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First $w\in {\ell }^2$, since ${\sum }_{n\ge 1}{4}^{-n}=1/3$. Any finite linear dependence among $w$ and the $e_m$ has the form $aw+{\sum }_{m\in F}{b}_m{e}_m=0$ for a finite index set $F$. At an index outside $F$, its coordinate equals $a{2}^{-n}$, so $a=0$. Each remaining coordinate then gives $b_m=0$. Therefore the set is linearly independent, and the prescribed values define a unique linear map on its algebraic span.

Take the tailsThey belong to $V$, satisfy $f(v_N)=1$, and haveThus $v_N\to 0$ while $f(v_N)$ does not approach $f(0)=0$. The map is not continuous, and part (b) also shows it is unbounded. This is a discontinuous functional detected by vanishing sequence tails, illustrating why the finite-dimensional conclusion in part (a) cannot be extended to arbitrary normed spaces.

If $f$ is a bounded linear operator, linearity gives $\Vert f(u)-f(v)\Vert \le C\Vert u-v\Vert$. It is therefore Lipschitz and continuous everywhere.

Conversely, suppose $f$ is continuous at zero. There is $\delta >0$ such that $\Vert u\Vert <\delta$ implies $\Vert f(u)\Vert <1$. For $u\ne 0$, apply this to $v=\delta u/(2\Vert u\Vert )$ and use homogeneity:The value at $u=0$ is zero by linearity. ConsequentlyThus boundedness and continuity are equivalent for linear maps between normed spaces. No finite dimensionality or completeness is needed for this equivalence.

Two norms $N_1,N_2$ on the same vector space are equivalent norms, or Lipschitz equivalent, if there are constants $0<c\le C<\infty$ such that $c{N}_1(v)\le N_2(v)\le C{N}_1(v)$ for every $v$. The constants are uniform over the whole space.

Let $N$ be any norm on ${\mathbb{R}}^n$. The triangle inequality and the Cauchy-Schwarz inequality giveThe reverse triangle inequality then gives $|N(x)-N(y)|\le C\Vert x-y{\Vert }_2$, so $N$ is Euclidean-continuous; this continuity was derived, rather than assumed. On the compact Euclidean unit sphere, $N$ attains a strictly positive minimum $c$, because it vanishes only at zero. Homogeneity yieldsFinally a linear map $A:{\mathbb{R}}^n\to {\mathbb{R}}^m$ satisfies $\Vert Ax{\Vert }_2\le ({\sum }_{i,j}|a_{ij}{|}^2{)}^{1/2}\Vert x{\Vert }_2$, again by Cauchy-Schwarz inequality. The equivalence just proved transfers this bound to any norms on the source and target. Hence every such linear map is continuous.

If $M=0$, take the empty filtration. Otherwise a nonzero quotient $Q=M/M_i$ is a finitely generated module. Because $R$ has the ascending chain condition on ideals, the nonempty collection of annihilators of nonzero elements of $Q$ has a maximal element. Part (b), applied to $Q$, supplies $q\ne 0$ with a prime ideal $P=\mathrm{Ann}(q)$.

The map $R\to Rq$, $r\mapsto rq$, is surjective with kernel $P$. The first isomorphism theorem gives $Rq\cong R/P$. Let $M_{i+1}$ be the inverse image of $Rq$ under $M\to Q$. ThenRepeat whenever the quotient remains nonzero. An infinite repetition would give a strictly ascending chain of submodules of the Noetherian module $M$, contradicting part (a). Thus the process stops after finitely many steps, necessarily at $M$.

This constructs the required prime filtration $0=M_0⊊\cdots ⊊M_l=M$, with each successive quotient $R/P_i$ for a prime $P_i$.

Let $P=\mathrm{Ann}(m)$ be maximal among annihilators of nonzero elements of $M$. It is a proper ideal, since $1m=m\ne 0$. Suppose $ab\in P$ and $b\notin P$. Then $bm\ne 0$ andThe inclusion holds because every $r\in P$ satisfies $r(bm)=b(rm)=0$. Maximality in the stated set forces $\mathrm{Ann}(bm)=P$, so $a\in P$. This proves the defining implication for a prime ideal:This is a maximal annihilator of a module element is prime argument. Maximality among element annihilators is sufficient; $P$ need not be a maximal ideal of $R$.

Work with the usual commutative unital Noetherian ring convention. First prove that every submodule of $R^d$ is a finitely generated module. For $d=1$, a submodule is an ideal, hence finitely generated by Noetherianity. For the induction step, project a submodule $N\subseteq R^d$ onto its last coordinate. Its image is a finitely generated ideal; choose lifts in $N$ of its generators. The projection kernel lies in $R^{d-1}$ and is finitely generated by induction. The kernel generators together with the chosen lifts generate $N$, since subtracting a combination of the lifts leaves an element of the kernel.

Now choose a surjection $R^d\to M$ from a finite generating set of the module $M$. For any submodule $U\subseteq M$, its inverse image in $R^d$ is finitely generated by the result just proved. The images of those generators generate $U$.

Every submodule of $M$ is finitely generated. In particular $M$ is a Noetherian module: the union of an ascending chain is a submodule, and its finite generators all belong to a single member of the chain, forcing eventual stabilization.

The adjoint operator ${\alpha }^{\ast }:W\to V$ is uniquely characterized byLet $A=\beta {\beta }^{\ast }+{\gamma }^{\ast }\gamma :V\to V$. Its quadratic form is nonnegative:If $Av=0$, both norms vanish. In particular $v\in \ker \gamma =\mathrm{im}\beta$, while ${\beta }^{\ast }v=0$ means $v$ is orthogonal to $\mathrm{im}\beta$, by the defining adjoint operator identity. A vector belonging to a vector subspace and its orthogonal complement must be zero. Hence $A$ is injective, and finite dimensionality makes it surjective.

ThusThe same calculation shows it is a positive-definite operator when $V\ne \{0\}$.

The algebraic dual map is $\delta ={\alpha }^{\vee }:W^{\ast }\to V^{\ast }$, defined by $\delta (\phi )=\phi \circ \alpha$. The annihilator of a vector subspace $U\subseteq V$ isEvery functional $\phi \circ \alpha$ vanishes on $\ker \alpha$, so $\mathrm{im}\delta \subseteq (\ker \alpha {)}^{\circ }$. Conversely, let $\lambda \in (\ker \alpha {)}^{\circ }$. Define a functional on $\mathrm{im}\alpha$ by $\tilde{\lambda }(\alpha v)=\lambda (v)$. It is well defined because two preimages differ by an element of $\ker \alpha$. Extend a basis of $\mathrm{im}\alpha$ to one of $W$ and extend the functional arbitrarily, for instance by zero on the extra basis vectors. The resulting $\phi \in W^{\ast }$ satisfies $\delta \phi =\lambda$.

This is the dual image and kernel annihilator identity. ThereforeThe basis extension theorem also gives $\dim U^{\circ }=\dim V-\dim U$. By the rank-nullity theorem,Apply rank-nullity theorem to both maps, using $\dim W^{\ast }=\dim W$, to obtainOver $\mathbb{C}$, this is the algebraic dual, without complex conjugation; the adjoint operator in part (b) uses the inner products instead.

The global time-dependent phase cancels from the probability current. Put $\chi =e^{ikx}+R{e}^{-ikx}$. Direct multiplication givesThe second term is real, because the expression in parentheses is purely imaginary. HenceThe first plane wave travels to the right with incident current $\hslash k/m$; the second travels to the left with reflected current $-(\hslash k/m)|R{|}^2$. Their wave interference changes the probability density but cancels from the net current. The current vanishes for $|R|=1$, is positive for $|R|<1$, and is negative for $|R|>1$. These plane waves are scattering states, not individually normalizable bound states.

For a particle of mass $m$, the probability density and one-dimensional probability current areThe Schrödinger equation and its complex conjugate, for real $V$, giveThe potential terms cancel precisely because $V$ is real. Thus ${\rho }_t+j_x=0$, the continuity equation for probability. Integrating over an interval says that its probability changes by the incoming current minus the outgoing current.

Let $p$ and $h$ count the pentagonal and hexagonal faces. Counting incidences of edges with vertices and faces givesThe Euler formula for a sphere also applies to this polygonal decomposition: triangulating a polygon by adding one diagonal increases both $E$ and $F$ by one, preserving $V-E+F$. Substitution givesThus there are exactly 12 pentagons, independently of the number of hexagons.

For a finite triangulation of a surface of the sphere, Euler formula for a sphere iswhere $V$, $E$ and $F$ count vertices, edges and triangular faces. Since each face has three edges and each edge borders two faces, also $3F=2E$.

The sequence has uniform convergence on $X$ if there is a function $f:X\to \mathbb{R}$ such thatThe same $N$ must work for every $x$. Equivalently, the supremum of the error over $X$ tends to zero. In pointwise convergence, $N$ is allowed to depend on $x$.

Put $x_0=1$. The constraints are equivalent to $y_i=x_i-x_{i-1}$, so the objective becomes the strictly convex quadraticIts derivative at an interior index gives $5x_i-2x_{i-1}-2x_{i+1}=0$, $1\le i<n$. The terminal derivative gives $3x_n-2x_{n-1}=0$. The characteristic equation $2r^2-5r+2=0$ has roots $2$ and $1/2$, soThe conditions $x_0=1$ and $3x_n=2x_{n-1}$ give $a+b=1$ and $b=2\cdot 4^{n}a$. ThusThese formulas also cover $n=1$, giving $x_1=2/3$, $y_1=-1/3$.

For a direct application of Lagrangian sufficiency theorem, use equality constraints $h_i=x_i-1-{\sum }_{k=1}^i{y}_k$ and multipliers ${\mu }_i=-2x_i^{\ast }$. The Lagrangian derivatives in $x_i$ vanish. Its derivative in $y_j$ is $4y_j^{\ast }+2{\sum }_{i=j}^n{x}_i^{\ast }$, which also vanishes: putting $S_j=2y_j^{\ast }+{\sum }_{i=j}^n{x}_i^{\ast }$, the interior recurrence gives $S_j-S_{j+1}=0$, and the terminal condition gives $S_n=0$. The Lagrangian has positive-definite quadratic part ${\sum }_i(x_i^2+2y_i^2)$, so this stationary point is its unique global minimum. The candidate is feasible, and the theorem proves it is the unique constrained optimal solution. Equivalently, strict convexity of the eliminated objective proves the same global conclusion.

For a minimization problem over a domain $D$ with inequality constraints $g_j(x)\le 0$ and equality constraints $h_{\ell }(x)=0$, letThe Lagrangian sufficiency theorem states: if $x_{\ast }$ is feasible, ${\lambda }_j\ge 0$, ${\lambda }_j{g}_j(x_{\ast })=0$, and $x_{\ast }$ globally minimizes $\mathcal{L}(\cdot ,\lambda ,\mu )$ on $D$, then $x_{\ast }$ globally minimizes the constrained objective. Equality multipliers have no sign restriction.

For any feasible $x$, the sign of the inequality multipliers gives $\mathcal{L}(x,\lambda ,\mu )\le f(x)$. Feasibility and complementary slackness give $\mathcal{L}(x_{\ast },\lambda ,\mu )=f(x_{\ast })$. ThereforeThis proves the theorem. No convexity is necessary if the global minimization of the Lagrangian is already known. In a differentiable convex optimization problem with convex inequalities and affine equalities, a stationary point of this convex Lagrangian is automatically a global minimizer, which gives the familiar sufficiency of the Karush-Kuhn-Tucker conditions.

The specified gamma distribution uses shape $k$ and rate $\lambda$. Multiplying its density by the Poisson likelihood gives a posterior density proportional to ${\theta }^{k+S-1}{e}^{-(\lambda +n)\theta }$. Thus the conjugate prior calculation givesFor squared-error loss, the posterior expected loss isIt is minimized at the posterior mean, so the Bayes estimator isFor a fresh conditionally independent Poisson observation, the posterior predictive probability of zero is the posterior expectation of $e^{-\theta }$. Evaluating the gamma integral givesThe samples are independent conditional on $\theta$; integrating out the uncertain common parameter produces the predictive probability above.