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On a compact metric space $X$, a conservative Feller semigroup is a strongly continuous semigroup of positive linear maps on $C(X)$ satisfying $P_{t}1=1$. Positivity means $f\ge 0⟹P_{t}f\ge 0$. These conditions imply sup-norm contraction. Conversely, for a contraction semigroup with $1\in D(L)$ and $L1=0$, differentiation on the generator domain gives $\frac{d}{dt}{P}_{t}1=P_{t}L1=0$, so $P_{t}1=1$. On the real space $C(X)$, if $0\le f\le 1$, then $\Vert 1-P_{t}f{\Vert }_{\infty }\le \Vert 1-f{\Vert }_{\infty }\le 1$, implying $P_{t}f\ge 0$. Scaling proves positivity for every nonnegative $f$. This is the unital contraction positivity criterion. For complex $C(X)$ the same conclusion follows because each functional $f\mapsto P_{t}f(x)$ has norm one and value one on $1$, hence is a positive linear functional.

For each $t,x$, the Riesz-Markov-Kakutani representation theorem represents that functional by a unique Borel probability measure $p_t(x,dy)$:Continuity of $P_{t}f$ gives weak continuity of $x\mapsto p_t(x,\cdot )$. Approximating indicators of open sets increasingly by continuous functions proves Borel measurability in $x$; a monotone-class argument then gives a Markov kernel. The semigroup property and uniqueness of representing measures yield $p_0(x,\cdot )={\delta }_x$ and the Chapman-Kolmogorov equationThese are the required transition probabilities.

An invariant probability measure for a semigroup satisfies ${\int }_X{P}_{t}fd\mu ={\int }_{X}fd\mu$ for every $t\ge 0$ and $f\in C(X)$. Equivalently its distribution is preserved by the Markov kernel. For $1\le p<\infty$, Jensen inequality gives $|P_{t}f{|}^p\le P_t(|f{|}^p)$ pointwise. Integrating and using invariance provesAlso $\Vert P_{t}f-f{\Vert }_{L^p(\mu )}\le \Vert P_{t}f-f{\Vert }_{\infty }\to 0$. Continuous functions are dense in these $L^p$ spaces for a finite Borel measure on a compact metric space, so the contraction extends to a strongly continuous semigroup on $L^p(\mu )$.

For real $f$ with $f,f^2\in D(L)$, the kernel form of Jensen inequality gives $P_t(f^2)-(P_{t}f{)}^2\ge 0$, with equality at $t=0$. Take its right derivative to obtain the generator square inequalityThis is nonnegativity of the associated carré du champ.

Each time-average functional ${\varphi }_n$ is positive with ${\varphi }_n(1)=1$, hence represents a Borel probability measure ${\mu }_n$. The compactness of probability measures on a compact metric space gives a subsequence converging weakly to a probability measure $\mu$. Explicitly, $C(X)$ is a separable Banach space, so Banach-Alaoglu theorem and a countable dense family give a subsequence converging on every continuous function, rather than merely a net. For fixed $t\ge 0$, the semigroup property gives the Krylov-Bogolyubov time-average argument:Its absolute value is at most $2t\Vert f{\Vert }_{\infty }/n$. Pass to the subsequential limit, observing that $P_{t}f\in C(X)$, to get ${\int }_X{P}_{t}fd\mu ={\int }_{X}fd\mu$. Thus the limiting measure is invariant.

Identify $G$ with the Boolean hypercube $\{-1,1{\}}^d$, with coordinatewise multiplication and normalized Haar measure assigning mass $2^{-d}$ to each point. Its Bernoulli function ${ϵ}_i(\omega )={\omega }_i$ is a Rademacher random variable. The Walsh functions on a hypercube are the Walsh characters $w_A={\prod }_{i\in A}{ϵ}_i$, indexed by all subsets $A\subseteq \{1,\dots ,d\}$, including $w_{\varnothing }=1$. Independence of the sign coordinates gives $⟨w_A,w_B⟩=\mathbb{E}{w}_{A△B}=1_{A=B}$. There are $2^d$ characters, equal to the dimension of $L^2(G)$, so they form an orthonormal basis.

Flipping coordinate $i$ negates $w_A$ exactly when $i\in A$. Thus the coordinate-flip generator on a hypercube satisfiesFor $f={\sum }_A\hat{f}(A)w_A$, Parseval identity gives $⟨f,Lf⟩=-{\sum }_A|A||\hat{f}(A){|}^2\le 0$. Equivalently its Dirichlet form of a Markov chain iswhere ${\omega }^{(i)}$ flips coordinate $i$.

Set $S(\omega )={\sum }_i{a}_i{ϵ}_i(\omega )$ and $F(\omega )=\Vert S(\omega )\Vert$. Since $F(-\omega )=F(\omega )$, its Walsh expansion contains only sets $A$ of even size. Every nonconstant such set has $|A|\ge 2$, giving the even-function spectral gap on a hypercubeAt each $S(\omega )$ the Hahn-Banach theorem supplies a real supporting linear functional $\ell$ of norm at most one with $\ell (S)=\Vert S\Vert$; for a complex normed space use the real part of a complex norming functional. At $S=0$ take $\ell =0$. Convexity of the norm gives $F({\omega }^{(i)})-F(\omega )\ge \ell (S({\omega }^{(i)})-S(\omega ))$. Since $LS=-S$, sum these inequalities to get $LF\ge -F$. Therefore $\mathcal{E}(F)=-\mathbb{E}[FLF]\le \mathbb{E}{F}^2$. Combining the two bounds yields the sharp Rademacher second-moment inequalityNo smoothness of the norm is required. The constant $2$ is sharp: two equal nonzero real coefficients give modulus $0$ or $2|a|$ with equal probabilities. This is the second-versus-first moment case of the Kahane-Khintchine inequality for arbitrary normed vector spaces.

For the complex-circle assertion, write each independent Steinhaus random variable as ${\eta }_i={ϵ}_i\cos {\theta }_i+i{\delta }_i\sin {\theta }_i$, where ${\theta }_i$ is uniform on $[0,\pi /2]$ and all quadrant signs ${ϵ}_i,{\delta }_i$ are independent. Conditional on the angles, ${\sum }_i{a}_i{\eta }_i$ is a Rademacher sum with $2d$ complex coefficients $a_i\cos {\theta }_i$ and $i{a}_i\sin {\theta }_i$. Its conditional second moment is ${\sum }_i|a_i{|}^2$, independent of the angles. The preceding inequality gives its conditional first moment at least $\sqrt{\frac{1}{2}{\sum }_i|a_i{|}^2}$. Average over the angles and square to obtain the Steinhaus first-moment lower bound

For nonzero $A$, a sufficient step-size condition for strong convergence of relaxed Landweber iteration isFor the unrelaxed iteration, this reads $\Vert A{\Vert }^2<2$. The data must also admit a minimum-norm least-squares solution $x^{†}$; for the exact problem, $y\in \mathrm{Ran}A$ is sufficient. More generally, the projected-data condition in part (a), or the Picard criterion ${\sum }_i|⟨y,u_i⟩{|}^2/{\sigma }_i^2<\infty$, ensures the required solution.

Every positive singular value satisfies $|1-\tau {\sigma }_i^2|<1$, so for fixed ${\sigma }_i>0$,This is the correct inverse coefficient. For exact data with a least-squares solution, the norm error isThe summable majorant is precisely the Picard criterion, so the dominated convergence theorem proves strong convergence, despite the absence of a uniform contraction constant when the singular values approach zero. Initialization at zero selects the solution orthogonal to $\ker A$. At $\tau =2/\Vert A{\Vert }^2$, a top singular mode of a nonzero compact operator has multiplier $-1$ and can oscillate indefinitely; the strict step bound avoids this failure.

Pointwise convergence of filters alone is not a full noisy-data inverse-problem regularization proof. For each finite $n$, put $t=\tau {\sigma }^2$; the step restriction gives $|1-t|\le 1$, andThis proves the uniform noise bound for relaxed Landweber iterationTogether with the exact-data convergence,whenever $n=n(\delta )\to \infty$ and $\delta \sqrt{n(\delta )}\to 0$. Equivalently, the a priori regularization parameter choice satisfies $\alpha (\delta )\to 0$ and $\delta /\sqrt{\alpha (\delta )}\to 0$. For example, $n(\delta )=\lfloor {\delta }^{-1}\rfloor$ works as $\delta \to 0$. The sharper familiar bound $\delta \sqrt{n\tau }$ is available when $\tau \Vert A{\Vert }^2\le 1$.

Thus finite iteration followed by an appropriate stopping rule is a convergent regularization of an inverse problem. Taking $n\to \infty$ for fixed noisy data is not the same operation and can amplify the small-singular-value noise without bound. If $A=0$, the zero starting iterate stays zero and is the minimum-norm least-squares solution for every datum; the step restriction is needed only for nonzero $A$.

Put $s=\sqrt{-14}$. Then ${\mathcal{O}}_K=\mathbb{Z}[s]$ and the field discriminant is $-56$. Reduction of $X^2+14$ modulo $2$ and $3$ givesTheir ideal norms are $2,3,3$, respectively. The Minkowski bound for ideal classes is $(2/\pi )\sqrt{56}<5$, so every ideal class has an integral representative of norm at most $4$. The only possible representatives are the unit ideal, ${\mathfrak{p}}_2$, ${\mathfrak{p}}_3$, ${\overline{\mathfrak{p}}}_3$, and the norm-$4$ ideal ${\mathfrak{p}}_2^2=(2)$. Thus there are at most four classes.

The class of ${\mathfrak{p}}_2$ is nontrivial, since a principal ideal of norm $2$ would require $x^2+14y^2=2$. Also ${\mathfrak{p}}_3^2$ is not principal: norm $9$ forces $x+ys=\pm 3$, whose ideal is ${\mathfrak{p}}_3{\overline{\mathfrak{p}}}_3$, not ${\mathfrak{p}}_3^2$.

Now $N(2+s)=18$. It is divisible by ${\mathfrak{p}}_2$ and ${\mathfrak{p}}_3$, but not by ${\overline{\mathfrak{p}}}_3$, as substitution of $s=-1$ modulo $3$ gives $1$. Unique ideal factorization and norms therefore give $(2+s)={\mathfrak{p}}_2{\mathfrak{p}}_3^2$. Consequently $[{\mathfrak{p}}_3{]}^2=[{\mathfrak{p}}_2{]}^{-1}=[{\mathfrak{p}}_2]$, a nontrivial class of order $2$. The class of ${\mathfrak{p}}_3$ has order $4$, and the four-class bound is attained:

Take $f_n(t)=\sin (2\pi nt)$. Its domain norm is $1$, whereas its image norm is $1+2\pi n$. Hence $T$ is unbounded, so it has no finite operator norm and is not a compact operator.

The integration functional satisfies $|Tf|\le \Vert f{\Vert }_{\infty }{\int }_0^1|h|$. If $h$ never vanishes, its sign is constant and $f=\mathrm{sgn}h$ attains equality. In general use the continuous functions $f_{\epsilon }=h/\sqrt{h^2+\epsilon }$, of supremum norm at most $1$. Thenby the dominated convergence theorem. Therefore $\Vert T\Vert ={\int }_0^1|h|$. Every bounded subset of its finite-dimensional target $\mathbb{R}$ is relatively compact, so $T$ is compact, including when $h=0$.

Hölder's inequality states that for conjugate exponents $1/p+1/q=1$, $f\in L^p$ and $g\in L^q$ imply $fg\in L^1$ andIt includes $p=1,q=\infty$ (and the reversed case), with the essential-supremum norm for $L^{\infty }$.

For the real coefficients of the stated self-adjoint Sturm-Liouville problem, define the weighted inner productSince $w>0$, a nonzero eigenfunction has positive squared norm. Let $Ly=-(p{y}^{\prime }{)}^{\prime }+qy$. Integration by parts and the Dirichlet boundary conditions giveTaking $u=v$ to be an eigenfunction shows that its eigenvalue is real, since $\lambda \int w|u{|}^2=\int (p|u^{\prime }{|}^2+q|u{|}^2)$ is real. For two eigenfunctions with eigenvalues $\lambda ,\mu$, the same identity givesHence distinct eigenvalues imply orthogonality in the weighted inner product. This is the orthogonality of Sturm-Liouville eigenfunctions. The calculation assumes the regularity needed for the stated self-adjoint boundary problem, so all boundary terms and integrals are defined.

Convergence of a numerical method on a fixed interval $[0,T]$ means that, as the step size tends to zero and the starting approximations tend to the exact initial values,For a vector-valued ordinary differential equation, replace the absolute value by a norm. The interval remains fixed as $h\to 0$; a small error at only one step is insufficient.

The order of a numerical method is measured by its defect on a smooth exact solution. With the unscaled residual convention for a linear multistep method, order at least $p$ means local residual $O(h^{p+1})$; equivalently, the residual divided by $h$ is $O(h^p)$. Order exactly $p$ means the next coefficient does not vanish in general. For a zero-stable method, under the usual smoothness and Lipschitz continuity assumptions and with starting errors $O(h^p)$, this yields global error $O(h^p)$ on the fixed interval. Local order by itself does not imply convergence without stability.

Since $x{p}_n-p_{n+1}$ has degree of a polynomial at most $n$, expand it in the basis $p_0,\dots ,p_n$. Multiplication by $x$ is self-adjoint for this weighted inner product: $⟨xp,q⟩=⟨p,xq⟩$. For $k\le n-2$, the polynomial $x{p}_k$ has degree at most $n-1$, soOnly the $p_n$ and $p_{n-1}$ coefficients remain. Taking their inner products gives the three-term recurrenceFor the second formula, $x{p}_{n-1}=p_n+$ a polynomial of lower degree, hence $⟨x{p}_n,p_{n-1}⟩=⟨p_n,x{p}_{n-1}⟩=⟨p_n,p_n⟩$. At $n=0$, take $p_{-1}=0$, ${\beta }_0=0$ and the same formula for ${\alpha }_0$. Positivity of the weight ensures that all the squared norms in the denominators are nonzero.

A norm on a real or complex vector space $V$ is a function $\Vert \cdot \Vert :V\to [0,\infty )$ with $\Vert v\Vert =0$ exactly when $v=0$, $\Vert \lambda v\Vert =|\lambda |\Vert v\Vert$, and the triangle inequality $\Vert v+w\Vert \le \Vert v\Vert +\Vert w\Vert$. These are positivity, absolute homogeneity and the triangle inequality.

For a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$, define the operator norm byIt is finite. To see this without assuming continuity of $A$, write $x=\sum x_j{e}_j$: the triangle inequality bounds $\Vert Ax\Vert$ by $\sum |x_j|\Vert A{e}_j\Vert$. Every norm in finite dimensions is equivalent to the Euclidean norm. One proof first uses this same coordinate bound to establish continuity of the norm on the Euclidean unit sphere; compactness then gives a strictly positive minimum there. Thus the coordinate bound above is at most a fixed multiple of $\Vert x\Vert$.

Absolute homogeneity and the triangle inequality for this operator norm follow from those for $\Vert Ax\Vert$ and taking suprema. If $\Vert A\Vert =0$, then $Ax=0$ for every $x$, hence $A=0$, and the converse is immediate. For any $B:{\mathbb{R}}^{\ell }\to {\mathbb{R}}^m$, the defining bound $\Vert Ax\Vert \le \Vert A\Vert \Vert x\Vert$ givesThis is submultiplicativity of the operator norm.

Apply the adjoint operator identity of part (c) with $\chi =\hat{A}\varphi$ and use part (a):For a nonzero eigenfunction, positivity of the norm in particular forces $E\ge \hslash \omega /2$. Hence the requested norm isIt vanishes exactly at the ground-state energy.