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Choose the real embeddings ${\sigma }_i$ and one from each complex pair ${\tau }_j$. The logarithmic embedding of number field units isIt is a group homomorphism from multiplication to addition. A unit has absolute norm one, so the coordinate sum is zero and its image lies in a Euclidean hyperplane of dimension $r+s-1$.

Here is the required discreteness argument. There are only finitely many algebraic integers of $K$ whose conjugate absolute values are bounded by a fixed constant: their monic minimal polynomials have degree at most $n$, and each integer coefficient is an elementary symmetric function of bounded roots, hence belongs to a fixed finite range. There are finitely many such polynomials and finitely many roots of each. A bounded set of logarithms bounds all conjugate absolute values, so every compact region in the logarithmic hyperplane meets the unit image in finitely many points. Thus the image is a discrete additive subgroup.

Use the allowed Euclidean-group theorem: a discrete subgroup of a finite-dimensional real vector space is free abelian on finitely many vectors, with rank at most the dimension. Therefore $\lambda ({\mathcal{O}}_K^{\times })$ has rank at most $r+s-1$. Its kernel is finite by the same bounded-conjugate argument, now with every absolute value one. Each element of a finite multiplicative group is a root of unity, and conversely every root of unity has all logarithms zero. Thus the kernel is exactly the finite group of roots of unity, and exactly the torsion subgroup of units.

Lift generators of the free logarithmic image to units; together with the finite kernel they generate every unit. HenceThis proves the requested upper bound without invoking the existence half of the Dirichlet unit theorem. The field $\mathbb{Q}(\sqrt{11})$ is a subfield of $\mathbb{R}$, and its only real roots of unity are $\boxed{1,-1}$.

For a nonsingular smooth flow generated by $F$ on a closed manifold, the Anosov flow condition is a continuous $D{\varphi }_t$-invariant splittingand constants $C,\lambda >0$ such that, for all $t\ge 0$,The norm can come from any auxiliary Riemannian metric; compactness makes the resulting definitions equivalent. The centre direction is the flow direction, with $F$ nowhere zero.

One construction is a suspension flow. The hyperbolic toral automorphism of ${\mathbb{T}}^2$ defined byhas eigenvalues ${\lambda }_{\pm }=(3\pm \sqrt{5})/2$, with $0<{\lambda }_{-}<1<{\lambda }_{+}$. Form the closed three-dimensional mapping torus $({\mathbb{T}}^2\times [0,1])/((x,1)\sim (Ax,0))$ and move in the second coordinate at unit speed. Each crossing of the section applies $A$. The two one-dimensional eigenspaces give invariant transverse bundles, and after $k$ crossings their derivatives multiply by ${\lambda }_{-}^k$ and ${\lambda }_{+}^k$. The difference between $k$ and elapsed time is bounded, so any smooth Riemannian metric gives the required uniform exponential estimates. The vertical direction supplies the centre line.

A different construction is the geodesic flow on the unit tangent bundle of a closed oriented hyperbolic surface. This bundle is a closed three-manifold. In curvature $-1$, a normal Jacobi field along a unit-speed geodesic satisfies $j^{\prime \prime }-j=0$. The initial data $(j,j^{\prime })=(a,-a)$ give $j(t)=a{e}^{-t}$, while $(a,a)$ give $j(t)=a{e}^t$. The linearized geodesic flow transports these initial data as $(j(t),j^{\prime }(t))$, so they give one stable and one unstable direction with precisely those exponential rates. The remaining tangent direction is the geodesic vector field. Thus this also constructs an Anosov flow, using geometric rather than return-map hyperbolicity.

Put $\rho =1/f$ and denote the flow generated by $fF$ by ${\psi }_t$. It follows the same oriented orbits as ${\varphi }_s$, with the clock determined implicitly byBecause $f$ is positive on a compact manifold, $0<f_{\min }\le f\le f_{\max }$ and $f_{\min }t\le s(t,x)\le f_{\max }t$ for $t\ge 0$. Negative times obey the corresponding absolute-value comparison. This makes it possible to transfer exponential estimates once the correct transverse graphs are found.

For $v\in E^s(x)$ defineThe integrand is bounded by a constant times $e^{-\lambda r}\Vert v\Vert$, since $d\rho$ is bounded and $v$ is stable. The integral converges uniformly on bounded bundle vectors, so it is continuous and linear in $v$. Consider the graphWe verify its invariance and contraction directly; the old Anosov stable bundle need not itself remain invariant under a positive time change of a smooth flow.

For a fixed old time $s$, setDifferentiation gives $D_{x}T(v)=I_s(x,v)$ and $D_{x}T(F)=\rho ({\varphi }_{s}x)-\rho (x)$. Differentiating the implicit clock at fixed $t$, then differentiating ${\varphi }_{s(t,x)}(x)$, therefore yields for $w=v+aF(x)$For $a={\lambda }_s(x,v)$ this becomesThe tail coefficient equals ${\lambda }_s({\varphi }_{s}x,D{\varphi }_{s}v)$ by the flow composition law, proving graph invariance. Both the first term and the tail have norm at most $C^{\prime }{e}^{-\lambda s}\Vert v\Vert$. Projection onto $E^s$ in the original splitting is uniformly bounded, hence $\Vert v\Vert \le C^{\prime \prime }\Vert w\Vert$. The clock comparison now givesFor the Anosov unstable bundle, use the convergent past integralThe same differentiation formula gives the tail $-\rho ({\varphi }_{s}x{)}^{-1}{\int }_{-\infty }^{s}d\rho (D{\varphi }_{r}v)dr$ for $s\le 0$, proving invariance and exponential contraction in negative new time.

Finally, these two graphs and $\mathbb{R}F=\mathbb{R}(fF)$ form a direct splitting. Projecting onto $TN/\mathbb{R}F$ identifies the graphs with the old complementary stable and unstable subspaces, so they intersect neither each other nor the centre. We have established the continuous invariant splitting and both uniform estimates, and therefore the flow of $fF$ is Anosov.

First the common boundary distance function determines the induced tangential Riemannian metric. For a boundary curve $c(t)$ with $c(0)=x$ and $c^{\prime }(0)=w\in T_x\partial M$, the local first-order distance expansion givesFor clarity, the upper bound follows from the length of the boundary curve. In a small coordinate neighbourhood the Riemannian metric is arbitrarily close to its value at $x$, so the length of any sufficiently short joining path is bounded below by that constant-metric displacement norm with an arbitrarily small relative error. A minimizing short path cannot leave that neighbourhood. These two bounds prove the limit. Equality of the boundary distance functions then gives $|w{|}_{g_1}=|w{|}_{g_2}$. The polarization identity yieldsThis alone fixes no choice of normal coordinates, so a boundary-fixing gauge is needed to compare the full tangent-space Riemannian metrics.

Let $n_i$ be the inward unit normals and letbe sufficiently small boundary normal coordinates. Their pullbacks have the form $C_i^{\ast }{g}_i=d{r}^2+h_i(r)$, with $h_i(0)=h$. On a collar define $\psi =C_2\circ C_1^{-1}$. It fixes the boundary, has $D\psi (w)=w$ for tangential $w$, and sends $n_1$ to $n_2$. For tangent-space vectors $w+a{n}_1$ and $z+b{n}_1$ at the boundary, therefore,This proves the desired full Riemannian metric equality locally at the boundary.

We also justify extending this local identification to a global diffeomorphism. Interpolate the positive Riemannian metrics by $g_s=(1-s)g_1+s{g}_2$, for $0\le s\le 1$. Let $C_s$ be their inward unit-normal exponential collars. Compactness of the boundary and parameter interval gives a common small collar width for all $s$. The intermediate Riemannian metrics need not be simple; only these local collars are used.

On the image of $C_s$, define the time-dependent vector fieldwhere $\chi$ is one near zero and vanishes before the outer edge of the collar; extend $Y_s$ by zero elsewhere. This is a smooth vector field on $M$, vanishing at the boundary because $C_s(x,0)=x$. Its time-dependent flow ${\Psi }_s$ is a global diffeomorphism fixing the boundary. For every sufficiently small fixed $r$, the curve $s\mapsto C_s(x,r)$ solves its equation, since $\chi (r)=1$. Uniqueness gives ${\Psi }_s(C_0(x,r))=C_s(x,r)$. Consequently ${\Psi }_1$ agrees near the boundary with $C_2{C}_1^{-1}$. Taking $\psi ={\Psi }_1$ proves

Suppose that $H$ is not a discrete subgroup. The identity is then an accumulation point of nonidentity subgroup elements: translate any failure of isolation to the identity. Choose $h_j\in H\setminus \{e\}$ with $h_j\to e$, and use a local exponential chart to write $h_j=\exp X_j$, where $X_j\ne 0$ and $X_j\to 0$ in the real Lie algebra $\mathfrak{g}$.

Choose a norm on this finite-dimensional vector space. By compactness of its unit sphere, pass to a subsequence with $X_j/\Vert X_j\Vert \to X$, where $\Vert X\Vert =1$. For an arbitrary real $t$, choose an integer $m_j$ nearest to $t/\Vert X_j\Vert$. Then $m_j\Vert X_j\Vert \to t$ andEvery left-hand element belongs to $H$, including for negative $m_j$. Closedness gives $\exp (tX)\in H$. Since $t$ was arbitrary and $X\ne 0$, the limit directions of a closed subgroup provide a nontrivial one-parameter subgroup in $H$. Its derivative at zero is $X$, so it cannot be the constant subgroup.

An extending absolute value on a field is non-Archimedean by Question 1(a). We prove finite-dimensional non-Archimedean norm equivalence over a complete field, using completeness rather than local compactness.

Fix a basis $e_1,\dots ,e_d$ of a finite-dimensional vector space over $K$, and let $\Vert \cdot \Vert$ be a norm homogeneous for the given absolute value on a field and satisfying the ultrametric inequality. Write $\Vert \sum a_i{e}_i{\Vert }_0={\max }_i|a_i|$. The upper boundis immediate. For the lower bound use induction on $d$. The one-dimensional case is homogeneity. For the induction step, the span $W$ of $e_1,\dots ,e_{d-1}$ is complete for its restricted norm by the induction comparison with the coordinate norm and completeness of $K$. A complete vector subspace of a metric space is closed, soFor $v=w+a{e}_d$ with $a\ne 0$, scaling by $a$ gives $\Vert v\Vert \ge |a|\delta$; the same coefficient bound is trivial when $a=0$. AlsoThe induction bound controls every coordinate of $w$ by a constant times $\Vert w\Vert$. Together with the bound on $a$, this proves $\Vert v{\Vert }_0\le C^{\prime }\Vert v\Vert$.

Two extending absolute values on a field on $L$ are norms of this kind on its finite-dimensional vector space over $K$. Hence there is $B>0$ with $|x{|}_2\le B|x{|}_1$ for every $x\in L$. Apply this to $x^N$ and use multiplicativity:Letting $N\to \infty$ gives $|x{|}_2\le |x{|}_1$. Interchanging the two norms gives equality. ThusThe proof covers the trivial base absolute value on a field too. It proves uniqueness, without assuming an existence theorem or compactness of the coordinate sphere $\{v:\Vert v{\Vert }_0=1\}$.

The relevant Banach-Alaoglu theorem says that for $1<p<\infty$ the closed bounded balls of $L^p({\mathbb{R}}^n)$ are compact in the weak topology. In particular, every bounded sequence has a weakly convergent subsequence whose limit has norm no larger than the common bound. Applying the preceding duality of Lp spaces also identifies this as weak-star compactness of $L^p=(L^q{)}^{\ast }$; its weak and weak-star test spaces are both $L^q$.

Here is a direct proof of both compactness and the subsequence assertion. Let $\Vert f_j{\Vert }_p\le M$ and choose a countable dense set $\{g_m\}$ in the separable Lp space $L^q$. Each scalar sequence $\int f_j{g}_m$ is bounded. Successive subsequence selection followed by the diagonal subsequence argument gives $f_{j_k}$ for which these pairings converge for every $m$. For arbitrary $g\in L^q$,First approximate $g$ by $g_m$ and then take $k,l$ large. This proves convergence of every pairing, and the limit $\Lambda (g)={\lim }_k\int f_{j_k}g$ is linear with $|\Lambda (g)|\le M\Vert g{\Vert }_q$. Apply the already proved duality of Lp spaces with $q$ in place of $p$: $\Lambda (g)=\int fg$ for an $f\in L^p$ with $\Vert f{\Vert }_p\le M$. ThusTo obtain compactness rather than only a sequential statement, on the radius-$M$ ball defineDensity and Hölder's inequality show that this is a metric space distance and that its topology is exactly the weak topology on this ball: every remaining test pairing is uniformly approximated there by a dense-set pairing. The subsequence argument proves sequential compactness for this metric. A sequentially compact metric space is compact: otherwise either an infinite separated sequence contradicts total boundedness or a Cauchy sequence without a limit contradicts sequential compactness; completeness and total boundedness give compactness by successive finite coverings. This completes the proof. The same argument is the Sequential Banach-Alaoglu theorem for a separable predual.

Write $A_q=⟨f{⟩}_{w,q}$ and $N_q=\Vert f{\Vert }_{w,q}$. The exponent on the finite-set expression is $-1/q^{\prime }$ in the PDF; the TeX aid drops its prime. Ignore zero-measure sets, whose integrals vanish. The equivalence isFor the first bound, let $E_a=\{|f|>a\}$ and choose a finite-measure subset $E\subset E_a$. Since ${\int }_E|f|\ge a|E|$,Exhaust $E_a$ by its intersections with bounded balls. If its measure is finite these give the desired bound $a|E_a{|}^{1/q}$ in the limit; if it is infinite they force $N_q=\infty$. Now take the supremum over $a$.

For the other direction it suffices to assume $0<A_q<\infty$; the zero and infinite cases are immediate. For a set $E$ with $0<|E|=m<\infty$, the layer cake representation givesSplit at $t_0=A_q{m}^{-1/q}$. The first contribution is $m{t}_0$, and the second is $A_q^q{t}_0^{1-q}/(q-1)$. Their sum is $q^{\prime }{A}_q{m}^{1/q^{\prime }}$. Multiply by $m^{-1/q^{\prime }}$ and take the supremum. This proves the equivalence even when the quantities are extended-valued. The integral expression is an actual norm on the weak Lq space, since the triangle inequality holds inside every set integral. The distribution-function expression is generally a quasi-norm. The factor $q^{\prime }$ is sharp: $f(x)=x^{-1/q}$ on $(0,\infty )$ has $A_q=1$ and attains $N_q=q^{\prime }$ on every initial interval. In higher dimensions, multiply this example by the indicator of a unit box in the remaining coordinates.

As usual, the bounded set is assumed Lebesgue measurable and an $L^q$ exponent is positive; the conventional Banach-space assertion concerns $1\le q<2^{\ast }=2n/(n-2)$. No boundary regularity of $\Omega$ is needed. Choose a smooth compactly supported cutoff $\eta$ equal to one on a ball containing $\Omega$. Weak convergence and the Uniform boundedness principle give a common $H^1$ bound for $f_j$, hence also for $g_j=\eta f_j$. These functions have a common compact support.

We prove their local $L^2$ compactness directly. Translation and the fundamental theorem along line segments give, by density from smooth functions,For a compactly supported mollifier ${\rho }_{\epsilon }$, integration of this estimate givesAt fixed $\epsilon$, Hölder's inequality bounds $\Vert g_j\ast {\rho }_{\epsilon }{\Vert }_{\infty }$ and $\Vert \nabla (g_j\ast {\rho }_{\epsilon }){\Vert }_{\infty }$ uniformly by the $L^2$ norm of the corresponding smoothing kernels. Their supports lie in one fixed compact set, so the Arzelà-Ascoli theorem makes this smoothed family precompact in $L^2$. The uniform $O(\epsilon )$ approximation makes the original family totally bounded in $L^2$: choose a finite net for the smoothed family and add the approximation error. Completeness gives relative compactness.

Every strongly convergent subsequence of $g_j$ has limit $\eta f$, since multiplication by $\eta$ preserves weak $L^2$ convergence. Consequently the whole sequence converges strongly: otherwise a subsequence with distance at least a fixed positive number would have a further convergent subsequence and contradict that unique limit. In particular $\Vert {\chi }_{\Omega }(f_j-f){\Vert }_2\to 0$.

The critical Sobolev inequality supplies a uniform bound in $L^{2^{\ast }}$. For clarity, this estimate follows from part 2(a), not from compactness at the critical exponent. For a compactly supported smooth $v$, integrating by parts against the vector field $(x-y)/|x-y{|}^n$ givesIts distributional divergence is the sphere-area multiple of a point mass; applying the Hardy–Littlewood–Sobolev inequality with $\alpha =1,p=2$ yields $\Vert v{\Vert }_{2^{\ast }}\le C\Vert \nabla v{\Vert }_2$. Density extends this to $H^1({\mathbb{R}}^n)$.

For $2<q<2^{\ast }$ choose $0<\theta <1$ with $1/q=\theta /2+(1-\theta )/2^{\ast }$. Hölder's inequality givesFor $1\le q<2$, use $\Vert v{\Vert }_{L^q(\Omega )}\le |\Omega {|}^{1/q-1/2}\Vert v{\Vert }_{L^2(\Omega )}$. The same integral estimate also works for $0<q<1$ if $L^q$ is understood with its usual quasi-norm. Thus ${\chi }_{\Omega }{f}_j\to {\chi }_{\Omega }f$ strongly for every positive $q<2^{\ast }$. We have never asserted that multiplying by the possibly nonsmooth indicator preserves $H^1$.

For a bounded connected domain with the interior cone property, and $1\le p<n$, the mean-zero form of the Poincaré inequality isNo zero boundary trace is imposed. Subtracting the mean is essential because nonzero constants have zero gradient.

Suppose the estimate failed. Subtract means and normalize to obtain $v_j\in W^{1,p}(\Omega )$ withThis sequence is bounded in the Sobolev space $W^{1,p}$. The allowed Rellich-Kondrachov compactness theorem on cone domains provides a subsequence converging strongly in $L^p(\Omega )$ to $v$. Its mean is zero and its norm is one. For any compactly supported smooth test function $\varphi$,Thus every weak derivative of $v$ vanishes. To show this implies constancy, mollify on any ball compactly contained in $\Omega$. The resulting smooth function has zero gradient and is constant on the smaller ball. Passing to its $L^p$ limit makes $v$ constant almost everywhere on that ball. Overlapping balls have the same constant, and connectedness connects any two points by a chain of such balls. Hence $v$ is constant almost everywhere throughout $\Omega$. Its zero mean makes it zero, contradicting norm one. This proves the stated Poincaré inequality. The frequently used stronger Sobolev-Poincare formulation also follows: with $p^{\ast }=np/(n-p)$, the cone-domain Sobolev embedding theorem gives $\Vert w{\Vert }_{p^{\ast }}\le C_S(\Vert w{\Vert }_p+\Vert \nabla w{\Vert }_p)$. Insert $w=v-v_{\Omega }$ and the estimate just proved to obtainThus both the same-exponent and critical-exponent mean-zero versions are covered. Connectedness cannot be dropped without subtracting a separate mean on every component. The usual meaning of the stated range is $1\le p<n$; $W^{1,p}$ with $p<1$ is not part of this Sobolev-space formulation.

Choose the scattering-potential convention $V(\mathbf{r})=k^2[n(\mathbf{r}{)}^2-1]$. Let $G_k(\mathbf{r},{\mathbf{r}}^{\prime })=e^{ik|\mathbf{r}-{\mathbf{r}}^{\prime }|}/(4\pi |\mathbf{r}-{\mathbf{r}}^{\prime }|)$, so $(\Delta +k^2)G_k=-\delta$. Since $(\Delta +k^2)\psi =-V\psi$, the outgoing Green representation is the Lippmann-Schwinger equationIterating it produces the Born seriesFor example its first two terms areThe first term describes one scattering event driven by the incident field. The second propagates its first scattered wave to another interaction before reaching the observer. Higher terms contain more successive interactions, including repeated visits to the same region; they account for multiple scattering rather than additional incident waves.

A sufficient convergence condition on an appropriate field norm is $\Vert T\Vert =\rho <1$. Then the first-Born remainder is bounded by ${\rho }^2\Vert {\psi }_i\Vert /(1-\rho )$, while $\Vert T{\psi }_i\Vert \le \rho \Vert {\psi }_i\Vert$. A simple concrete sufficient condition on the bounded scatterer, using the supremum norm, isThis makes the field inside the inhomogeneity close to the incident field. Physically weak index contrast with small accumulated phase shift, such as $k\ell |n-1|\ll 1$ for a smooth weak scatterer of thickness $\ell$, is a common first-Born regime, provided resonant enhancement and multiple reflections are negligible. Weak local contrast alone is not sufficient for an arbitrarily thick scatterer. If the opposite sign convention for $V$ is used, the integral operator changes sign and the same consistent iteration applies.

Using the singular value system with $A{u}_j={\sigma }_j{v}_j$ and $A^{\ast }{v}_j={\sigma }_j{u}_j$, the formal minimum-norm inverse isIt defines an element of the unknown Hilbert space only if ${\sum }_j|(y,v_j){|}^2/{\sigma }_j^2<\infty$, the Picard criterion. A null-space component is undetermined by the data.

Adding an error along a single left singular vector changes precisely one coefficient:Although this perturbation has data norm $\delta$, its amplification factor is $1/{\sigma }_j$. If ${\sigma }_j\to 0$, these factors are unbounded. More explicitly choose $y^{(j)}={\sigma }_j{v}_j$, tending to zero, while the minimum-norm inverse is $u_j$, of norm one. Thus the inverse is discontinuous at zero. It may also fail to exist for data outside its nonclosed range, so merely minimizing a residual does not guarantee that the formal inverse exists. This is ill-posedness by unbounded small-singular-value amplification, not a claim that every finite-dimensional or bounded forward map has an unstable inverse.