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Put $b=\mathrm{Im}\tau >0$. On $|\mathrm{Im}z|\le R$, the modulus of the $n$th summand is at mostThe resulting Gaussian majorant is summable, independently of the unbounded real part of $z$. The Weierstrass test proves uniform convergence on every such strip. Since the summands are entire, locally uniform convergence implies $\psi$ is entire. It is not identically zero: integrating $\psi (x)e^{-\pi ix}$ over $0\le x\le 1$ extracts its nonzero $n=0$ Fourier coefficient.

Logarithmic differentiation of the two supplied transformation laws givesThese hold as meromorphic identities, with poles at zeros of $\psi$. For a translated fundamental parallelogram with no boundary zeros, opposite edges parallel to $\tau$ cancel by the first identity. The other edge pair contributes $2\pi i$ by the second, so the argument principle counts exactly one zero, including multiplicity, in that parallelogram.

Changing the summation index $n\mapsto -n-1$ shows $\psi (-z)=-\psi (z)$, hence $\psi (0)=0$. Quasiperiodicity makes every lattice point a zero. Every generic fundamental parallelogram contains exactly one lattice point, which already accounts for its one zero; varying its position rules out any additional zeros and proves the lattice zeros are simple. ThereforeIn particular the centred parallelogram in the question has no lattice points on its boundary and its unique zero is zero. Using generic translated contours first avoids assuming this boundary fact before determining the zeros.

For the proposed sum, the first period always holds, while $f(z+\tau )-f(z)=-2\pi i{\sum }_j{\lambda }_j$. Thus it is $\Lambda$-periodic exactly when ${\sum }_j{\lambda }_j=0$. Meromorphicity follows from the meromorphic logarithmic derivatives. Finally differentiation removes the additive shift in $\ell$, so $\frac{d}{dz}[{\psi }^{\prime }(z-a)/\psi (z-a)]$ has both periods for every $a$.

Let $B_X$ be the open unit ball. Since $T(X)={\bigcup }_{n\ge 1}T(n{B}_X)$ is not meagre in $Y$, at least one closure $\overline{T(n{B}_X)}$ has nonempty interior. A ball contained in that closure, subtracted from itself, gives a ball about zero contained in $\overline{T(2n{B}_X)}$. Scaling yields some $\delta >0$ withNo completeness of $Y$ is used here; the hypothesis is second category in $Y$, not merely in the relative image topology.

If $\Vert y\Vert <\delta /2$, choose $x_1\in \frac{1}{2}{B}_X$ with $\Vert y-T{x}_1\Vert <\delta /4$. Then choose $x_2\in \frac{1}{4}{B}_X$ so the new residual is below $\delta /8$, and continue. The series ${\sum }_j{x}_j$ converges in the Banach space $X$, has norm less than one, and continuity of $T$ gives $T({\sum }_j{x}_j)=y$. Thus $B_Y(0,\delta /2)\subseteq T(B_X)$.

The linear subspace $T(X)$ contains a ball about zero, so it is all of $Y$: $T$ is surjective. Translated and scaled unit-ball images show that every open-set image is open. If $T$ is injective, its inverse is linear and the ball inclusion gives $\Vert T^{-1}y\Vert \le (2/\delta )\Vert y\Vert$, by scaling and taking a limiting bound. Thus $T^{-1}$ exists and is bounded.

For the first nonlinear counterexample use $f(x)=e^x$: its image $(0,\infty )$ is of second category but is not all of $\mathbb{R}$. For the second use the continuous surjection $f(x)=x+1$ for $x\le -1$, $f(x)=0$ for $-1\le x\le 1$, and $f(x)=x-1$ for $x\ge 1$. The open interval $(-1/2,1/2)$ maps to the nonopen singleton $\{0\}$. Linearity is essential to the conclusions.

Use simplicial homology with a fixed coefficient group or ring. The Mayer–Vietoris sequence isThe inclusion maps define $\alpha ([z])=([z],-[z])$ and $\beta ([a],[b])=[a+b]$. To define $\delta$, split a simplicial cycle $c\in C_{i+1}(X)$ as $c=a+b$ with $a$ a chain in $A$ and $b$ in $B$. Since $\partial a=-\partial b$, this boundary lies in $C_i(A\cap B)$, and $\delta [c]=[\partial a]$. At degree zero the unreduced sequence ends with $H_0(A)\oplus H_0(B)\to H_0(X)\to 0$.

If $[z]=\delta [c]$, its representative is a boundary in $A$ and the negative of a boundary in $B$, so $\alpha [z]=0$. Thus $\mathrm{im}\delta \subseteq \ker \alpha$.

Conversely if $\alpha [z]=0$, choose $a\in C_{i+1}(A)$ and $b\in C_{i+1}(B)$ with $\partial a=z=\partial b$. The chain $c=a-b$ is a simplicial cycle in $X$. Its splitting into the $A$ chain $a$ and $B$ chain $-b$ gives $\delta [c]=[z]$. Hence $\mathrm{im}\delta =\ker \alpha$, the requested exactness at $H_i(A\cap B)$.

Write $p=8k+7$. In Gauss's lemma, for the multiplier 2, precisely the integers $j$ with $p/4<j\le (p-1)/2$ give residues $2j>p/2$. Their number isIt is even, so the Legendre symbol $(2/p)=1$: 2 is a quadratic residue modulo $p$. For completeness, the sign rule follows by replacing those large residues by their negatives; the resulting absolute residues permute $1,\dots ,(p-1)/2$. Multiplying them and cancelling their nonzero product gives $2^{(p-1)/2}\equiv (-1{)}^{2k+2}=1(\mathrm{mod}p)$, equivalent to the residue assertion by Euler's criterion.

Now take the prime $p=2q+1$. The condition $q\equiv 3(\mathrm{mod}4)$ gives $p\equiv 7(\mathrm{mod}8)$, so $2^q\equiv 1(\mathrm{mod}p)$. Thus $p$ divides $2^q-1$. For integers $q>3$, $2^q-1>2q+1$ (true at 4 and preserved on increasing $q$), so this divisor is proper and $2^q-1$ is composite. At $q=3$, the divisor is the number itself: $2^3-1=2\cdot 3+1=7$. The congruence still holds, but it does not establish compositeness.

For a complex irreducible representation, every central element acts as a scalar by Schur lemma. If the representation is faithful, these scalars embed $Z$ in ${\mathbb{C}}^{\ast }$. Every finite subgroup of ${\mathbb{C}}^{\ast }$ is cyclic: if its order is $m$, all its elements satisfy $z^m=1$, and the $m$th roots of unity form a cyclic group. Thus $Z$ is cyclic.

The trivial group satisfies the conclusion immediately. Otherwise let $G$ be a finite $p$-group with cyclic centre, and decompose a faithful representation as a direct sum of irreducibles using Maschke's theorem. The nontrivial cyclic $p$-group $Z$ has a unique subgroup $C$ of order $p$. Some component does not kill $C$, since otherwise the whole representation would kill it. If that component had a nontrivial kernel $N$, the kernel would be a normal subgroup and the allowed intersection theorem gives $N\cap Z\ne 1$. Every nontrivial subgroup of cyclic $Z$ contains $C$, contradicting the choice of component. That irreducible component is faithful.

Finally the regular representation is faithful, since left multiplication by a nonidentity element does not fix the basis vector at the identity. A cyclic centre therefore supplies a faithful irreducible component of it. The first argument proves the converse, giving a finite $p$-group has a faithful irreducible complex representation exactly when its centre is cyclic.

An automorphism sends the primitive root of unity ${\xi }_n$ to another primitive root of unity, hence uniquely to ${\xi }_n^a$ with $a\in (\mathbb{Z}/n\mathbb{Z}{)}^{\ast }$. Define $\chi (\sigma )=a$. Composition multiplies exponents, and an automorphism with exponent one fixes the generator and thus all of $L$. Therefore $\chi$ is an injective group homomorphism and $G$ is abelian.

The extension is finite Galois: it is the splitting field of $X^n-1$, whose roots are all present and distinct. Existence of a root of order $n$ in positive characteristic implies that the characteristic does not divide $n$. The Fundamental theorem of Galois theory identifies an intermediate field with $L^H$. Since $G$ is abelian, $H$ is normal; the normal-subgroup criterion gives $L^H/K$ Galois, with group $G/H$.

For a non-surjective nontrivial example, take $K=\mathbb{Q}(i)$ and $n=8$. Its cyclotomic extension has degree two, and its image is $\{1,5\}$ inside the four-element group $(\mathbb{Z}/8\mathbb{Z}{)}^{\ast }$. For $K=\mathbb{Q}$ and prime $p$, ${\Phi }_p(X)=1+\cdots +X^{p-1}$ is irreducible: ${\Phi }_p(X+1)$ is Eisenstein at $p$. Thus the extension has degree $p-1$, equal to the unit group's order, proving surjectivity.

For $n=7$, let $\zeta ={\xi }_7$. The cyclic Galois group of order six has exactly one subgroup of each order dividing six, so there are exactly four fields:Here ${\sigma }_a(\zeta )={\zeta }^a$. To identify the quadratic field without guessing, put $A=\zeta +{\zeta }^2+{\zeta }^4$ and $B={\zeta }^3+{\zeta }^5+{\zeta }^6$. Multiplication gives $A+B=-1$ and $AB=2$, so $(A-B{)}^2=-7$. The order-three subgroup fixes $A$, and $A-B\notin \mathbb{Q}$, proving the quadratic identification. The real field is cubic because complex conjugation is ${\sigma }_6$, with fixed subgroup of order two.

Let $R(s,t)$ be the two-colour clique Ramsey number. At a vertex of $K_{R(s-1,t)+R(s,t-1)}$, either its red neighbourhood has at least $R(s-1,t)$ vertices or its blue neighbourhood has at least $R(s,t-1)$. In the first case a red $(s-1)$-clique extends through the vertex, or there is a blue $t$-clique; the second case is symmetric. HenceWith $R(1,t)=R(s,1)=1$, induction and Pascal's identity give $R(s,t)\le \left(\genfrac{}{}{0ex}{}{s+t-2}{s-1}\right)$. Taking $s=t$ provesThe four-vertex graph is the paw graph, a triangle with a pendant edge; the copy need not be induced. For the lower bound, colour two disjoint triangles in $K_6$ red and all edges between them blue. Red components have only three vertices and blue is bipartite, so neither colour contains a paw. Thus $R(H)\ge 7$.

For the upper bound, $R(3,3)\le 6$ guarantees a monochromatic triangle in $K_7$, say red on a vertex set $T$. A red edge from $T$ to any outside vertex immediately supplies a red paw. Otherwise all edges from $T$ to the remaining four vertices are blue. If those four vertices contain a blue edge $ab$, then $a,b$ and any vertex of $T$ form a blue triangle, and another vertex of $T$ joined to $a$ supplies its pendant edge. If they contain no blue edge, they are a red $K_4$, which contains a red paw. Therefore

A deduction from premises $\Gamma$ in propositional logic is a finite sequence of formulas, each an allowed logical axiom instance, a member of $\Gamma$, or obtained from earlier formulas by an inference rule, such as modus ponens. For example a Hilbert system can use the schemas $p\to (q\to p)$, $[p\to (q\to r)]\to [(p\to q)\to (p\to r)]$, and $(\neg q\to \neg p)\to (p\to q)$, with modus ponens as its rule. Its last formula is the conclusion. A finite deduction uses only finitely many premises. The completeness theorem for propositional logic says $\Gamma \models v$ implies $\Gamma ⊢v$; together with soundness, syntactic consistency is equivalent to existence of a satisfying valuation.

For compactness, if $\Gamma$ had no model, completeness would give a deduction of a contradiction. That deduction uses a finite subset of $\Gamma$, which therefore also has no model. Contrapositively, if every finite subset has a model, so does $\Gamma$.

Fix the given valuation on $Q$ and let $L_v$ be its collection of true literals. Suppose $\{s\}\cup L_v$ were inconsistent. A finite subset of its literals, with conjunction $\ell \in L(Q)$, would satisfy $s⊢\neg \ell$ by the deduction theorem and classical logic. But $\neg \ell \in U$ whereas the given valuation makes $\ell$ true, contradicting that it satisfies all of $U$. Hence $\{s\}\cup L_v$ is consistent.

If $U\cup \{\neg t\}$ had a model, restrict that valuation to $Q$ and apply the preceding consistency result. Completeness supplies an extension to $P\cup Q$ making $s$ true and preserving those $Q$ values. Combine its $P$ values with the original model's $Q\cup R$ values. Disjointness of $P,Q,R$ makes this a well-defined valuation satisfying $s$ and $\neg t$, contradicting soundness of $s⊢t$. Thus $U\cup \{\neg t\}$ is inconsistent.

Compactness now supplies $u_1,\dots ,u_k\in U$ with $u_1,\dots ,u_k⊢t$. Put $u={\bigwedge }_j{u}_j$. ThenThis proves propositional interpolation. The empty conjunction is the truth constant; if the language omits truth and falsehood symbols, use an equivalent tautology when $Q$ is nonempty, or adjoin those constants for the empty-common-language case.

For the Laplace-type contour ansatz, differentiating under the integral and putting $P(t)=t^2-t-2$ gives the differential-equation residualIntegration by parts turns this into $[e^{zt}Pf{]}_{\partial \gamma }+{\int }_{\gamma }{e}^{zt}[(2-t)f-(Pf{)}^{\prime }]dt$. Thus it vanishes whenever the endpoint contribution vanishes and $(Pf{)}^{\prime }=(2-t)f$. The latter condition is $P{f}^{\prime }=3(1-t)f$, solved byup to a constant multiplier. Closed contours avoiding the poles make the endpoint contribution zero.

Choose separate positively oriented small circles around 2 and $-1$. The residue theorem gives, after division by $2\pi i$,ThereforeThe Wronskian of the two basis functions is $-9z{e}^z$, nonzero for $z\ne 0$. They span the two-dimensional solution space on every regular domain, and both extend entire across the apparent singular point at zero. Their initial values satisfy $w(0)=A+B$ and $w^{\prime }(0)=2(A+B)$. Thus every solution has $w^{\prime }(0)=2w(0)$, and the requested combination $w(0)=0$, $w^{\prime }(0)\ne 0$ is impossible. No additional logarithmic solution is lost: the two independent functions already span the solution space away from zero.

The diffusive flux is $\mathbf{J}=-D\nabla C$ by Fick's first law. Conservation in any fixed control volume $V$ gives $\frac{d}{dt}{\int }_{V}C=-{\int }_{\partial V}\mathbf{J}\cdot \mathbf{n}+{\int }_{V}F$. The divergence theorem and arbitrariness of $V$ yieldAfter the initial point injection, $F=0$ and $D=kC$ with $k>0$, so radial symmetry gives $C_t=k{r}^{-2}(r^{2}C{C}_r{)}_r$. Total mass is $4\pi {\int }_0^{\infty }{r}^{2}Cdr=4\pi M$.

A spreading radius $R$ and typical concentration obey $C{R}^3\sim M$ and $R^2\sim kCt$. Thus $R^5\sim Mkt$. In the proposed ansatz this givesDirect substitution, including the time derivative of $\xi$, givesMultiplying by ${\xi }^2$ makes the left side $-({\xi }^{3}f{)}^{\prime }/5$. Regularity and zero radial flux at the origin set the integration constant to zero, so $f{f}^{\prime }=-\xi f/5$. On the positive region $f^{\prime }=-\xi /5$. Nonnegativity, compact support and continuity at its edge giveIts normalization is $1={\int }_0^{{\xi }_0}{\xi }^{2}fd\xi ={\xi }_0^5/75$. Therefore the spherical nonlinear-diffusion source profile isThe concentration and flux vanish at the free boundary, so there is no distributional flux jump despite the derivative kink. The solution satisfies the diffusion equation weakly across that boundary and classically inside its positive region. Its support shrinks to the origin as $t\downarrow 0$, and testing against a continuous compactly supported function gives the limit $4\pi M$ times its value at zero. This verifies the prescribed point injection as well as the mass normalization.