For a uniform contraction from a comparison function on a bounded set, the closed set $M$ is complete and has diameter at most $2R$. Suppose first $R>0$. The limit condition supplies $r_0>0$ and $q_0<1$ such that $\psi (r)/r\le q_0$ for $0<r\le r_0$. On the remaining compact interval $[r_0,2R]$, if nonempty, the continuous ratio $\psi (r)/r$ attains a maximum $q_1<1$, since it is strictly below one at every point. Thus there is one $q<1$ with $\psi (r)\le qr$ for every distance $0\le r\le 2R$ arising in $M$.

Consequently $d(Tx,Ty)\le qd(x,y)$ throughout $M$. Banach's theorem applies and gives a unique fixed point in $M$. No linear structure or compactness of the ball is used; the compact object in the argument is only the scalar distance interval. If the closed ball is the singleton at radius zero, its self-map fixes that point. The estimate on the ratio near zero is what prevents the compact-interval supremum argument from approaching one at its missing endpoint.

A weak solution belongs to $V=H_0^1(\Omega )$ and satisfieswith the nonlinear right-hand side defining a bounded functional on $V$. Under the stated global Lipschitz hypothesis this requirement is automatic: $|f(s)|\le |f(0)|+ϵ|s|$ and boundedness of $\Omega$ give $f(u)\in L^2$ whenever $u\in V$. Continuity of $f$ alone, without growth control, would not guarantee this for an arbitrary Sobolev function.

For the semilinear Poisson contraction threshold, use the gradient norm on $V$, and let $C_P$ satisfy the Poincaré inequality $\Vert v{\Vert }_2\le C_P\Vert \nabla v{\Vert }_2$. For $g\in L^2$, Lax-Milgram gives the zero-Dirichlet Poisson solution $Pg\in V$. Testing with $Pg$ yields $\Vert \nabla Pg{\Vert }_2\le C_P\Vert g{\Vert }_2$. Define $T(v)=P(f(v))$, a self-map of the complete space $V$. For $v,w\in V$,Therefore $ϵC_P^2<1$ is a sufficient condition for Banach's theorem to give a unique fixed point and hence a unique weak solution. There is no need to assume $f(0)=0$ or to restrict to a small ball: the Lipschitz bound makes the map defined on all of $V$. The same estimates giveWith the optimal Poincaré constant, the threshold is $ϵ<{\lambda }_1$ for the First Dirichlet eigenvalue; the stated sufficient condition does not depend on knowing that optimal constant.

Banach's fixed-point theorem states that a map $T:Y\to Y$ on a nonempty complete metric space with $d(Tx,Ty)\le qd(x,y)$ for a constant $0\le q<1$ has a unique fixed point, approached from every starting point by iteration.

Choose $x_0$ and let $x_{n+1}=T{x}_n$. Induction gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$. For $m>n$, the triangle inequality and geometric series giveThus the sequence is Cauchy and converges by completeness to some $x_{\ast }$. The contraction estimate implies continuity, so $T{x}_{\ast }=\lim T{x}_n=\lim x_{n+1}=x_{\ast }$. If $y_{\ast }$ were another fixed point, $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$ would force equality of the points. Passing $m\to \infty$ also yields the quantitative error estimate

Weak convergence $u_j\rightharpoonup u$ means $F(u_j)\to F(u)$ for every $F\in X^{\ast }$. Weak-star convergence $T_j\stackrel{\ast }{\rightharpoonup }T$ means $T_j(x)\to T(x)$ for every $x\in X$, viewing $X$ as the predual. The Sequential Banach-Alaoglu theorem for a separable predual says that every norm-bounded sequence in $X^{\ast }$ has a weak-star convergent subsequence. Separability is what turns the bounded weak-star compactness statement into this sequential form.

For the weak-star limit of periodic rescalings, apply it with $X=L^1(\mathbb{R})$, which is separable, and $X^{\ast }=L^{\infty }(\mathbb{R})$. The rescaled functions have uniform essential-supremum bound $\Vert \theta {\Vert }_{\infty }$. Let $q=\theta -\bar{\theta }$ and $F(t)={\int }_0^{t}q(s)ds$. Since $q$ is one-periodic with zero mean, $F$ is a bounded one-periodic absolutely continuous primitive. Thus for every finite interval,Every weak-star subsequential limit $h$ therefore satisfies ${\int }_a^b(h-\bar{\theta })=0$ for all $a<b$. To apply the supplied uniqueness fact without assuming global integrability, fix $R>0$ and set $v=(h-\bar{\theta })1_{[-R,R]}$. This is in $L^1(\mathbb{R})$, and its integral over any interval is zero, since intersecting with $[-R,R]$ again gives an interval. Hence $v=0$ almost everywhere. Taking integer $R\to \infty$ yields $h=\bar{\theta }$ almost everywhere on the whole line.

Finally, if the full sequence did not converge weak-star to this constant, some $g\in L^1$ and some subsequence would have its pairing with $\theta (jx)-\bar{\theta }$ bounded away from zero. Sequential compactness gives a further weak-star convergent subsequence, whose limit must be the constant just identified, a contradiction. ThereforeThis is averaging against integrable test functions, not a claim of pointwise or strong norm convergence.

The weak derivative condition isBoth integrals exist since $u,v$ are locally integrable and the test function has compact support. Derivatives are defined up to almost-everywhere equality. The space $W^{1,p}(\Omega )$ consists of $u\in L^p(\Omega )$ whose first weak derivatives $D_{i}u$ all lie in $L^p(\Omega )$. One standard norm isA sum in the last formula gives an equivalent norm. For restriction to $E$, extend any $\varphi \in C_c^{\infty }(E)$ by zero to $\Omega$. Its support is compactly contained in $E$, so the extension is smooth. The identity over $\Omega$ then reduces exactly to the identity over $E$, proving locality of the weak derivative.

For the jump contribution to a piecewise weak derivative, integration by parts on the two sides of the joining point gives the distributional derivativeThe coefficient is the right trace $1+a$ minus the left trace $2$. If a locally integrable weak derivative existed, locality on each open side would identify it almost everywhere with the displayed regular part. Substituting that into the global test identity would force $(a-1)\varphi (1)=0$ for every test function, hence $a=1$. Equivalently a nonzero Dirac point mass cannot be represented by a locally integrable derivative.

For $a=1$ the point-mass term vanishes, and both the function and its piecewise derivative are bounded on the finite interval. Thus $u\in W^{1,\infty }$ and hence in every listed $W^{1,p}$. Therefore membership holds exactly for $a=1$. Changing a single point value is irrelevant to Sobolev membership; it cannot remove mismatched one-sided traces when $a\ne 1$.

By Lagrange's theorem, the index of a subgroup is $[G:H]=75/25=3$. Thus the set of left cosets has three elements, and the preceding group action yields a group homomorphism $\Phi :G\to S_3$ after labelling them.

Its image is a subgroup of $S_3$, so its order divides $6$. By the first isomorphism theorem, $|\mathrm{im}\Phi |=|G|/|\ker \Phi |$ also divides $75$. Hence its order divides $\gcd (6,75)=3$, giving possibilities $1$ and $3$. The action on cosets is transitive: $g_2{g}_1^{-1}$ sends $g_{1}H$ to $g_{2}H$. It cannot be the trivial action on a three-element set, so the image does not have order $1$. ThereforeIt follows that $|\ker \Phi |=75/3=25$. Part (b)(i) gives $\ker \Phi \le H$, and $|H|=25$; equal finite orders force equality. A kernel of a group homomorphism is a normal subgroup, soThis is an instance of the fact that a subgroup of index the smallest prime divisor is normal.

Define $g\cdot (g_{1}H)=(g{g}_1)H$ on the set of left cosets. It is well-defined: if $g_{1}H=g_{2}H$, then $g_2=g_{1}h$ for some $h\in H$, so $(g{g}_2)H=(g{g}_{1}h)H=(g{g}_1)H$. The identity fixes each coset, and associativity in $G$ givesHence this is a left group action, and part (a) supplies its group homomorphism $\Phi :G\to \mathrm{Sym}(S)$.

An element of the kernel of a group homomorphism $\Phi$ fixes every coset, in particular $H$ itself. Thus $g\in \ker \Phi$ implies $gH=H$, which implies $g\in H$. ThereforeThe subgroup symbol is justified because a kernel of a group homomorphism is a subgroup, and the argument proves containment.

The group action law givesThus ${\varphi }_g$ is a bijection with inverse ${\varphi }_{g^{-1}}$, hence a permutation of $S$. Also, for every $s$,Therefore the assignment $\Phi (g)={\varphi }_g$ takes values in the symmetric group $\mathrm{Sym}(S)$ and satisfiesIt is a group homomorphism, with the group product on the right interpreted as composition of permutations.

A left group action of $G$ on a set $S$ is a function $\varphi :G\times S\to S$ satisfyingfor all $s\in S$ and $g,h\in G$. Writing $g\cdot s=\varphi (g,s)$, these are $e\cdot s=s$ and $(gh)\cdot s=g\cdot (h\cdot s)$. No commutativity is assumed, so the order of composition matters.

By the preceding relabelling formula, every conjugate of $(123)$ is a three-cycle, and any three-cycle on four symbols can be obtained by a suitable relabelling. ThereforeThere are eight elements, agreeing with the count $\left(\genfrac{}{}{0ex}{}{4}{3}\right)\cdot 2$.

Likewise, conjugation preserves the pattern of two disjoint transpositions, and every partition of four symbols into two pairs is realized by relabelling. ThusThese have the respective cycle types $(3,1)$ and $(2,2)$ in the symmetric group $S_4$.

Conjugation of a permutation relabels each symbol by the inverse conjugating permutation. Specifically,To verify the direction of the inverse, apply the left side to ${\tau }^{-1}(a_j)$: $\tau$ first takes it to $a_j$, the original cycle takes it to $a_{j+1}$, and ${\tau }^{-1}$ takes it to ${\tau }^{-1}(a_{j+1})$, with cyclic indexing. Any symbols fixed by the original permutation are similarly relabelled and remain fixed. More generally, conjugate every cycle of a disjoint-cycle decomposition in this way; disjointness is preserved because ${\tau }^{-1}$ is a bijection.

If $g\in Z(G)$, it commutes with each $x\in G$, so $x^{-1}gx=g$. Its conjugacy class is therefore $C(g)=\{g\}$.

Conversely, if $C(g)=\{g\}$, then $x^{-1}gx=g$ for every $x$. Multiplication by $x$ on the left gives $gx=xg$ for every $x$, which is exactly membership in the center of a group. Thus

The center of a group consists of its elements commuting with every element. First check it is a subgroup. The identity element is central. If $a,b\in Z(G)$ and $x\in G$, then $b^{-1}$ commutes with $x$ andThus $a{b}^{-1}\in Z(G)$, proving closure under the subgroup criterion. Moreover, if $z\in Z(G)$, then $xz{x}^{-1}=z$ for every $x\in G$. Hence conjugation fixes the center elementwise, soTherefore the center of a group is a normal subgroup.

A subgroup $N$ of $G$ is a normal subgroup if conjugation by every element preserves it:Equivalently, $gN=Ng$ for every $g\in G$. Here equality means equality of sets; it does not require each individual element of $N$ to commute with every element of $G$.

A group homomorphism from an additive cyclic group ${\mathbb{Z}}_n$ is determined by the image $a$ of $1$, subject exactly to $a^n=e$ in multiplicative notation, or $na=0$ in an additive target.

For ${\mathbb{Z}}_{18}\to {\mathbb{Z}}_{15}$, the relation is $18a\equiv 0(\mathrm{mod}15)$, equivalently $3a\equiv 0(\mathrm{mod}15)$. The possibilities are $a=0,5,10$, and each defines a homomorphism between cyclic groups. Thus there are three such maps, an instance of the number of homomorphisms between finite cyclic groups.

For ${\mathbb{Z}}_9\to S_4$, the image of $1$ must be a permutation whose order of a group element divides $9$. A permutation on four symbols can have order $1,2,3$ or $4$, so the allowed images are the identity and the three-cycles. There are $\left(\genfrac{}{}{0ex}{}{4}{3}\right)(3-1)!=8$ three-cycles: choose their three symbols and one of the two cyclic orientations. Hence there are nine maps.

Finally, each transposition in $S_5$ has order $2$, so its image in ${\mathbb{Z}}_5$ must have order dividing both $2$ and $5$, and is therefore the identity. Every permutation is a product of transpositions, so every element must map to the identity. There is one map. In the order asked, the counts are

Let $n=o(g)$. A group homomorphism preserves powers and the identity element, soThe preceding divisibility property givesNow assume $\varphi$ is surjective and choose $h\in H$ of order $m$. Take a preimage $g$ of $h$, and write $o(g)=mr$, as the divisibility result permits. The element $g^r$ has order exactly $m$: its $m$th power is $e$, while $(g^r{)}^s=e$ implies $mr\mid rs$, hence $m\mid s$. Therefore $G$ contains an element of order $m$, namely $g^r$.

The preimage $g$ itself need not have order $m$; taking the suitable power is the essential step. The constructed element need not map to the original $h$. This proves that element orders in a finite quotient occur in the original group.

Lagrange's theorem states that for a subgroup $H$ of a finite group $G$,where $[G:H]$ is the index of a subgroup, the number of its left cosets. In particular, $|H|$ divides $|G|$.

The cyclic subgroup $⟨g⟩$ consists of the distinct elements $e,g,\dots ,g^{n-1}$, where $n=o(g)$. Every power reduces to one of these by the Euclidean division, and a repetition among them would contradict minimality of $n$. Thus $|⟨g⟩|=o(g)$. Applying Lagrange's theorem gives

The order of a group element $o(g)$ is the least positive integer $n$ for which $g^n=e$, whose existence follows from part (i). If $g^k=e$, use the Euclidean division to write $k=qn+r$, where $0\le r<n$. ThenMinimality of $n$ forces $r=0$. ThusThe argument also works for negative $k$, using integer powers and the same remainder convention.

Among the $|G|+1$ powers $e,g,g^2,\dots ,g^{|G|}$ in the finite group $G$, the pigeonhole principle forces two to be equal: $g^i=g^j$ with $0\le i<j\le |G|$. Multiply by $g^{-i}$ to obtainThis proves the existence of a positive power equal to the identity element, without presupposing the order of a group element.

The image of a linear map $\mathrm{im}(ST)=S(\mathrm{im}T)$ lies inside $\mathrm{im}S$. Therefore $r(ST)\le r(S)$. Also $S$ restricted to $\mathrm{im}T$ cannot have rank of a linear map greater than its domain's dimension, so $r(ST)\le \dim \mathrm{im}T=r(T)$. ThusEquivalently, apply the rank-nullity theorem to that restriction: $r(ST)=r(T)-\dim (\mathrm{im}T\cap \ker S)\le r(T)$. This proof keeps track of the intermediate vector subspace, rather than assuming an invertible factor.