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A renewal process counts arrivals $N(t)=\max \{j:A_j\le t\}$, where $A_0=0,A_j=S_1+\cdots +S_j$ and the holding times are independent, identically distributed positive random variables. Their sum tends to infinity almost surely, so the count is finite on bounded time intervals. Use $N(s)=X_s$ in the question's notation. The index $J=N(s)+1=\min \{j:A_j>s\}$ is a stopping index for the holding-time sequence. Conditional on $J=j$ and the first $j$ holding times, every later $S_{j+k}$ retains its original probability distribution. Summing over $j$ provesThe straddling time $T_1=S_J$ is different. Let $\overline{F}(t)=P(S_1>t)$ and let $U$ be the renewal measure $U(B)={\sum }_{j\ge 0}P(A_j\in B)$. Independence of the next holding time from its start givesFor two numbers in $[0,1]$, their minimum is at least their product. Since $\overline{F}(\max (t,a))=\min (\overline{F}(t),\overline{F}(a))$, subtracting $\overline{F}(t)$ times the second integral provesThese formulae handle holding-time atoms with the stated renewal convention $A_j\le s<A_{j+1}$; sums may be interchanged by nonnegativity.

For rate-$\lambda$ exponentials, the renewal process is Poisson. The elapsed age $B_s=s-A_{N(s)}$ has probability distribution $\min (E,s)$ with $E\sim \mathrm{Exp}(\lambda )$: for $0\le b<s$, $P(B_s>b)$ is the probability of no arrivals in $(s-b,s]$, namely $e^{-\lambda b}$, with an atom $e^{-\lambda s}$ at $s$. The forward residual $R_s$ is an independent rate-$\lambda$ exponential by independent future increments. Thus $T_1=B_s+R_s$. Conditioning on the age gives the exact survival lawFor $s,t>0$ the extra term is strictly positive, proving strict inequality. For fixed $t$ and $s\to \infty$, the limit is $e^{-\lambda t}(1+\lambda t)$, which equals the survival probability of the sum of two independent rate-$\lambda$ exponentials. This is the finite-time form of the stochastic length bias of a renewal interval.

A family $\mathcal{R}$ is uniformly integrable ifThis implies integrability and a uniform bound on $E|X|$, since $E|X|\le K+E[|X|1_{\{|X|>K\}}]$. Convergence in probability means $P(|X_n-X|>\epsilon )\to 0$ for every $\epsilon >0$; convergence in $L^1$ means $E|X_n-X|\to 0$. The latter implies the former by Markov's inequality. Conversely, convergence in probability together with uniform integrability of $\{X_n\}$ implies that $X$ is integrable and $X_n\to X$ in $L^1$. Moreover an $L^1$-convergent sequence is uniformly integrable. This is the criterion that excludes rare very large values from carrying a nonvanishing first-moment error.

For the sum family, use the pointwise boundIndeed, if the left indicator is one then the larger of $|X|,|Y|$ exceeds $K/2$, and twice that larger number bounds $|X+Y|$. Taking expectations and the supremum over ${\mathcal{R}}_1,{\mathcal{R}}_2$ bounds the sum-family tail by the sum of their two uniformly vanishing tails. Thus the whole family of sums is uniformly integrable, with no independence assumption.

The plane planar isoperimetric inequality says that a simple closed curve of length $L$ enclosing area $A$ satisfies $L^2\ge 4\pi A$, with equality for circles.

For sufficiently small $r$, below the injectivity radius at $p$, geodesic polar coordinates give the surface metric $d{r}^2+j(r,\theta {)}^{2}d{\theta }^2$. The radial variation field is a Jacobi field, soIts integral equation is $j=r-{\int }_0^r(r-u)K({\gamma }_{\theta }(u))j(u,\theta )du$. Continuity of curvature and $j(u,\theta )=u+o(u)$, uniformly in direction, giveThe Gauss lemma makes the angular metric factor both the circle line element and the polar area factor. ThereforeSquaring the length expansion and subtracting givesThis identifies the requested remainder and proves its ratio to $r^4$ tends to zero. For $K(p)>0$, sufficiently small geodesic disks have $L^2<4\pi A$, the opposite strict inequality to the planar bound. Their positive curvature allows more area for the same perimeter; the Euclidean inequality is not a universal inequality for curved surfaces.

A function element is a pair $(f,D)$ where $D$ is a nonempty disk and $f$ is holomorphic there; the germ at its center captures its local information. Elements continue one another when their functions agree on an overlapping connected neighborhood. A complete analytic function is a maximal collection reachable from one element by chains of analytic continuation, including all further continuations of its members.

If $f^n=p$ in the initial disk, the identity theorem propagates this identity across every overlapping disk in a continuation chain. Induction along the chain proves every element of the complete analytic function also satisfies ${\tilde{f}}^n=p$.

Put $F(z,w)=w^n-p(z)$. Its gradient is $(-p^{\prime }(z),n{w}^{n-1})$. At a point of the curve with $w\ne 0$, the second entry is nonzero; at $w=0$, the root of $p$ is simple and the first entry is nonzero. Thus the holomorphic implicit function theorem supplies charts. Its precise local statement is that a holomorphic equation with nonzero derivative in one variable can locally be solved uniquely and holomorphically for that variable as a function of the other. For $w\ne 0$, use $z$ as coordinate and write $w=w(z)$; at a simple root use $w$ as coordinate and write $z=z(w)$. The transition maps are holomorphic with nonzero derivatives, giving a Riemann surface. The two projections are holomorphic because they are holomorphic expressions in either kind of chart.

Away from the zeros of $p$, the projection onto $z$ is an unramified $n$-sheeted covering. A germ of a root corresponds to $(z,f(z))$ and its analytic continuations follow this covering. Near a simple root, $z-z_0$ is a nonzero analytic factor times $w^n$; a circuit about $z_0$ cyclically permutes all $n$ branches. For nonconstant square-free $p$, these circuits connect all sheets. Hence the germ surface $S(F)$ is the curve with the points over the roots removed, and $C$ is its branched completion obtained by adding those ramification points. An ordinary function element in the variable $z$ cannot include such a point: the zero order of a holomorphic $n$th power is divisible by $n$, whereas $p$ has zero order one. If a convention already includes these branch points in the Riemann surface of a multivalued function, that completed surface is $C$ itself. For a nonzero constant $p$, the curve has $n$ separate components, and the chosen element determines just one of them. The identically zero polynomial is excluded by the non-singularity premise.

A linear operator between Banach spaces is a compact operator when it maps the unit ball into a relatively compact set, equivalently when every bounded sequence of inputs has an image subsequence converging in norm. Such an operator is bounded. For a bounded linear operator on a complex Banach space, the resolvent set consists of the complex $\lambda$ for which $T-\lambda I$ is bijective with bounded inverse; its complement is the spectrum. The point spectrum consists of those $\lambda$ for which $T-\lambda I$ has a nonzero kernel, namely the eigenvalues.

On a Hilbert space, $T$ is self-adjoint if $T=T^{\ast }$, or equivalently $⟨Tx,y⟩=⟨x,Ty⟩$ for all vectors. The given prescription extends uniquely toIt is bounded with norm one. The real diagonal entries prove self-adjointness by the inner-product identity. Define the finite-rank truncation $T_{m}x={\sum }_{j\le m}{x}_j{e}_j/j$. Then $\Vert T-T_m\Vert =1/(m+1)\to 0$. To see compactness directly, extract successively convergent subsequences of the finitely many first coordinates from a bounded input sequence; the uniform tail bound just displayed makes the resulting image subsequence Cauchy. Hence $T$ is compact and self-adjoint.

Each $1/j$ is an eigenvalue with eigenspace $\mathbb{C}{e}_j$. If $Tx=\lambda x$, each coordinate satisfies $(1/j-\lambda )x_j=0$, so there are no other eigenvalues; in particular zero is not an eigenvalue. But $\Vert T{e}_j\Vert =1/j\to 0$, which contradicts a bounded inverse at zero. Thus zero is in the spectrum.

For $\lambda \notin \{0,1,1/2,1/3,\dots \}$, its distance from this closed set is positive. The diagonal formulathen defines a bounded inverse. This proves the complete description

The Seifert-van Kampen theorem states that if a space is covered by path-connected open sets $U,V$ with path-connected intersection containing a basepoint, then its fundamental group is the free product ${\pi }_1(U)\ast {\pi }_1(V)$ modulo the normal subgroup generated by identifying the two images of each loop in $U\cap V$. More generally it is the corresponding pushout of groups.

A wedge of two circles has fundamental group $F(a,b)$, the free group on their two loops: apply the theorem to thickened circle neighborhoods whose intersection is contractible and use ${\pi }_1(S^1)=\mathbb{Z}$. Represent the torus by a square with opposite edges identified. Remove an interior disk; the remainder deformation retracts onto this wedge. Replacing the disk attaches a two-cell whose boundary follows $ab{a}^{-1}{b}^{-1}$. A disk neighborhood and the punctured torus have annular intersection, and van Kampen kills precisely that boundary loop. HenceA torus with a point removed retracts to a torus with a small open disk removed, and then to the same wedge. ThusFor the connected sum, remove a disk from each torus and glue their boundary circles with reversed orientations. The two pieces have free groups $F(a,b)$ and $F(c,d)$ and their overlap is an annulus. Its boundary generators correspond to $[a,b]$ and $[c,d{]}^{-1}$, where $[a,b]=ab{a}^{-1}{b}^{-1}$. ThereforeThe boundary relation is essential: this group is neither the free group on four generators nor the abelian group ${\mathbb{Z}}^4$.

The ring of integers is ${\mathcal{O}}_k=\mathbb{Z}[\sqrt{7}]$, since $7\equiv 3(\mathrm{mod}4)$, and its discriminant is $28$. Rationalizing givesBoth are algebraic integers, so $\epsilon$ is a unit. Set $\mathfrak{p}=(3+\sqrt{7})$. Since $(3+\sqrt{7}{)}^2=2\epsilon$,Also $N(\mathfrak{p})=|N(3+\sqrt{7})|=2$. It is the unique ideal of ideal norm two: any quotient of size two is ${\mathbb{F}}_2$, and the image of $\sqrt{7}$ must satisfy $z^2=1$, hence be one. The kernel of a ring homomorphism onto ${\mathbb{F}}_2$ is uniquely $(2,1+\sqrt{7})$, which is $\mathfrak{p}$.

The given Minkowski constant means every ideal class has an integral representative of ideal norm at most $\frac{1}{2}\sqrt{28}=\sqrt{7}<3$. Such a ideal norm is one or two. The norm-one ideal is the whole ring, and the sole norm-two ideal is principal. Thus the class number is one.

If $x^2-7y^2=2$, the principal ideal $(x+y\sqrt{7})$ has ideal norm two and hence equals $\mathfrak{p}$. Its generator differs from $3+\sqrt{7}$ by a unit. Conversely multiplication by a norm-one unit preserves the field norm two. With the assumed fundamental unit, all units are $\pm {\epsilon }^n$; in fact a norm-minus-one unit is impossible since $x^2\equiv -1(\mathrm{mod}7)$ has no solution. Therefore the complete answer isFor $n=1$ and the positive sign, the product is $45+17\sqrt{7}$, giving $(x,y)=(45,17)$; $45^2-7\cdot 17^2=2$ checks the result.

Let $g$ generate the cyclic group of nonzero residues, whose order $p-1=2^{2^n}$ is a power of two. Write $a=g^j$. Its multiplicative order is $(p-1)/\gcd (j,p-1)$, so it is a primitive root precisely when $j$ is odd. The quadratic residues are exactly the even powers of $g$. Thus the primitive roots are precisely the quadratic nonresidues, with nonzero residue classes understood.

For $n\ge 1$, $p\equiv 1(\mathrm{mod}4)$, so quadratic reciprocity gives $(7/p)=(p/7)$. Since $2^3\equiv 1(\mathrm{mod}7)$ and $2^n$ alternates between $1$ and $2$ modulo three, $p$ is respectively $3$ or $5$ modulo seven. The nonzero squares modulo seven are $1,2,4$. Consequently $(7/p)=-1$, andThere is a genuine endpoint exception if the printed form permits $n=0$: then $p=3$ and $7\equiv 1(\mathrm{mod}3)$, which has order one and is not primitive. The first equivalence remains true for $p=3$, but the final assertion requires excluding that prime.

First suppose $G$ is 3-connected and noncomplete, and write $\Delta =\Delta (G)$. There is an induced path $a,v,b$ with $a,b$ nonadjacent: otherwise the first three vertices of any shortest path between nonadjacent vertices would contradict that assertion. Since $G$ is 3-connected, $G-\{a,b\}$ is a connected graph. Choose a spanning tree of that graph rooted at $v$. Color $a,b$ first with the same color, then color the other vertices in reverse tree order, leaving $v$ last. Every vertex except $v$ has its tree parent uncolored and thus at most $\Delta -1$ previously colored neighbors. At $v$, the two neighbors $a,b$ share a color, so again at most $\Delta -1$ colors are forbidden. This constructs a proper $\Delta$-coloring and proves Brooks' bound for 3-connected noncomplete graphs. A complete graph is exactly the excluded case; a 3-connected graph cannot be an odd cycle.

For the partition claim, assume $d_1,d_2$ are nonnegative integers with $d_1+d_2=\Delta -1$. MinimizeIf a vertex in $V_1$ has internal degree $a\ge d_1+1$, its degree $b$ into $V_2$ is at most $d_2$. Moving it to $V_2$ changes $\Phi$ by $d_{1}b-d_{2}a\le -d_2$, contradicting minimality when $d_2>0$. The analogous move handles a violation in $V_2$ when $d_1>0$. For a zero coefficient, break ties among minimizers by minimizing the total number of internal edges. If $d_2=0<d_1$, minimality of $\Phi =d_{1}e(G[V_2])$ ensures $V_2$ is independent; a violating vertex in $V_1$ has all its $\Delta =d_1+1$ neighbors in $V_1$ and no neighbor in $V_2$, so moving it reduces the tie-break quantity without increasing $\Phi$. The reversed argument handles $d_1=0$. If both are zero then $\Delta =1$, and the graph is a union of isolated vertices and edges, admitting two independent parts. ThusFor $\Delta \ge 5$, put $d_1=3,d_2=\Delta -4$. A maximum-degree-three graph containing no $K_4$ is 3-colorable. Here is why the previously proved 3-connected case suffices: connected graphs of degree at most two are paths or cycles and use at most three colors. Split a larger graph at cut vertices and permute colors when gluing. In a 2-connected cubic graph with a two-vertex cut $\{u,v\}$, a cut edge $uv$ gives colorings of the sides glued with their distinct endpoint colors. If $u,v$ are nonadjacent, the cut components have positive numbers of edges to both endpoints and the total endpoint degrees are at most three. There are at most three components. With three components, adding $uv$ to each side preserves maximum degree three and gives the endpoints degree two. No new $K_4$ is created, so induction supplies side colorings with distinct endpoint colors, which can be permuted to agree. With two components, the components incident twice to $u$ and twice to $v$ either agree or differ. If they differ, adding $uv$ to each side preserves maximum degree three; each augmented side is smaller and cannot be $K_4$, because one endpoint then has degree only two, so induction gives distinct-endpoint colorings. If they agree, call the side with only one edge from each endpoint the small side. Any coloring of the other side can be matched on it: for equal endpoint colors, contract $u,v$ in the small side (the merged vertex has degree at most two); for distinct colors add $uv$ (the endpoints again have degree two). Induction applies in both cases. A 2-connected graph with no two-vertex cut is 3-connected and was handled above. This proves the needed subcubic coloring fact without quoting the unproved full theorem.

Use three colors on $V_1$ and at most $d_2+1=\Delta -3$ fresh colors on $V_2$, by greedy coloring. This gives $\chi (G)\le \Delta$. More usefully, repeatedly remove degree-at-most-three parts: the partition supplies a four-degree reduction for a cost of three colors. If $f(\Delta )=\frac{3}{4}\Delta +\frac{3}{2}$, then $3+f(\Delta -4)=f(\Delta )$. Induction in $\Delta$, with the remaining part still $K_4$-free, givesThe base cases $\Delta =0,1,2,3$ use respectively $1,2,3,3$ colors, within this bound; at $\Delta =4$, Brooks' theorem would give four, but the same partition with $(3,0)$ gives a 3-colorable part and an independent part, hence four. Thus the induction requires no unproved general Brooks theorem.