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Use the Poisson integral to choose a smooth harmonic function $G$ on the unit disk whose boundary value is $g$. Seek $v$ in the form $v=w+G$, which is the standard homogenization of Dirichlet boundary data. Then $w\in H_0^1(\Omega )$ and, since $\Delta G=0$,Part (d) gives a unique such $w$, and henceis the unique weak solution with boundary value $g$.

Put $H=H_0^1(\Omega )$ with its usual Hilbert space inner product, and defineThe Cauchy-Schwarz inequality shows that this is a bounded bilinear form. For $u\in H$, Young inequality givesand thereforeBecause $|A{|}_2<2$, this is a coercive bilinear form.

The map $u\mapsto {\int }_{\Omega }uf$ is a bounded linear functional on $H$. By the Riesz representation theorem, it equals $⟨u,F{⟩}_H$ for some $F\in H$. Applying part (b) produces one and only one $v\in H$ such thatThis is precisely the weak formulation of the stated Dirichlet problem, obtained by integration by parts in the Laplace operator term. Thus $v$ is its unique weak solution; this is the constant-drift massive-Laplacian Dirichlet problem.

The zero-boundary Sobolev space isHere $H^1(\Omega )$ consists of the $L^2$ functions whose first weak derivatives also belong to $L^2$, with normOn a sufficiently regular domain, the closure condition is equivalent to having zero boundary trace.

By part (a), $B$ is a bounded bilinear form: $|B(u,v)|\le M\Vert u\Vert \Vert v\Vert$. For each fixed $v$, the map $u\mapsto B(u,v)$ is consequently a bounded linear functional. The Riesz representation theorem supplies a unique vector $Lv$ such thatUniqueness of the representing vector makes $L$ linear, and $\Vert Lv\Vert \le M\Vert v\Vert$ makes it a bounded linear operator.

The coercivity assumption implieshence $\Vert Lv\Vert \ge C\Vert v\Vert$. Moreover, using the adjoint operator,so the same argument gives $\Vert L^{\ast }v\Vert \ge C\Vert v\Vert$. The invertibility from lower bounds on an operator and its adjoint therefore shows that $L$ is invertible and $L^{-1}$ is bounded.

Set $v_f=L^{-1}f$. Then $B(u,v_f)=⟨u,L{v}_f⟩=⟨u,f⟩$ for every $u$. If another $v$ has this property, then $L(v-v_f)=0$, and the injectivity of $L$ gives $v=v_f$. Thus the unique vector is $v_f=L^{-1}f$.

This is the boundedness criterion for a bilinear form. Since $B$ is continuous at $(0,0)$, there are $r,s>0$ such that $|B(x,y)|\le 1$ whenever $\Vert x\Vert <r$ and $\Vert y\Vert <s$. For nonzero $u,v$, apply this toLinearity in each argument then givesThe same inequality is immediate if either vector is zero, so one may take $M=4/(rs)$.

For $x\in X$,The map $S$ defined by $(Sx{)}_n=x_{n-1}$ satisfies $ST=TS=I$, so $T$ is a surjective isometry and hence a unitary operator.

If $Tx=\lambda x$, thenFor $\lambda \ne 0$, this gives $x_n={\lambda }^n{x}_0$ for every integer $n$. When $|\lambda |>1$ the positive tail is not square summable, when $|\lambda |<1$ the negative tail is not square summable, and when $|\lambda |=1$ neither tail is square summable unless $x_0=0$. The case $\lambda =0$ also forces every coordinate to vanish. Thus

The reverse triangle inequality gives, for every $x$,Hence $T-\lambda I$ is bounded below whenever $|\lambda |\ne 1$, so such $\lambda$ do not belong to the approximate point spectrum.

Now let $|\lambda |=1$ and define the unit vectorThe equation $x_{n+1}^{(N)}=\lambda x_n^{(N)}$ fails only at the two endpoints, soTherefore the bilateral shift operator has

If $T$ is invertible, the bounded inverse theorem makes $T^{-1}$ bounded, andMoreover $T^{\ast }$ is invertible with inverse $(T^{-1}{)}^{\ast }$, so the same argument shows that $T^{\ast }$ is bounded below.

Conversely, supposefor some $c,d>0$. The first bound makes $T$ injective. It also makes $\mathrm{ran}T$ closed: if $T{x}_n$ is Cauchy, then $x_n$ is Cauchy, and continuity gives its limit in the range. The second bound gives $\ker T^{\ast }=\{0\}$. By image-kernel orthogonality for an adjoint,so the range is dense. Being both closed and dense, it equals $H$. Thus $T$ is bijective, and the first lower bound gives $\Vert T^{-1}\Vert \le c^{-1}$. Therefore

The real Riesz representation theorem states that for every bounded linear functional $f:H\to \mathbb{R}$ on a real Hilbert space there is a unique $y\in H$ such thatand $\Vert f\Vert =\Vert y\Vert$.

If $f=0$, take $y=0$. Otherwise $Y=\ker f$ is a closed proper subspace, so the supplied orthogonal decomposition by a closed subspace gives a nonzero $z\in Y^{\perp }$. Since $f(z)\ne 0$, for every $x\in H$,Taking its inner product with $z$ yieldsand thereforeThus $y=f(z)z/\Vert z{\Vert }^2$ represents $f$. If both $y$ and $y^{\prime }$ represent it, then $⟨x,y-y^{\prime }⟩=0$ for every $x$; taking $x=y-y^{\prime }$ proves uniqueness. Finally, Cauchy-Schwarz inequality gives $\Vert f\Vert \le \Vert y\Vert$, while evaluating at $x=y/\Vert y\Vert$ when $y\ne 0$ gives the reverse inequality. Hence

Now suppose ${\varphi }^2={\mathrm{Id}}_X$. The map is fixed-point-free, so $L(\varphi )=0$. On each rational homology group,projects onto the invariant subspace and has integral trace. Henceis an integer. Therefore

Suppose ${\varphi }^3={\mathrm{Id}}_X$. Since $\varphi$ has no fixed points, neither does ${\varphi }^2$: if ${\varphi }^2(x)=x$, then applying $\varphi$ gives $\varphi (x)={\varphi }^3(x)=x$. The Lefschetz theorem therefore gives

For each $q$, the averaging operatoris the projection of $H_q(X;\mathbb{Q})$ onto its $⟨\varphi ⟩$-invariant subspace, so its trace is an integer. Taking the alternating sum of these traces givesThe left side is an integer. By Euler-characteristic divisibility from a fixed-point-free cyclic action,

Taking norms ingivesso $s=n$.

Suppose for contradiction that ${\mathcal{O}}_K=\mathbb{Z}[\alpha ]$, and let $f$ be the degree-$n$ minimal polynomial of $\alpha$. Part (a) identifies the prime ideals above $p$ with the distinct irreducible factors of $\overline{f}$. Since there are $n$ distinct prime ideals, all with norm $p$, the norm formula forces $\overline{f}$ to have $n$ distinct irreducible factors of degree one. But ${\mathbb{F}}_p[X]$ has only the $p$ distinct monic linear polynomialsThis is impossible when $n>p$. The completely split small-prime obstruction to monogenicity therefore gives

Evaluation at $\alpha$ gives the ring isomorphismand reducing modulo $p$ givesFor each $i$,This quotient is a field because ${\overline{g}}_i$ is irreducible. Hence $P_i$ is a nonzero prime ideal; it is nonzero because it contains $p$. Its ideal norm is

The ideals are distinct. Indeed, if $P_i=P_j$ with $i\ne j$, then both ${\overline{g}}_i$ and ${\overline{g}}_j$ would vanish in the same quotient. Since distinct monic irreducibles are coprime, a Bezout identity between them would make $1$ vanish there, a contradiction.

Because the chosen lifts satisfyfor some $H\in \mathbb{Z}[X]$, evaluation at $\alpha$ givesEvery generator of the product ideal ${\prod }_i{P}_i^{e_i}$ either contains a factor $p$ or is the displayed product, soOn the other hand, multiplicativity of ideal norms and ${\sum }_i{e}_i\mathrm{deg}{\overline{g}}_i=\mathrm{deg}f=[L:\mathbb{Q}]$ giveTwo finite-index ideals with one contained in the other and equal index are equal. Thus the Dedekind factorization theorem is proved in this monogenic case: