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For $n\ge 3$, extend the restriction of the continuous function $f$ to $[1/n,1-1/n]$ to a continuous function $f_n$ on $[0,1]$ by making it constant on each of the two remaining intervals. The Weierstrass approximation theorem supplies a polynomial $P_n$ with $\Vert P_n-f_n{\Vert }_{\infty ,[0,1]}<1/n$. Every compact set $K\subset (0,1)$ is eventually contained in $[1/n,1-1/n]$, where $f_n=f$. Hence ${\sup }_K|P_n-f|<1/n$ eventually. This proves uniform convergence on every compact subset, without assuming any endpoint limits of $f$. The first two polynomials can be chosen arbitrarily.

The equilibrium-delay theorem for a renewal process with positive holding time $S$, finite mean $\mu$, and distribution function $F$ says that the first future renewal must have the equilibrium residual-life distribution,independently of subsequent ordinary holding times. This is the residual lifetime seen at a stationary observation time, rather than an ordinary inter-renewal lifetime.

Here $\mu =3/\lambda$ and $1-F(t)=e^{-\lambda t}(1+\lambda t+{\lambda }^2{t}^2/2)$. Thus the required initial-delay density isEquivalently, choose with equal probability a sum of one, two, or three independent rate-$\lambda$ exponential random variables. The three Erlang distributions have densities that average to the displayed density. A concrete stationary construction chooses the phase modulo three of a stationary Poisson clock uniformly; the number of exponential stages until its next third-stage renewal is then uniform on $\{1,2,3\}$.

Convolve three rate-$\lambda$ exponential distributions. The first two have convolution density ${\lambda }^{2}x{e}^{-\lambda x}{1}_{x>0}$; convolving once more givesThus $X$ is an Erlang distribution of shape three, equal in law to the sum of three independent exponentials. Addition of independent means and variances yieldsFor $t\ge \lambda$ the moment-generating function is infinite.

Realize each holding time as a consecutive block of three exponential holding times from one Poisson process $N_t$. The renewal count is then $X_t=\lfloor N_t/3\rfloor$ pathwise under this coupling. If $R_t$ is the remainder of $N_t$ modulo three, $N_t=3X_t+R_t$ and $\mathbb{E}{R}_t=p_1(t)+2p_2(t)$. Taking expectations givesThis proves the formula without assuming that the renewal count itself is Poisson.

Let $S_1,S_2,\dots$ be independent exponential random variables of rate $\lambda$, let $T_n=S_1+\cdots +S_n$ and $T_0=0$, and define $N_t=\max \{n:T_n\le t\}$. The strong law of large numbers gives $T_n/n\to 1/\lambda$, so there is no finite-time explosion. This defines the Poisson process of rate $\lambda$.

For $n\ge 1$, repeated convolution gives $T_n$ the gamma distribution with density ${\lambda }^n{s}^{n-1}{e}^{-\lambda s}/(n-1)!$. Independence of $S_{n+1}$ givesFor $n=0$, the probability is $\mathbb{P}(S_1>t)=e^{-\lambda t}$. Hence $\boxed{N_t\sim \mathrm{Poisson}(\lambda t)}$ for all $t\ge 0$, including the degenerate law at $t=0$.

Write the plane as $\{x:\nu \cdot x=c\}$ for a fixed unit vector $\nu$, and put $h(x)=\nu \cdot x-c$. Suppose $X$ misses the plane. Choose a point of $X$ and a number $A$ larger than its distance to the plane. The hypothesis $D_n\to \infty$ makes $\{x\in X:|h(x)|\le A\}$ bounded, and closedness of $X$ makes it compact. Thus $|h|$ attains its global minimum at some $p\in X$, with $|h(p)|>0$.

Locally the sign of $h$ is constant, and $h$ has a local minimum or maximum at $p$. Its surface gradient vanishes, so the tangent plane of $X$ there is parallel to the given plane and its normal is $\pm \nu$. For a surface tangent vector, the intrinsic Hessian of $h$ is the normal component of the second fundamental form, namely ${\mathrm{Hess}}_{X}h=\pm b$ at $p$. The Hessian is semidefinite because this is a local extremum. Since $X$ is a minimal surface, ${\mathrm{tr}}_{g}b=2H=0$. A semidefinite symmetric form with zero trace must be zero, so $b(p)=0$. This makes $p$ a planar point, contradicting the assumption. Therefore $\boxed{X\cap \Pi \ne \varnothing }$.

The boundaryless hypothesis ensures that the minimizing point is an interior surface point. The condition on $D_n$ ensures that the infimum is attained; disjointness from a plane alone would not do so.

The Theorema Egregium states that the Gaussian curvature of a surface in ${\mathbb{R}}^3$ is determined entirely by its first fundamental form; in particular a local isometry preserves it.

Let $r(u^1,u^2)$ be a regular parametrization, $g_{ij}=r_i\cdot r_j$, $n$ a unit normal, and $b_{ij}=r_{ij}\cdot n$ the second fundamental form. Differentiating the metric and decomposing $r_{ij}$ into tangent and normal parts gives the Gauss formulaDifferentiating $n\cdot r_j=0$ gives $n_i=-b_i{}^k{r}_k$, the Weingarten formula. Compute two successive ambient derivatives of a tangent vector and subtract them. Ambient derivatives commute, so their tangential parts yield the Gauss equationwhere $S$ is the shape operator and $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$ is the intrinsic curvature of the connection with the displayed Christoffel symbols. Taking $X={\partial }_1,Y={\partial }_2,Z={\partial }_2$ and the inner product with ${\partial }_1$ givesBut $K=\det S=\det (b)/\det (g)$, soBoth the connection and its curvature on the right are computed from $g$ and its derivatives alone. This proves the theorem, rather than merely using its name as a substitute for the argument.

Two function elements $(U,f)$ and $(V,g)$ define the same germ of a holomorphic function at $p$ when $p\in U\cap V$ and they agree on some neighborhood of $p$. Write the germ as $[f{]}_p$. Basic open sets of the space of germs of holomorphic functions arefor open $U$ on which $f$ is holomorphic. If two such sets meet, equality of the germs gives a smaller open sheet contained in their intersection, so these sets form a basis. The projection $\pi ([f{]}_p)=p$ maps each sheet homeomorphically onto $U$. Compose this with a local coordinate on $X$ to define the complex charts on the germ space. Transition maps are the corresponding transition maps on $X$, hence holomorphic; $\pi$ is a local biholomorphism and is onto, since even constant functions provide germs over every point.

The germ space is Hausdorff. Germs over different points are separated by base neighborhoods. For distinct germs over the same point, choose representatives on one small connected disk. If their two sheets met anywhere in that disk, the representatives would agree on an open set and hence everywhere on the disk by the identity theorem, a contradiction. These sheets thus separate the germs. Together with the allowed second-countability fact, the charts make each component a Riemann surface.

Here “covering in the sense of complex analysis” means this locally biholomorphic, possibly incomplete analytic germ projection. It does not mean an evenly covered topological covering map: an analytic branch can cease to exist at a singularity, so the restriction to a component need not project onto all of $X$.

The evaluation map $E([f{]}_p)=f(p)$ is holomorphic, since on the sheet ${\tilde{U}}_f$ it is just $f$. Starting from one germ, continuation through overlapping disks produces exactly the germs in its connected component. Indeed such continuations give paths in the germ space; conversely, cover a path by finitely many successive sheets to obtain a chain of function elements. Components are path-connected because the surface is locally path-connected. No two distinct germs can be added to that component without being obtainable by continuation. It therefore represents the maximal analytic continuation, or complete holomorphic function, of any of its germs. Conversely every such maximal continuation embeds into the germ space by taking its local germs; completeness makes its image the entire component. These constructions are inverse, giving the requested bijection between complete holomorphic functions and connected components, with functions understood up to their natural equivalence as analytic continuations, rather than merely as single-valued functions defined on all of $X$.

The lacunary series converges locally uniformly on $|z|<1$: for $|z|\le r<1$, $\sum r^{2^n}\le {\sum }_{j\ge 1}{r}^j<\infty$. It therefore defines a holomorphic function there. Let ${\zeta }^{2^N}=1$. For $n\ge N$, $(r\zeta {)}^{2^n}=r^{2^n}$ is positive real. As $r\uparrow 1$, the sum of any fixed number of these tail terms tends to that number, while the earlier finite sum stays bounded. Thus $\mathrm{Re}f(r\zeta )\to +\infty$. A analytic continuation through $\zeta$ would be bounded in a small closed neighborhood and cannot have this radial divergence.

Roots of unity of orders $2^N$ are dense on the unit circle. If an analytic continuation existed through any boundary point, its neighborhood would contain one of these singular roots, and the identity theorem would make it agree with $f$ on the inside portion. This contradicts the divergence just proved. Hence the entire unit circle is a natural boundary. The exponent printed in the PDF is $2^n$, not $2n$; the latter would give a rational function and would not have this property.

Set $g_n=f-f_n$. These are nonnegative continuous functions, decrease pointwise, and converge pointwise to zero. For $\epsilon >0$, the open sets $U_n=\{x:g_n(x)<\epsilon \}$ increase and cover $K$. A finite subcover exists by compactness; because the sets are increasing, one $U_N$ already covers $K$. Hence $0\le g_n(x)<\epsilon$ for every $x\in K$ and every $n\ge N$, and $\Vert f-f_n{\Vert }_{\infty }\le \epsilon$. Therefore $\boxed{\Vert f_n-f{\Vert }_{\infty }\to 0}$. This is Dini's theorem; continuity of the limit is essential to the openness used in the proof.

Assume first that $\overline{\mathcal{F}}$ is compact in the supremum norm. A finite norm-net shows that the norms of its members are bounded, so pointwise boundedness follows. Given $x$ and $\epsilon >0$, choose a finite $\epsilon /3$-net $g_1,\dots ,g_N$ in the closure. Their continuity gives a common neighborhood $U$ of $x$ where $|g_j(y)-g_j(x)|<\epsilon /3$ for every $j$. For any $f\in \mathcal{F}$, choose $g_j$ with $\Vert f-g_j{\Vert }_{\infty }<\epsilon /3$; the triangle inequality then gives $|f(y)-f(x)|<\epsilon$. Thus the family is equicontinuous.

Conversely, suppose pointwise boundedness and equicontinuity. For a prescribed $\epsilon >0$, choose neighborhoods $U_x$ where every $f$ varies from $f(x)$ by less than $\epsilon /3$. Compactness supplies a finite subcover $U_{x_1},\dots ,U_{x_N}$. Pointwise boundedness makes the set of vectors $(f(x_1),\dots ,f(x_N))$ bounded in ${\mathbb{R}}^N$. Partition a bounding box into finitely many boxes of coordinate diameter less than $\epsilon /3$, and choose one function from each occupied box. If $f,g$ belong to the same box and $y\in U_{x_j}$, thenThe chosen functions are therefore a finite $\epsilon$-net in the supremum norm. Thus $\mathcal{F}$ is totally bounded. Its closure is complete and totally bounded, hence compact. To recall the last metric-space fact, successively selecting subsequences in balls of radii $2^{-j}$ produces a Cauchy diagonal subsequence from every sequence; completeness supplies its limit. This proves both directions of the Arzelà-Ascoli theorem.

For a compact space $K$, the Arzelà-Ascoli theorem says that a family $\mathcal{F}\subset C(K)$ is relatively compact in the supremum norm if and only if it is pointwise bounded and equicontinuous. In the usual formulation $K$ is compact Hausdorff; the elementary argument below also works for any compact $K$ for real-valued continuous functions.

Pointwise bounded means ${\sup }_{f\in \mathcal{F}}|f(x)|<\infty$ for every $x\in K$. Equicontinuity means that for every $x\in K$ and $\epsilon >0$ there is a neighborhood $U$ of $x$ such that $|f(y)-f(x)|<\epsilon$ for every $y\in U$ and $f\in \mathcal{F}$. Relative compactness means compactness of the closure in $C(K)$, where $\Vert f{\Vert }_{\infty }={\sup }_K|f|$. This is a complete metric vector space, since uniform limits of continuous functions are continuous. Equivalently, every sequence in the family has a uniformly convergent subsequence, with limit in $C(K)$, though not necessarily in the original family.

For a finite chain complex of finite-dimensional vector spaces, its Euler characteristic is $\chi (C)={\sum }_n(-1{)}^n\dim C_n$. For finitely generated free abelian groups, replace dimension by rank. Write $Z_n=\ker d_n$ and $B_n=\mathrm{im}{d}_{n+1}$. The identities $\mathrm{rank}{C}_n=\mathrm{rank}{Z}_n+\mathrm{rank}{B}_{n-1}$ and $\mathrm{rank}{Z}_n=\mathrm{rank}{H}_n+\mathrm{rank}{B}_n$ show by cancellation that $\chi (C)={\sum }_n(-1{)}^n\mathrm{rank}{H}_n$. An exact sequence thus has alternating rank sum zero.

Starting at the left, the three displayed sequences have alternating rank sums respectively $0,1,1$. Therefore, given that one is exact, the first sequence is the exact one. For completeness its necessary image ranks, successively, are $11,13,7,6,14,11$, all nonnegative and consistent with the adjacent module ranks. Split each middle free group as the incoming image plus a complementary outgoing image, and use identity maps on these summands, to realize an exact sequence of the first rank pattern. Vanishing of the alternating sum alone is not a general test for the exactness of unspecified maps; the supplied existence assertion is what selects the sequence.

A chain homotopy from $g$ to $f$ consists of homomorphisms $s_n:C_n\to C_{n+1}^{\prime }$ such thatfor every $n$. If $z\in C_n$ is a cycle, $d_{n}z=0$, so $f_n(z)-g_n(z)=d_{n+1}^{\prime }{s}_n(z)$ is a boundary. Hence $f_n(z)$ and $g_n(z)$ represent the same homology class. Every homology class has a cycle representative, proving $\boxed{f_{\ast }=g_{\ast }}$ in every degree.

The character of the dual representation is ${\chi }_{V_n^{\ast }}(g)={\chi }_n(g^{-1})=\overline{{\chi }_n(g)}$, using unitarizability. The displayed sine quotient is real, so this character equals ${\chi }_n$. The theorem that characters determine finite-dimensional semisimple complex representations of a compact group therefore gives $\boxed{V_n\cong V_n^{\ast }}$. This is complex-linear self-duality; it does not assert that every $V_n$ has an invariant real structure.

The same Weyl integration formula for SU2 gives $⟨{\chi }_n,{\chi }_m⟩={\delta }_{nm}$. More strongly, these characters are complete in the space of square-integrable class functions: multiplication by $\sin \theta$ identifies that weighted space with $L^2(0,\pi )$, and sends ${\chi }_n$ to $\sin ((n+1)\theta )$. The sine functions form a complete orthogonal basis of $L^2(0,\pi )$, as follows from the Fourier series theorem applied to odd extensions.

Suppose a finite-dimensional continuous irreducible group representation were not isomorphic to any $V_n$. Character orthogonality would make its character orthogonal to every ${\chi }_n$. Completeness forces its character to vanish in $L^2$, and continuity then forces it to vanish everywhere. This is impossible at the identity, where the character equals the positive dimension of the representation. Hence $\boxed{\text{every such irreducible representation is some }{V}_n}$. Since the dimensions $n+1$ differ, the index is unique.

The $n+1$ monomials $z_1^{n-j}{z}_2^j$ form a basis, so $\boxed{\dim V_n=n+1}$. For irreducibility, use the following compact-group facts: every finite-dimensional continuous representation is unitarizable by averaging an inner product; such representations decompose into irreducibles; their irreducible characters are orthonormal under normalized Haar measure. In particular the squared character norm is the sum of squares of the irreducible multiplicities.

The Weyl integration formula for SU2 gives, for a class function $F$,Using the character already found,Thus exactly one irreducible constituent occurs with multiplicity one: $V_n$ is irreducible for every $n\ge 0$.

The first assertion is false. The complete graph $K_4$ has $\chi (K_4)=4$, but ${\chi }^{\prime }(K_4)=3$: color each of its three perfect matchings with its own color. It is connected and has more than two vertices.

The second assertion is false. Take the crown graph with bipartition $a_1,\dots ,a_5$ and $b_1,\dots ,b_5$, joining $a_i$ to $b_j$ exactly when $i\ne j$. Its chromatic number is two. In the ordering $a_1,b_1,a_2,b_2,\dots ,a_5,b_5$, the greedy coloring algorithm assigns color $i$ to both $a_i,b_i$: both see all previously used colors, but they are not adjacent to each other. It therefore uses five colors, exceeding $2\chi (G)=4$. Taking more pairs makes the ratio arbitrarily large.

The third assertion is true. When an edge is colored greedily, at most $2\Delta -2$ adjacent edges have previously received colors. Thus the greedy coloring algorithm uses at most $2\Delta -1\le 2{\chi }^{\prime }(G)-1<2{\chi }^{\prime }(G)$ colors whenever the graph has an edge. The edgeless case is trivial. This bound holds for every edge ordering.

The Vizing theorem states that for every finite simple graph of maximum vertex degree $\Delta$, its edge chromatic number satisfies $\boxed{\Delta \le {\chi }^{\prime }(G)\le \Delta +1}$. The lower bound follows because all edges incident to a maximum-degree vertex need different colors. The simplicity hypothesis matters: unrestricted multigraphs need not satisfy this upper bound.

The Brooks' theorem states that a finite connected simple graph with maximum vertex degree $\Delta$ has chromatic number at most $\Delta$, unless it is a complete graph or an odd cycle graph; those exceptions have chromatic number $\Delta +1$.

First, if the graph is not regular, choose a vertex $v$ of degree less than $\Delta$ and a spanning tree rooted at $v$. Order vertices with every child before its parent, and apply the greedy coloring algorithm. Every vertex except $v$ has an as-yet uncolored parent, so has at most $\Delta -1$ colored neighbors; $v$ also has fewer than $\Delta$ neighbors. Thus $\Delta$ colors suffice. This justifies reducing to the regular case.

Now let $G$ be regular, 3-connected and not complete. A shortest path between two nonadjacent vertices contains an induced two-edge path $a,v,b$ with $a,b$ nonadjacent. By 3-connectivity, $G-\{a,b\}$ is connected. Color $a,b$ with the same color. Choose a spanning tree of the remaining graph rooted at $v$ and color its vertices in child-before-parent order. Each vertex other than $v$ again has an uncolored parent and thus at most $\Delta -1$ already colored neighbors. At the last vertex $v$, the two neighbors $a,b$ share a color, so its $\Delta$ neighbors use at most $\Delta -1$ distinct colors. A color is therefore available at every step, proving $\boxed{\chi (G)\le \Delta }$ in the required 3-connected noncomplete case. A 3-connected graph has $\Delta \ge 3$, so the odd-cycle exception cannot arise here.

For $P=\beta {\rho }^{4/3}$, the sound speed obeys $c_s^2=(4\beta /3){\bar{\rho }}^{1/3}$. Using $a=(t/t_0{)}^{2/3}$ and $\bar{\rho }=(6\pi G{t}^2{)}^{-1}$ givesThe mode equation is an Euler differential equation. Substituting ${\delta }_k=t^n$ produces $n(n-1)+(4/3)n-2/3+{\bar{v}}_s^2{k}^2=0$. Thus, for distinct roots,A genuinely growing mode occurs only for ${\bar{v}}_s^2{k}^2<2/3$; at $k=0$ the exponents are $2/3$ and $-1$. At ${\bar{v}}_s^2{k}^2=2/3$ the larger root is zero. Between $2/3$ and $25/36$, both roots are negative. At the repeated-root value ${\bar{v}}_s^2{k}^2=25/36$, the independent solutions are $t^{-1/6}$ and $t^{-1/6}\log t$. Beyond it, write $\nu =\sqrt{{\bar{v}}_s^2{k}^2-25/36}$ to obtain $t^{-1/6}[A\cos (\nu \log t)+B\sin (\nu \log t)]$. Thus the phrase “growing and decaying” applies to the long-wavelength regime; pressure-stabilized modes decay instead.