$, so inversion gives $u=W_t\ast g$. Thus the Kirchhoff formula isThis extends to arbitrary smooth $g$, without any growth restriction: for each finite spacetime neighborhood, multiply $g$ by a compactly supported smooth cutoff equal to one on every sphere occurring in that neighborhood. The formula is unchanged there, so its verified differential equation remains valid locally. Its initial values follow from $u(t,x)=tg(x)+o(t)$ and differentiation under the sphere integral. Since the neighborhood was arbitrary, it is a smooth solution everywhere for $t\ge 0$.

Choose the Fourier transform conventionIt maps the Schwartz space to itself. For a tempered distribution $u$, define the transform by bilinear distributional duality:For an ordinary integrable function this agrees with the first formula by interchanging its integrals. The absence of complex conjugation is the distribution-pairing convention.

Use the Runge exhaustion of a slit disk, with the principal argument $-\pi <\arg z<\pi$:These are nested compact subsets of $\Omega$. Their complements are connected: the inner disk and exterior region join through the omitted sector around the negative real radius. Every compact subset of $\Omega$ has positive distance from zero, the unit circle and the slit, so it is contained in $K_m$ for all sufficiently large $m$.

Since $f$ is holomorphic on $\Omega$, it is holomorphic on a neighborhood of each $K_m$. The polynomial Runge theorem gives $P_m$ with ${\sup }_{K_m}|P_m-f|<1/m$. Given any compact $K\subset \Omega$, eventually $K\subseteq K_m$, so the same estimate holds on $K$. ThereforeThe nested exhaustion makes this one sequence work simultaneously on all compact subsets.

The polynomial Runge theorem states that, for a compact set $K\subset \mathbb{C}$ with connected complement and a function $f$ holomorphic on a neighborhood of $K$, for every $\epsilon >0$ there is a polynomial $P$ such thatThe connected-complement hypothesis permits polynomial approximation rather than approximation by rational functions with poles in bounded complementary components.

Assume the riskless gross return $q=1+r>0$, as required by the positive-wealth model, and a finite moment $m_b=\mathbb{E}{X}_0^b$. DefineThis maximum exists: the control interval is compact and dominated convergence gives continuity, using the integrable bound $(X_0+q{)}^b\le X_0^b+q^b$ for $0<b\le 1$.

The terminal value is $V_N(w)=U(w)$. If $V_{n+1}(w)=a_{n+1}U(w)$, the Bellman equation and independence of the next return giveBackward induction therefore provesAny maximizing proportion can be used each day; allowing adapted choices cannot exceed the Bellman bound. For $\alpha =1$, $C_{\ast }=m_b$, exactly agreeing with part (b). If $m_b=\infty$ and at least one day remains, the allowed choice $\theta =1$ already gives infinite expected utility, so the value is infinite rather than a finite coefficient. This is the necessary moment qualification to the finite optimization formula.

Put $b=1-R>0$ and $m_b=\mathbb{E}{X}_0^b$. Homogeneity of the constant relative risk aversion utility gives $U(z_N)=U(z_n){\prod }_{j=n}^{N-1}{X}_j^b$. ThusAs before this is an extended-valued conditional expectation if the required moment is infinite and days remain.

The product $z_n=z_0{\prod }_{j=0}^{n-1}{X}_j$ uses only past multipliers, so all $X_n,\dots ,X_{N-1}$ are independent of it. With $m=\mathbb{E}{X}_0$, independence givesThe formula is finite when $m<\infty$; since the variables are positive, it also has the extended value $+\infty$ for $m=\infty$ and $n<N$. At $n=N$, it simply reads $z_N$.

For equilibrium, let $t\to \infty$ in part (e). Monotone convergence of the tail integral and the finite mean giveand consequentlyThis limiting argument uses the equilibrium age and residual-service distribution produced by the queue, not just a fresh independent copy of $S$ for each mosquito present. Replacing their residual service by fresh service would generally be wrong unless the service law is memoryless.

After closing, the count can only decrease. Thus $V(t)>v$ is the event that at least one mosquito remains at $t+v$. The zero probability of a Poisson count yields the clearance time after closing an infinite-server queue:The clearance time has an atom at zero of mass $e^{-m_t}$, corresponding to an empty room when the window closes.

Only arrivals at times $u\le t$ remain after the window closes. Such an arrival survives until $t+v$ exactly when its service time exceeds $t+v-u$. The same independent Poisson thinning givesAt $v=0$ this agrees with part (b), while for large $v$ the parameter tends to zero.

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.