A subset is compact if every cover of it by open sets of the ambient space has a finite subcover; equivalently use open sets of its subspace topology. A cover of $A\cup B$ separately covers $A$ and $B$. Compactness of each supplies two finite collections, whose union is finite and covers $A\cup B$. If $A$ is closed in a compact space, add the open complement of $A$ to any open cover of $A$. Compactness supplies a finite subcover of the whole space, and discarding the complement leaves a finite cover of $A$.

For the enlarged space, every old open set is open in $C$ and has the same trace on $X$. The trace on $X$ of a new open set $V\cup \{\infty \}$ is $V$, which is open in $X$ because its complement is closed. Thus the induced topology on $X$ is exactly the original one.

Given an open cover of $C$, choose a member containing infinity. It has the form $V\cup \{\infty \}$ with $K=X\setminus V$ compact. The remaining cover restricted to $K$ has a finite subcover. Together with the chosen infinity-neighborhood, these finitely many sets cover $C$. Hence $C$ is compact, with no Hausdorff assumption needed.

Finally put $a=f(\infty )$. Continuity makes $U=f^{-1}((a-ϵ,a+ϵ))$ open and containing infinity. Therefore $U=V\cup \{\infty \}$ with $W=X\setminus V$ compact and closed in $X$. For every $x\in X\setminus W=V$ we haveThis expresses convergence to the value at infinity outside compact closed subsets.

The monotone convergence theorem states that if $0\le h_n\uparrow h$ pointwise almost everywhere and the functions are measurable, then $\int h_n\uparrow \int h$, allowing the value infinity.

Interpret the displayed densities as zero outside $(0,1)$, so no undefined logarithm is evaluated there. Each $f_k$ is nonnegative and measurable. With $x=e^{-t}$,where the last equality is integration by parts. This is finite, proving Lebesgue integrability.

The nonnegative partial sums satisfy, for $0<x<1$,They vanish outside the interval as well. Monotone convergence therefore givesThe series is finite by comparison with the convergent series ${\sum }_{j\ge 1}{j}^{-2}$, whose convergence follows from the integral test. Since $g\ge 0$, this finite integral proves absolute Lebesgue integrability as required.

A primitive root modulo $p$ is an integer whose residue generates the multiplicative group ${\mathbb{F}}_p^{\times }$, equivalently has order $p-1$.

The Chinese remainder theorem states that, for pairwise coprime positive integers $m_1,\dots ,m_r$, the system $x\equiv a_i(\mathrm{mod}{m}_i)$ has a unique solution modulo $M={\prod }_i{m}_i$. For a constructive proof put $M_i=M/m_i$. Bézout's identity gives an integer $u_i$ with $M_i{u}_i\equiv 1(\mathrm{mod}{m}_i)$. Then $x={\sum }_i{a}_i{M}_i{u}_i$ has all the required residues. If two solutions exist, their difference is divisible by every $m_i$; pairwise coprimality makes it divisible by their product, proving uniqueness modulo $M$.

Here $3,5,7$ divide $p-1$, and none is zero modulo $p$. Choose a primitive root $g$ and exponents $a,b,c$ withUse the theorem with moduli $3,5,7$ to choose $r$ such that $r\equiv a(\mathrm{mod}3)$, $r\equiv b(\mathrm{mod}5)$, $r\equiv c(\mathrm{mod}7)$. Define nonzero residuesThe exponents are integers; negative ones are interpreted using inverses in ${\mathbb{F}}_p^{\times }$. Taking integer representatives now givesThe theorem aligns the exponent modular congruences, which is why a primitive root converts the three power conditions into a linear modular congruence problem.

Right multiplication gives a right action $(Hx)\cdot g=Hxg$. To obtain a homomorphism into permutations with the usual function-composition convention, use the associated left actionThis is well-defined: replacing $x$ by $hx$, $h\in H$, does not change the image coset. Moreover $\rho (g)\rho (k)(Hx)=Hx{k}^{-1}{g}^{-1}=\rho (gk)(Hx)$, and each map is bijective with inverse $\rho (g^{-1})$.

An element $g$ is in the kernel precisely when $Hx{g}^{-1}=Hx$ for every $x\in G$. This says $x{g}^{-1}{x}^{-1}\in H$, equivalently $g\in x^{-1}Hx$. HenceThis is the largest normal subgroup of $G$ contained in $H$.

Now the index of $H$ is $q$, so $G/\ker \rho$ embeds into $S_q$. Its order therefore divides $q!$. Since $p>q$ is prime, $p$ does not divide $q!$. From $|G|=pq$ it follows that $p$ divides $|\ker \rho |$. But $\ker \rho \subseteq H$ and $|H|=p$, forcing $\ker \rho =H$. Therefore $H$ is normal.

For the last case let $H=⟨b⟩$. Normality makes conjugation by $a$ an automorphism, so $a^{-1}ba=b^r$ with $r\in \{1,2,3,4\}$. Applying it twice and using $a^2=1$ gives $b=b^{r^2}$, so $r^2\equiv 1(\mathrm{mod}5)$. Thus $(r-1)(r+1)\equiv 0(\mathrm{mod}5)$, andBoth possibilities are allowed; the order assumptions alone do not force the inverse case.

A real inner product is a bilinear symmetric form satisfying $(v,v)>0$ for every nonzero $v$, and $(0,0)=0$. Its norm is $\Vert v\Vert =\sqrt{(v,v)}$. An orthogonal polynomial family has nonzero members of exact degrees $0,1,\dots ,k$, with $({\varphi }_i,{\varphi }_j)=0$ for different indices. Exact degree is part of this definition; merely belonging to the indicated polynomial vector spaces would not suffice.

Distinct degrees make ${\varphi }_0,\dots ,{\varphi }_{\ell -1}$ linearly independent. There are $\ell$ of them in the $\ell$-dimensional space ${\Pi }_{\ell -1}$, so they form a basis. Any lower-degree polynomial is their linear combination, and bilinearity therefore gives $({\varphi }_{\ell },p)=0$ for every $p\in {\Pi }_{\ell -1}$.

Use the orthogonal-projection characterization: $p_{\ast }$ is the best approximation to $v$ in a finite-dimensional subspace $W$ exactly when $v-p_{\ast }$ is orthogonal to $W$. Here $v={\varphi }_{\ell }$ is already orthogonal to $W={\Pi }_{\ell -1}$, so its unique best approximation is zero. For $\ell \ge 1$, ${\varphi }_{\ell -1}$ is nonzero and therefore cannot be that best approximation. Directly,For the specified weight, symmetry makes every odd moment vanish, whileChoose ${\varphi }_0=1$ and ${\varphi }_1=x$. A monic quadratic orthogonal to $x$ has no linear term by symmetry, and orthogonality to 1 requires $\int |x|(x^2-c)dx=1/2-c=0$. Thus a valid answer isThe weight vanishes only at one point, so a nonzero polynomial has positive squared norm; the form is indeed positive definite on the polynomial vector space. Nonzero rescalings of individual family members give equally valid unnormalized answers.

The elementary elimination step replaces row $i$ by row $i$ minus a multiple $m_{ij}$ of an earlier pivot row $j$. With coordinate vectors $e_i$, its premultiplication matrix isand its inverse is $I+m_{ij}{e}_i{e}_j^T$, since $(e_i{e}_j^T{)}^2=0$. Choose $m_{ij}$ as the current entry below the pivot divided by the current pivot. Required pivots are assumed nonzero, consistently with elimination without row interchanges.

For dimension 4, first eliminate positions $(2,1),(3,1),(4,1)$, then $(3,2),(4,2)$, then $(4,3)$. ThusThe rightmost operation acts first, and every multiplier is computed from the matrix at its own stage. For general $n$, the numbers of eliminations in successive columns are $n-1,n-2,\dots ,1$, totaling $n(n-1)/2$; a zero multiplier simply gives an identity operation.

The product $E$ is unit lower triangular and $U=EA$ is upper triangular. Its inverse $L=E^{-1}$ is unit lower triangular, soThe stored multiplier in position $(i,j)$ is the corresponding subdiagonal entry of $L$, as follows by reversing the pivot-row subtractions column by column.

If $A$ is upper Hessenberg, only the first subdiagonal can be nonzero below the diagonal. At stage $j$ only position $(j+1,j)$ needs elimination. Subtracting row $j$ cannot create lower entries further down because its entries before its pivot are already zero. Hence only $n-1$ operations are needed. The inverse factors occur in increasing column order, and their cross products vanish, givingwhich is lower bidiagonal; $U$ is upper triangular, possibly dense above its diagonal.

If the Hessenberg matrix is also symmetric, it is tridiagonal. At each elimination step the pivot row has no entries past its immediate superdiagonal, so the updated matrix retains that upper bandwidth. Thus $L$ is lower bidiagonal and $U$ is upper bidiagonal, with at most two nonzero entries in each row, as required.