Again the series converges absolutely. $T=2{\sum }_j{B}_j/3^j$ has the Cantor distribution. Every particular digit sequence has probability zero, so it has no atoms; its support is the Cantor set, which has Lebesgue measure zero, so it cannot have a Lebesgue density.

Absolute convergence gives $S_n\to S$ almost surely. Put $B_j=(X_j+1)/2$. Then $S=2{\sum }_j{B}_j/2^j-1$. Independent fair binary digits make the sum uniform on $[0,1]$, so $S$ is uniform on $[-1,1]$.

Characteristic-function convergence theorem says $X_n\Rightarrow X$ iff ${\Phi }_{X_n}(t)\to {\Phi }_X(t)$ for all $t$, where the limiting function is continuous at zero.

$X_n\Rightarrow X$ means $Ef(X_n)\to Ef(X)$ for every bounded continuous $f$, equivalently convergence of distribution functions at continuity points of $X$.

Define $F:TX\to \mathbb{R}$ by $F(p,v)=I_p(v,v)$. On $F^{-1}(1)$, the vertical derivative in direction $v$ is $2I_p(v,v)=2$, so 1 is regular. Since $\dim TX=4$, the preimage theorem makes the unit tangent bundle $SX$ a 3-manifold.

A value $y$ is regular when $d{f}_x:T_{x}X\to T_{y}Y$ is surjective for every $x\in f^{-1}(y)$. The preimage theorem says $f^{-1}(y)$ is then a submanifold of dimension $\dim X-\dim Y$.

For a local parametrization $\varphi :U\subset {\mathbb{R}}^k\to X$, the map $(q,w)\mapsto (\varphi (q),d{\varphi }_{q}w)$ parametrizes $TX$ locally. Its inverse uses the base chart and derivative, and its rank is $2k$. These charts make $TX$ a $2k$-manifold.

$H$ is a closed linear subspace of reflexive $L^p$. A minimizing sequence is bounded by the triangle inequality. Reflexivity supplies a weakly convergent subsequence with limit $f_0\in H$; each norm $\Vert f-g_i{\Vert }_p$ is weakly lower semicontinuous, so their sum attains the infimum at $f_0$.

For separable Banach $X$, the closed unit ball of $X^{\ast }$ is weak-star compact. Embed it in the product of compact discs by evaluations on a countable dense subset; diagonal subsequences converge on that subset, boundedness extends the limit to all $X$, and the limit functional remains in the ball. Metrizability of the bounded weak-star topology turns sequential compactness into compactness. Scaling gives the general Banach-Alaoglu theorem.

No. The point evaluations ${\delta }_x$ satisfy $\Vert {\delta }_x-{\delta }_y\Vert =2$ for $x\ne y$ (choose a continuous function taking values $1,-1$ at the two points). Thus $C[0,1{]}^{\ast }$ is nonseparable, whereas ${\ell }_1$ is separable.

The series defining ${\varphi }_y$ converges and $\Vert {\varphi }_y\Vert \le \Vert y{\Vert }_1$. For each finite set of $k_n$, normality and Tietze extension give a continuous $f$ of norm at most one taking prescribed signs there. Letting the finite set grow proves $\Vert {\varphi }_y\Vert =\Vert y{\Vert }_1$. Thus $y\mapsto {\varphi }_y$ is an isometric embedding of ${\ell }_1$ into $C(K{)}^{\ast }$.

The canonical embedding puts $X$ isometrically, hence with closed range, inside $X^{\ast \ast }\simeq {\ell }_2$. Every closed subspace of a Hilbert space is Hilbert and, in the infinite-dimensional separable case here, is isomorphic to ${\ell }_2$. Thus $X\simeq {\ell }_2$.

If $S:X\to Y$ is an isomorphism, then $S^{\ast }:Y^{\ast }\to X^{\ast }$ is an isomorphism with inverse $(S^{-1}{)}^{\ast }$. If $S$ and $S^{-1}$ are isometries, the same norm calculation makes the dual maps isometries.

$X^{\ast }$ is the Banach space of bounded linear functionals. For $T:X\to Y$, define $T^{\ast }:Y^{\ast }\to X^{\ast }$ by $T^{\ast }f=f\circ T$. It is linear and $\Vert T^{\ast }f\Vert \le \Vert T\Vert \Vert f\Vert$, so bounded (indeed $\Vert T^{\ast }\Vert =\Vert T\Vert$).

Dedekind's criterion factors $(p)$ according to the factorization of $X^3-7$ modulo $p$ (the monogenic assumption removes the index obstruction). With $\alpha =\sqrt[3]{7}$,Minkowski's bound is below 11, so classes are generated by prime ideals over $2,3,5,7$. Relations from the displayed factorizations and principal ideals reduce them to $\mathfrak{p}=(3,\alpha -1)$ with ${\mathfrak{p}}^3$ principal. Part (a) shows $\mathfrak{p}$ is not principal, so $\mathrm{Cl}(K)\simeq \mathbb{Z}/3\mathbb{Z}$.

The left side is the norm form $N_{K/\mathbb{Q}}(x+y\alpha +z{\alpha }^2)$ for ${\alpha }^3=7$. Reducing a hypothetical equation modulo 7 and then descending through the forced divisibilities of $x,y,z,t$ gives an infinite descent, so only the zero solution exists. If $N(\theta )=3$, clearing denominators minimally and applying the norm equation would produce a forbidden nonzero integer solution; hence no such $\theta$ exists.

Mackey and induced representation sum over $H\backslash G/K$. Coprimality makes every $H\cap gK{g}^{-1}$ trivial, each double coset has size $|H||K|$, and each local inner product is $(\dim V)(\dim W)$. Therefore

The relations are ${\sigma }^n={\tau }^2=1$ and $\tau \sigma \tau ={\sigma }^{-1}$, so the generated group is $D_{2n}$. The functions $XY$ and $X^n+Y^n$ are fixed. Conversely $X^n,Y^n$ are roots of $T^2-(X^n+Y^n)T+(XY{)}^n$, and adjoining $X$ then $Y$ gives degree at most $2n$. Artin gives degree exactly $|G|=2n$, proving $M^G=\mathbb{C}(X^n+Y^n,XY)$.