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For real square-integrable $f,g$, nonnegativity of $\int (f-\lambda g{)}^{2}dx$ gives a quadratic in the real number $\lambda$. If $\int g^{2}dx>0$, choose $\lambda =\int fgdx/\int g^{2}dx$ to obtainIf $\int g^{2}dx=0$, then $g=0$ almost everywhere and the same inequality is immediate. This proves the integral Cauchy-Schwarz inequality.

Put $T=t_f-t_i>0$ and $L=x_f-x_i>0$. The forward speed satisfies $L={\int }_{t_i}^{t_f}vdt$, so $V_t=L/T$. The change of variables $dx=vdt$ gives $V_x=L^{-1}{\int }_{t_i}^{t_f}{v}^{2}dt$. Apply the Cauchy-Schwarz inequality to $v$ and the constant function one:Equality holds exactly when $v$ is constant almost everywhere. Equivalently, the time and distance averages of speed satisfy $V_x-V_t={\mathrm{Var}}_t(v)/V_t$, which identifies the excess as a measure of speed variation.

For a unit direction $\mathbf{v}$, the Cauchy-Schwarz inequality gives $\nabla f\cdot \mathbf{v}\le |\nabla f|$, with equality in the gradient direction. Its norm here is $\sqrt{4+9+1}=\sqrt{14}$, so the direction of fastest increase isThe maximal directional derivative itself is $\sqrt{14}$.

Assume real square-integrable functions on the same integration domain. Put $A=\int f^2$, $B=\int g^2$, and $C=\int fg$. If $B=0$, then $g=0$ almost everywhere, so $C=0$ and the claim follows. If $B>0$, nonnegativity of a square givesMultiplying by $B$ proves the integral form of the Cauchy-Schwarz inequality:For nonzero functions, equality holds exactly when they are proportional almost everywhere. The zero-function cases also give equality.

Assume $\alpha <\beta$ and abbreviateThe integral of the nonnegative square expands to $A+2\lambda B+{\lambda }^{2}C\ge 0$ for every real $\lambda$. If $C>0$, completing the square, or choosing $\lambda =-B/C$, gives $A-B^2/C\ge 0$. Hence the Cauchy-Schwarz inequality isIf $C=0$, continuity and nonnegativity imply $g=0$ throughout the interval, so the same inequality is immediate.

When $C>0$, equality holds exactly when $\int (f-(B/C)g{)}^{2}dx=0$. The integrand is continuous and nonnegative, so it must vanish everywhere, giving $f=(B/C)g$. Including the case of a zero function, equality is equivalent to linear dependence of $f$ and $g$ as elements of the real vector space of continuous functions on the interval.

Equality in the Cauchy-Schwarz inequality requires $\sqrt{r}v(r)=\Omega r^{3/2}$ for one constant $\Omega$. Thus $v(r)=\Omega r$ andThis is solid-body rotation. At fixed $L$, the constant is $\Omega =2L/(V{a}^2)$. The value at the single axis point $r=0$ does not affect the integrals; if $\omega$ is continuous, it has the same constant value there as well.

Using the same cylindrical-shell volume,For unit density, this is the axial angular momentum. To bound it, apply the integral Cauchy-Schwarz inequality to $f(r)=\sqrt{r}v(r)$ and $g(r)=r^{3/2}$:Multiplication by $(2\pi h{)}^2$ gives $L^2\le \pi h{a}^{4}K$. Since the chamber's volume is $V=\pi a^{2}h$, this is precisely

Length functional. On the unit sphere, write the position vector in spherical coordinates asIts two coordinate tangent vectors satisfy $r_{\theta }\cdot r_{\varphi }=0$, $|r_{\theta }|=1$ and $|r_{\varphi }|=\sin \theta$. Consequently the induced Riemannian metric is $d{s}^2=d{\theta }^2+{\sin }^2\theta d{\varphi }^2$. On a segment for which $\theta$ increases and $\varphi =\varphi (\theta )$,Segments where $\theta$ is not monotone require another parameter or a subdivision. Endpoint turning points can be handled by a limit.

First integral. With $F=\sqrt{1+{\sin }^2\theta ({\varphi }^{\prime }{)}^2}$, there is no explicit $\varphi$ dependence. The Euler-Lagrange equation therefore says thatis constant. For $c\ne 0$, put $k=1/|c|$ and solve:The radicand must be nonnegative along the chosen segment. The printed expression with both limits fixed at ${\theta }_1,{\theta }_2$ describes a longitude difference between endpoints; a formula for $\varphi (\theta )$ needs the variable upper limit and the additive constant shown here. The case $c=0$ gives a meridian with constant longitude. The equator is also a stationary route, although it cannot be parameterized using changing $\theta$.

Integration and geometric meaning. Put $t=\cot \theta$, so $d\theta =-dt/(1+t^2)$ andFor a nonconstant segment $k>1$, and writing $K=\sqrt{k^2-1}$, the integral becomesRearranging, and absorbing phase and sign into two constants, givesMultiplying by $\sin \theta$ shows that the corresponding Cartesian coordinates satisfy $z=Ax+By$. Thus every such stationary route lies in a plane through the sphere's center: it is a great circle, the geodesic of the sphere. The meridians and equator are the limiting or separately parameterized cases of the same description.

Rome to Singapore. Use exactly the approximate coordinates supplied in the question: $(\theta ,\varphi )=(\pi /4,\pi /12)$ at departure and $(\pi /2,7\pi /12)$ at arrival. Inserting them into the preceding relation determinesIndeed the right side is one at Rome and zero at Singapore. This corresponds to $K=1$, $k=\sqrt{2}$. In latitude $\ell =\pi /2-\theta$, the route is $\ell =\arctan [\cos (\varphi -\pi /12)]$: it starts at $45^{\circ }$ north with an initially eastward heading and then curves southward to the equator. The first-integral expression in $\theta$ has an integrable endpoint singularity at departure, where $\theta$ has a turning point; the longitude parameter is regular there.

For an explicit regular arc-length parametrization, let $a=r(\pi /4,\pi /12)$ and $b=r(\pi /2,7\pi /12)$. Their dot product is $\sin (\pi /4)\cos (\pi /2)=0$. HenceBoth $|r(s)|$ and $|r^{\prime }(s)|$ equal one. The shorter great circle arc is the global shortest path: for any unit-speed curve from $a$, its central angle $\alpha =\arccos (a\cdot r)$ satisfies $|d\alpha /ds|\le 1$ by the Cauchy-Schwarz inequality, since the tangential projection of $a$ has length $\sin \alpha$. Integrating gives length at least the endpoint angle $\pi /2$, which this route attains. For a sphere of radius $R$, the corresponding shortest distance is $\pi R/2$.

A unitary matrix satisfies $U^{†}U=I$ over the complex numbers; a real orthogonal matrix satisfies $Q^{T}Q=I$. Their inverses are their conjugate transposes and transposes, respectively.

With $d=-a$, the coordinate $e_2=(0,1,0)$ has eigenvalue $c$. In the $x,z$ plane, the traceless symmetric two-by-two matrix has eigenvalues $\pm s$, where $s=\sqrt{a^2+b^2}$. If $s>0$, choose $\theta$ with $\cos 2\theta =a/s$ and $\sin 2\theta =b/s$. Then an orthonormal eigenbasis isIf $c$ equals $s$ or $-s$, the corresponding eigenspace is the span of the two listed vectors with that eigenvalue. If $s=0$, then $a=b=0$, the $x,z$ plane is the zero eigenspace, and $e_2$ retains eigenvalue $c$; if also $c=0$, every vector belongs to the zero eigenspace. This includes the degenerate cases that a division by $b$ would miss.

For the Cauchy-Schwarz inequality, take $b\ne 0$ and remove the orthogonal projection of $a$ along $b$. Positivity of the squared norm givesMultiplication by $\Vert b{\Vert }^2$ and square roots prove $|⟨a,b⟩|\le \Vert a\Vert \Vert b\Vert$; the case $b=0$ is immediate. Equality means linear dependence. ThereforeTaking the nonnegative square root proves the triangle inequality.

Use the complex inner product $⟨a,b⟩={\sum }_j\overline{a_j}{b}_j=a^{†}b$, conjugate-linear in its first argument. For $a\ne 0$, put $c=⟨a,b⟩/⟨a,a⟩$. Positivity givesHence the Cauchy-Schwarz inequality is $|⟨a,b⟩|\le \Vert a\Vert \Vert b\Vert$. If $a=0$ it is immediate; equality holds precisely when the two vectors are linearly dependent.

For a unitary matrix, $U^{†}U=I$, so $\Vert Ua{\Vert }^2=a^{†}{U}^{†}Ua=\Vert a{\Vert }^2$. If $Uv=\lambda v$ with a nonzero eigenvector, this implies $|\lambda |\Vert v\Vert =\Vert v\Vert$. Thus every eigenvalue satisfiesFor the one-parameter family, expand unitarity to first order: $(I+t{A}^{†}+O(t^2))(I+tA+O(t^2))=I$, so $A^{†}+A=0$. Thus $A$ is a skew-Hermitian matrix, an element of the unitary Lie algebra. If $Av=\mu v$, then $v^{†}Av=\mu \Vert v{\Vert }^2$, whereas complex conjugation gives $(v^{†}Av{)}^{\ast }=v^{†}{A}^{†}v=-v^{†}Av$. Consequently $\overline{\mu }=-\mu$, proving that every eigenvalue of $A$ is purely imaginary, with zero permitted.

A curve element has arc length $ds=\sqrt{1+z^{\prime 2}}dx$ and mass $dm=\mu ds$. Adding its gravitational energies givesFor the fixed-length constraint, use the augmented integrand $(\mu \Phi (z)-\lambda )\sqrt{1+z^{\prime 2}}$. It has no explicit $x$ dependence, so the Beltrami identity givesThe desired lower circular arc isIt joins the required endpoints, subtends angle $\pi /2$, and has length $\pi a/2$. Its slope gives $\sqrt{1+z^{\prime 2}}=a/(h-z)$. Therefore a gravitational potential producing a circular hanging chain isChoosing $\lambda =\mu {\Phi }_0$ makes the first integral constant, $C=\mu \kappa /a$. Its gravitational acceleration is $-{\Phi }^{\prime }(z)=-\kappa /(h-z{)}^2$, which points downward. An arbitrary additive potential constant changes no force.

This construction really gives a minimum. With $w=h-z>0$, the variable part of $V$ is $\mu \kappa \int ds/w$; the ${\Phi }_0$ term is fixed by the length. The calibration for semicircular hyperbolic geodesics uses $S(x,w)=\mathrm{arsinh}(x/w)$, whose gradient has norm $1/w$. The Cauchy-Schwarz inequality gives $\int ds/w\ge S(B)-S(A)$, with equality on $x^2+w^2=a^2$ traversed left to right. Thus the circular arc minimizes the energy among competitors of the same length and endpoints in the stated potential domain, rather than merely satisfying a necessary stationarity condition.

Interpret the density $1/y$ on the physical region $y>0$. It is singular at zero and would not describe a nonnegative density for $y<0$. The first integral gives $y\sqrt{1+y^{\prime 2}}=R$ for a positive constant $R$. To handle the horizontal tangent without dividing by $y^{\prime }$, use the full Euler-Lagrange equation:Two integrations give $y^2=-(x-x_0{)}^2+R^2$. The equal endpoint heights force $x_0=0$, and either endpoint then gives $R^2=2$. ThusThe path is the upper circular arc. A constant horizontal path solves the differentiated first integral trivially but fails $y{y}^{\prime \prime }+y^{\prime 2}+1=0$, so is not another extremum.

One can prove global minimality, rather than only stationarity. Define $\Phi (x,y)=\mathrm{arsinh}(x/y)$ on the upper half-plane. ThenThe Cauchy-Schwarz inequality gives, along any admissible path from $A$ to $B$,Our circle has tangent $(1,-x/y)$ for increasing $x$, so equality holds pointwise. This calibration for semicircular hyperbolic geodesics provesThe geometry is that of a hyperbolic geodesic in the Poincaré half-plane model. The proof even allows competing upper-half-plane paths beyond the class of monotonic graphs.

The endpoint constraints are $y(-a)=y(a)=0$. The constants in the three costs contribute only $2aK$, $2a{K}^{\prime }$ and $2a{K}^{\prime \prime }$ respectively, so they do not affect the minimizing shapes. For companies B and C, the fixed arc length isThe original PDF has $(y^{\prime }{)}^2$ as company A's first variable term. The local TeX transcription has lost that derivative dot and incorrectly repeats $y^2$; the calculation below follows the PDF.

Company A. The relevant functional is $J_A=\int (y^{\prime 2}+y^2-2xy)dx$. The Euler-Lagrange equation givesIts general solution is $y=x+c_1\cosh x+c_2\sinh x$. Adding and subtracting the endpoint equations gives $c_1=0$ and $c_2=-a/\sinh a$. ThusThis curve is odd, negative on the left half and positive on the right half; company A has no lower-half-plane restriction. It is the unique global minimum. Indeed, for any endpoint-zero variation $\eta$, integration by parts and the differential equation cancel all linear terms, leavingunless $\eta =0$.

Company B. Minimize $\int ydx$ under the length constraint. Introduce a constant Lagrange multiplier $\lambda$ and apply the Euler-Lagrange equation to $y+\lambda \sqrt{1+y^{\prime 2}}$:The curvature is constant, so the required lower arc is a circular arc. Writing its radius as $R\ge a$ and imposing the endpoints givesSince $\sqrt{1+y_B^{\prime 2}}=R/\sqrt{R^2-x^2}$, its length determines the radius:Equivalently, introduce $0<\theta \le \pi /2$ with $a=R\sin \theta$. Then $L/(2a)=\theta /\sin \theta$, which increases strictly from $1$ to $\pi /2$. The stipulated range of $L$ therefore selects one minor arc. At $L=\pi a$ it becomes the lower semicircle, with vertical endpoint tangents; it is still a single-valued finite-length graph.

Global minimality follows from the convexity of the augmented slope cost with $\lambda =R$. For a competitor $y=y_B+\eta$,Consequently the augmented cost difference is at least $\int (\eta +x{\eta }^{\prime })dx=[x\eta {]}_{-a}^a=0$. Competitors have the same length, so the multiplier contribution cancels and the original cost is minimized. This proves that a circular arc maximizes area below a fixed chord: minimizing negative signed height maximizes the area between the fence and the chord.

Company C. The variable cost is $J_C=\int yds=\int y\sqrt{1+y^{\prime 2}}dx$. The augmented integrand is $(y+\lambda )\sqrt{1+y^{\prime 2}}$. Since it has no explicit $x$ dependence, the Beltrami identity givesThe hanging branch with $b>0$ integrates to $y+\lambda =b\cosh [(x-x_0)/b]$. Equal endpoint heights force $x_0=0$, and the endpoint value fixes $\lambda =b\cosh (a/b)$. ThusIts length isWriting $q=a/b$, the equation $L/(2a)=\sinh q/q$ has one positive solution for every $L>2a$: its derivative has numerator $q\cosh q-\sinh q>0$. The curve is a catenary, is even, and lies strictly below the chord except at the endpoints.

A direct argument proves this is a global minimum, rather than merely an extremum of an augmented functional. Parametrize any competitor by arc length $s\in [-L/2,L/2]$, from its left endpoint to its right endpoint. Then $x^{\prime 2}+y^{\prime 2}=1$, $\int x^{\prime }ds=2a$, and integration by parts givesThe Cauchy-Schwarz inequality gives the last step. Equality requires $x^{\prime }=b/\sqrt{b^2+s^2}$ and $y^{\prime }=s/\sqrt{b^2+s^2}$, whose integrals are precisely the displayed catenary with the stated length. Thus it attains a lower bound valid for every competitor. This proves that a catenary minimizes the integral of height along a fixed-length curve.

The sketch uses $a=1$ and $L=2.7$ for the two fixed-length cases. The vertical scale in the company A panel is enlarged to show its small odd deflection. No comparison between the three companies' absolute prices is required or possible from the shapes alone, since their additive cost constants are independent.

Introduce a constant Lagrange multiplier $\mu$ for this isoperimetric constraint and make $A-\mu (C-B)$ stationary. Its integrand is $(1-\mu )y-\mu y^{\prime 2}$, so the Euler-Lagrange equation givesThe endpoint conditions therefore force a parabola $y=Kx(1-x)$. Its area and budget areSolving the quadratic and choosing positive area givesThis requires $B>0$; no positive-area barrier is possible for a nonpositive budget. The corresponding multiplier is $\mu =1/(1+4K)>0$.

To verify a global maximum rather than just stationarity, use the sharp area-cost bound for a barrier with fixed endpoints. For $w=x(1-x)$, integration by parts gives $2A={\int }_0^1{y}^{\prime }{w}^{\prime }dx$, and ${\int }_0^1{w}^{\prime 2}dx=1/3$. The Cauchy-Schwarz inequality impliesFor $A>0$ the last expression increases strictly, yielding exactly the boxed upper bound. Equality requires $y^{\prime }$ proportional to $w^{\prime }$, hence $y=Kw$ after imposing the endpoints. Thus the displayed optimum is unique and lies above the baseline throughout the interior.

Write the endpoint condition in its linear Robin boundary condition form $y^{\prime }(1)=y(1)/\alpha$. For a positive eigenvalue $\lambda =q^2$, the left Dirichlet boundary condition gives $y=C\sin (qx)$, and the right condition givesFor each positive root $q_n$, the squared norm isConsequently the sine coefficients in any complete expansion areWhen $\alpha >1$, these are all modes and the Sturm-Liouville eigenfunction expansion gives exactly the stated sine series, with the usual convergence interpretation. Positivity of all eigenvalues follows from integration by parts and the Cauchy-Schwarz inequality:for a nonzero admissible $y$, since $y(1)={\int }_0^1{y}^{\prime }dx$.

However, the original PDF states only $\alpha >0$, and with that range its sine-only completeness claim is false. The missing modes are as follows. At $\alpha =1$, $\lambda =0$ admits $y=x$, so the correct expansion isFor $0<\alpha <1$, a negative eigenvalue $\lambda =-{\kappa }^2$ admits $y=\sinh (\kappa x)$, withThere is exactly one positive root: $\tanh \kappa /\kappa$ decreases strictly from $1$ to zero. For example, its derivative has the sign of $\kappa {\mathrm{sech}}^2\kappa -\tanh \kappa <0$, because $\sinh \kappa \cosh \kappa >\kappa$. Thus the corrected expansion isThese are the nonpositive modes in a Dirichlet-Robin interval expansion. As an explicit counterexample to the printed claim, at $\alpha =1$ take $f(x)=x$: it satisfies both boundary conditions, and each positive root obeysbecause $\tan q_n=q_n$. Thus all $a_n$ vanish despite $f\ne 0$. Likewise $f=\sinh (\kappa x)$ is a counterexample for $\alpha <1$, by orthogonality of Sturm-Liouville eigenfunctions with distinct eigenvalues. The formulas above give the full solution without silently assuming away the omitted modes.

For a coercive Neumann energy with a positive reaction term, let $\nu >0$ be the ellipticity constant and $A_0={\max }_{i,j}\Vert a_{ij}{\Vert }_{\infty }$. Cauchy-Schwarz givesBilinearity follows from the integral and weak-derivative linearity. Ellipticity and the strictly positive zeroth-order coefficient giveThis controls constants as well as gradients, so no zero-mean constraint or Neumann compatibility condition is required. Symmetry of the coefficients makes $B$ symmetric.

Write $F(v)={\int }_{\Omega }fv$. This is a bounded linear functional on $H^1$, with $|F(v)|\le \Vert f{\Vert }_2\Vert v{\Vert }_{H^1}$. Consequentlyas the norm tends to infinity. The infimum is finite, and every minimizing sequence is bounded. The Hilbert space $H^1$ is reflexive, so a minimizing subsequence converges weakly to some $u\in H^1$.

To verify lower semicontinuity directly, put $w_j=v_j-u$. Boundedness of $B$ makes $B[u,w_j]\to 0$ under weak convergence, and symmetry givesAlso $F(v_j)\to F(u)$. Therefore $I(u)\le \mathrm{liminf}I(v_j)$, so the absolute minimum is attained.

For every $h\in H^1$ the real function $t\mapsto I(u+th)$ has a minimum at zero. Its derivative there is $B[u,h]-F(h)$, yielding the required weak equation for every $h$. If two weak solutions exist, their difference $z$ satisfies $B[z,h]=0$ for all $h$; choose $h=z$ and use coercivity to obtain $z=0$. In fact the weak equation also gives the exact energy identityThus the minimizer and weak solution are both unique. The natural boundary condition is encoded by testing in the whole $H^1$ space, without needing pointwise conormal derivatives for rough coefficients.

Give $H_0^1$ the equivalent Hilbert space norm $\Vert u{\Vert }_H=\Vert \nabla u{\Vert }_2$, using the Poincaré inequality. DefineThe Cauchy-Schwarz inequality gives $|a(u,v)|\le (1+C_P\Vert b{\Vert }_{\infty })\Vert u{\Vert }_H\Vert v{\Vert }_H$ and $|\ell (v)|\le C_P\Vert f{\Vert }_2\Vert v{\Vert }_H$.

For $\phi \in C_0^{\infty }(\Omega )$, the definition of weak divergence, applied to ${\phi }^2/2$, givesThe left side is continuous in the $H_0^1$ norm. Approximate any $u\in H_0^1$ by test functions and pass to the limit on the left to obtain $\int (b\cdot \nabla u)u\ge 0$. This density argument is essential: local integrability of $\mathrm{div}b$ alone does not justify treating $(\mathrm{div}b)u^2$ as integrable for arbitrary $u\in H_0^1$.

Consequently $a(u,u)\ge \Vert u{\Vert }_H^2$, so $a$ is a coercive bilinear form. The Lax-Milgram theorem supplies a unique weak solution with $\boxed{\Vert \nabla u{\Vert }_2\le C_P\Vert f{\Vert }_2}$. The reusable estimate is coercivity from a drift with nonpositive weak divergence.

A weak solution belongs to the zero-boundary Sobolev space $H_0^1(\Omega )$ and satisfiesThe boundedness of $b$ and the Cauchy-Schwarz inequality make the drift term integrable. This weak formulation moves one derivative from the Laplacian onto the test function, while retaining the first-order drift term.

The graph area functional is finite on $H^1$, since $\sqrt{1+|\nabla u{|}^2}\le 1+|\nabla u|$ and $\Omega$ has finite volume. For $p=\nabla u$, $q=\nabla v$, the difference quotient of $\sqrt{1+|p+tq{|}^2}$ is bounded in absolute value by $|q|$. This is integrable by the Cauchy-Schwarz inequality. The dominated convergence theorem therefore givesThis defines a bounded linear functional of $v$, since $|⟨A(u),v⟩|\le |\Omega {|}^{1/2}\Vert \nabla v{\Vert }_2$.

For $a(p)=p/\sqrt{1+|p{|}^2}$ the Jacobian matrix isThe inequality follows from the Cauchy-Schwarz inequality. Integrating this derivative along the segment between any two gradients shows $(a(p)-a(r))\cdot (p-r)\ge 0$. Integrating over $\Omega$ proves that $A$ is a monotone operator.

It is not coercive on $H^1(\Omega )$. The constant functions $u_k\equiv k$ satisfy $\Vert u_k{\Vert }_{H^1}=k|\Omega {|}^{1/2}\to \infty$ but $A(u_k)=0$. Thus $⟨A(u_k),u_k⟩/\Vert u_k{\Vert }_{H^1}=0$.

For $g=\nabla f(x)\ne 0$, the unit steepest descent direction is $-g/\Vert g{\Vert }_2$ by the Cauchy-Schwarz inequality. The usual unnormalized direction is $\boxed{p=-g}$; choosing its scale separately as a line-search step gives the same ray. It satisfies $\boxed{g^{T}p=-\Vert g{\Vert }_2^2<0}$, so it is a descent direction. The first-order Taylor expansion then guarantees decrease for sufficiently small positive steps. The subsequent bounds use this unnormalized negative-gradient convention.

The Gauss-Seidel method updates coordinates successively using the newest available entries:It is well-defined for every initial vector exactly when every $a_{ii}\ne 0$, equivalently when the triangular matrix $D+L$ is invertible. Nonsingularity of $A$ alone does not ensure this. The matrix form isIf the iterates converge to $z$, passing to the limit gives $(D+L)z=b-Uz$, hence $Az=b$. Nonsingularity identifies $z$ as the unique solution $x$.

Now assume $A$ is a symmetric positive-definite matrix. Testing with the coordinate vector gives $a_{ii}=e_i^{T}A{e}_i>0$, so $D$ is positive definite. The form $⟨v,w{⟩}_A=v^{T}Aw$ is an inner product: bilinearity and symmetry are immediate and positivity is the hypothesis. Its Cauchy-Schwarz inequality givesproving the triangle inequality; definiteness and absolute homogeneity are immediate. Thus $\Vert v{\Vert }_A=\sqrt{v^{T}Av}$ is the energy norm.

Subtract the iteration from the exact equation to obtain $(D+L)e_{k+1}=-L^T{e}_k$. HenceUsing the preceding iteration also gives $(D+L)e_k=-L^T{e}_{k-1}$ and therefore $A{e}_k=L^T(e_k-e_{k-1})$ for $k\ge 1$, as required.

Put $d=e_k-e_{k+1}$. Expanding the two quadratic forms and using the first identity gives the exact Gauss-Seidel energy decrease identity:Here $2d^{T}Ld=d^T(L+L^T)d$. If $d=0$, then $A{e}_k=0$, so $e_k=0$. Otherwise $d^{T}Dd>0$. Thus $\boxed{\Vert e_k{\Vert }_A>\Vert e_{k+1}{\Vert }_A}$ unless the exact solution has already been reached.