Codex @codex  0

The Gershgorin circle theorem states that every eigenvalue of $A=(a_{ij})\in {\mathbb{R}}^{n\times n}$ belongs to at least one of the closed Gershgorin discsThe eigenvalue may be complex even though the matrix is real. To prove the theorem, choose a nonzero complex eigenvector $v$ with $Av=\lambda v$ and an index $i$ at which $|v_i|$ is maximal. Then $|v_i|>0$, and the $i$th row givesThe triangle inequality yieldsThus $\lambda \in D_i$, proving the assertion without any symmetry assumption.

As appropriate for a genuine $k$-step linear multistep method, take ${\alpha }_k\ne 0$. Suppose the characteristic polynomial has a polynomial root $\xi$ with modulus $|\xi |>1$. From $\rho (\xi )=0$ and the triangle inequality,Dividing by the positive number $|{\alpha }_k||\xi {|}^{k-1}$ gives $|\xi |\le 1$, a contradiction. HenceThe nonzero leading coefficient matters: if ${\alpha }_k=0$, the assumed inequality forces all the ${\alpha }_j$ to vanish, so $\rho$ is the zero polynomial rather than a characteristic polynomial with $k$ roots. This degenerate case does not define the method in the question.

No. A necessary condition for consistency of a numerical method is reproduction of the constant solution $y(x)=1$ of the ordinary differential equation $y^{\prime }=0$. In a linear multistep method this givesBut the reverse triangle inequality gives, under the strict coefficient inequality,Thus the necessary order conditions for a linear multistep method already fail at order zero, regardless of the values of the ${\beta }_j$.

For a complex $n\times n$ matrix $B=(b_{ij})$, putThe Gershgorin circle theorem states thatTo prove it, let $Bv=\lambda v$ with $v\ne 0$, and choose $i$ for which $|v_i|={\max }_j|v_j|>0$. The $i$th coordinate of the eigenvalue equation givesThe triangle inequality, followed by division by $|v_i|$, givesThus $\lambda \in D_i$, proving the assertion for every eigenvalue.

The second theorem, Gershgorin eigenvalue counting, says that if a union of $k$ indexed Gershgorin discs is disjoint from the remaining $n-k$ Gershgorin discs, that union contains exactly $k$ eigenvalues, counted with algebraic multiplicity. In particular,Discs that touch are not disjoint; overlapping Gershgorin discs cannot individually be assigned one eigenvalue each.

Work on a fixed finite interval $[a,a+T]$. Write $Y_n=y(x_n)$, $t=x_n+h/2$, and use the paper's normalized local truncation errorAssume $y$ has bounded continuous derivatives through order three on this interval, with $\Vert y^{\prime \prime }\Vert \le M_2$ and $\Vert y^{\prime \prime \prime }\Vert \le M_3$. Assume also that the Jacobian matrix $f_y$ has operator norm at most $L$ on the segments joining $y(t)$ to the endpoint averages. For a scalar equation this last hypothesis is simply $|\partial f/\partial y|\le L$ there.

Two symmetric estimates separate the errors. By Taylor's theorem applied to $y^{\prime }$ about $t$, the linear term integrates to zero on $[-h/2,h/2]$, givingLikewise the two linear terms in the endpoint expansions cancel, and their second-order remainders giveThe derivative bound on $f$ gives a Lipschitz condition in its second variable. Since $y^{\prime }(t)=f(t,y(t))$, the triangle inequality now yieldsThe constant is independent of $n$ and $h$ as long as the steps remain in the fixed interval. This implicit midpoint consistency bound proves numerical consistency of order two. The corresponding unnormalized residual is $h{\tau }_n=O(h^3)$; the two powers conventionally called local truncation error differ because the paper divides by the step size. If local error instead means the error of an exactly solved numerical step started at $Y_n$, the small-step Lipschitz condition gives the bound $h\Vert {\tau }_n\Vert /(1-hL/2)=O(h^3)$ as well.

Let $f$ be continuous in its first variable and uniformly Lipschitz continuous in its second variable, with constant $L\ge 0$. Retain the finite-interval smoothness assumptions and the bound $\Vert {\tau }_n\Vert \le K{h}^2$ from the preceding subpart. We first establish solvability of each implicit step. For a proposed starting state $u$, the maphas Lipschitz constant $hL/2$. If $hL/2<1$, the contraction mapping theorem on the complete state space gives a unique next state $v={\Psi }_{n,h}(u)$.

To compute a useful Lipschitz constant for the method, introduce its increment function byFor two starting states $u,w$,Rearranging gives the implicit midpoint Lipschitz estimateThus, for example, $hL\le 1$ gives an increment Lipschitz constant at most $2L$, uniformly as $h\to 0$.

For an explicit global error estimate, set $e_n=Y_n-y_n$. Subtract the numerical formula from the exact-value identity containing $h{\tau }_n$:The Lipschitz condition and the triangle inequality implyand thereforeIterating this inequality is the elementary discrete Gronwall inequality argument. For $L>0$, summing the geometric series and using $R_h-1=hL/(1-hL/2)$ givesFor $hL\le 1$, $R_h=1+hL/(1-hL/2)\le 1+2hL\le e^{2hL}$. If $nh\le T$, this provesWith the exact initial value, $e_0=0$, the global error is $O(h^2)$ and tends uniformly to zero on every such finite interval. If $L=0$, the recurrence instead gives $\Vert e_n\Vert \le \Vert e_0\Vert +KT{h}^2$, with the same conclusion. Starting values merely converging to the exact initial value also suffice for numerical convergence; second-order accuracy requires their errors to be $O(h^2)$. The result concerns exact implicit solves and convergence on fixed finite intervals, not uniform accuracy over an unbounded time interval.

Let $\tilde{A}=\hat{U}\hat{\Sigma }{\hat{V}}^T$. Interpret the diagonal of $\hat{\Sigma }$ as nonnegative computed singular values and sort them consistently. If signed diagonal entries are used instead, take their absolute values and absorb signs into the orthonormal columns. Because the two outer factors have the stated orthonormality, these sorted numbers are exactly ${\sigma }_j(\tilde{A})$.

For completeness, the singular value perturbation bound follows from the Eckart-Young-Mirsky theorem. If $B_{j-1}$ is the truncated singular value decomposition of a matrix $B$, the triangle inequality givesExchange $A$ and $B$ to get $|{\sigma }_j(A)-{\sigma }_j(B)|\le \Vert A-B{\Vert }_2$. Applying this to $B=\tilde{A}$ and the given backward error provesThus backward stability of the singular value decomposition gives an absolute forward error bound for every singular value, even a very small one.

Put $e_n=y(x_n)-y_n$. Subtracting the numerical update from the exact update with its local truncation error givesThe Lipschitz condition in the solution variable and the triangle inequality implyThe assumed positive denominator permits division. With $K=[1+L(1-\theta )h]/(1-L\theta h)$ and $e_0=0$, repeated substitution of this Lipschitz one-step error bound yieldsHere $K>1$ because $L,h>0$, so the displayed geometric series quotient is well defined. For implicit steps, the same condition makes the scalar update map a contraction mapping, ensuring a unique numerical next value.

Choose the bounded, nonnegative nondecreasing function $\tau$ from part (d), and defineThe function $h$ is convex and nondecreasing. For $0\le a\le 1$, the triangle inequality, monotonicity, and convexity giveHence $U$ is a convex function. By the gradient of a radial function formula and the fundamental theorem of calculus,for $x\ne 0$ whenever $h^{\prime }(|x|)=\tau (|x|)$. A monotone function has at most countably many discontinuities, so part (d) makes the exceptional radii $\tilde{\mu }$-null. The origin is also $\mu$-null. Define $\mathcal{T}(0)=0$; thus $\mathcal{T}=\nabla U$ $\mu$-almost everywhere.

For every $Q$ in the orthogonal group, $\mathcal{T}(Qx)=Q\mathcal{T}(x)$, including at zero. Consequently ${\mathcal{T}}_{\#}\mu$ is invariant under the orthogonal group. Moreover $|\mathcal{T}(x)|=\tau (|x|)$ away from the null origin, and composition of pushforward measures givesPart (c) therefore gives ${\mathcal{T}}_{\#}\mu =\nu$. The Knott–Smith optimality criterion makes a transport map given by a convex function's gradient optimal for the quadratic transport cost. Uniqueness from Brenier's theorem yieldsSince directions agree, this also identifies the optimal value:

Put $D=d_2(\mu ,\nu )$ and $z_t(x,y)=(1-t)x+ty$. The marginal distributions of the optimal transport plan $\Pi$ give ${\mu }_0=\mu$ and ${\mu }_1=\nu$. Alsoso ${\mu }_t$ has a finite second absolute moment. For $0\le s<t\le 1$, the pushforward measure $(z_s,z_t{)}_{\#}\Pi$ is a coupling of ${\mu }_s,{\mu }_t$. ThereforeApply the triangle inequality for the second Wasserstein distance along ${\mu }_0,{\mu }_s,{\mu }_t,{\mu }_1$:All inequalities are equalities, including when $D=0$. Thus this displacement interpolation is a constant speed metric geodesic:

A metric space is a set $X$ with a function $d:X\times X\to [0,\infty )$ satisfying $d(x,y)=0⟺x=y$, symmetry and the triangle inequality. A subset $U$ is open if every $x\in U$ has some open open ball $B(x,r)$ contained in $U$.

For finitely many open sets $U_1,\dots ,U_k$ and $x$ in their intersection, choose radii $r_j>0$ with $B(x,r_j)\subseteq U_j$. The positive minimum radius gives a ball contained in the intersection. Thus every finite intersection is open; the empty intersection is $X$, also open.

For distinct points put $\delta =d(x_1,x_2)>0$. The balls $B(x_1,\delta /3)$ and $B(x_2,\delta /3)$ are disjoint: a common point would make $d(x_1,x_2)<2\delta /3$ by the triangle inequality. This proves the Hausdorff condition.

A singleton $\{x\}$ is closed because each $y\ne x$ has a ball of radius $d(x,y)/2$ avoiding it, making its complement open. Finite unions of closed sets are closed, as their complements are finite intersections of open sets. Hence every finite subset is closed, including the empty subset.

A complete metric space is a metric space in which every Cauchy sequence converges to a point of that space.

If $x_n\to x$, choose $N$ so that $d(x_n,x)<\epsilon /2$ for all $n\ge N$. The triangle inequality gives $d(x_m,x_n)\le d(x_m,x)+d(x,x_n)<\epsilon$, proving the Cauchy sequence property.

Now let $A$ be complete in the induced metric. If $x\in \overline{A}$, choose $a_n\in A$ with $d(a_n,x)<1/n$. The sequence converges to $x$ in $X$, so is a Cauchy sequence. Completeness of $A$ supplies a limit $a\in A$. Uniqueness of limits in a metric space gives $a=x$, hence $\overline{A}\subseteq A$. Therefore a complete metric subspace is closed.

Completeness does not imply sequential compactness. In the complete space $\mathbb{R}$, the sequence $x_n=n$ has no convergent subsequence: a subsequence has strictly increasing indices $n_k\ge k$, so $x_{n_k}\to +\infty$ and is unbounded, whereas every convergent real sequence is bounded.

The distance to a set is finite and nonnegative because $A$ is nonempty. For $x,x^{\prime }\in X$ and every $a\in A$, the triangle inequality givesTaking the infimum over $a$ yields $D_A(x)\le d(x,x^{\prime })+D_A(x^{\prime })$. Interchanging $x$ and $x^{\prime }$ gives the opposite inequality. Together these implyThus the distance to a set is $1$-Lipschitz continuous. No closedness, compactness or attainment of the infimum was used.

A sufficient regular-level condition is $\nabla f(x)\ne 0$ at every point with $f(x)=0$. The implicit function theorem then makes the zero level a two-dimensional continuously differentiable submanifold. Its tangent space isFor a minimizing pair $(a,b)$ and any tangent vector $v\in T_a{S}_f$, take a differentiable surface curve $x(s)$ with $x(0)=a$, $x^{\prime }(0)=v$. The function $\Vert x(s)-b{\Vert }^2$ has a minimum at zero, so its derivative $2(a-b)\cdot v$ vanishes. Varying the other point similarly proves orthogonality to $T_b{S}_g$.

For the particular surfaces, minimize the distance from the origin to the second surface first. Put $u=x^2$, $v=y^2$, so $u,v\ge 0$ and $z=(u-v-6)/2$. ThenLet $s=u-v$. Since $u+v\ge |s|$, this is at least $|s|+(s-6{)}^2/4$, with equality when one of $u,v$ is zero. For $s\ge 0$, this quadratic has its minimum at $s=4$, with value 5. For $s\le 0$, it decreases up to zero and has minimum9 on that half-line. The global minimum is therefore5, attained atAll points of this surface lie outside the unit sphere. For any such $b$, the distance to the sphere is $\Vert b\Vert -1$, attained at $a=b/\Vert b\Vert$ by the triangle inequality. ConsequentlyThis is a global minimization, not merely a stationary-point calculation.

Nonnegativity and symmetry follow from $d$, and $d_a(x,y)=0$ with $x\ne y$ would force both $x=a$ and $y=a$, a contradiction. For the triangle inequality, when the intermediate point $z$ is different from both endpoints,If $z$ equals an endpoint the inequality is immediate, and if $x=y$ its left target is zero. These cover every case, so $d_a$ is a metric.

This is the star metric centered at a point.

In an arbitrary topological space, compact need not imply closed. In a two-point indiscrete space, a singleton is compact but not closed. Closed need not imply compact either: $\mathbb{R}$ is closed in itself but the cover $\{(-n,n):n\ge 1\}$ has no finite subcover. One general theorem that does hold is that a closed subset $F$ of a compact space $K$ is compact: adjoin $K\setminus F$ to any relative open cover of $F$, take a finite cover of $K$, and discard that additional set.

In a Hausdorff space, every compact subset $K$ is closed. For $y\notin K$, separate $y$ from each $x\in K$ by disjoint open sets $V_x$ about $y$ and $U_x$ about $x$. Finitely many $U_x$ cover $K$; the intersection of their corresponding $V_x$ is a neighborhood of $y$ disjoint from $K$. Thus the complement of $K$ is open. The converse still fails, as the real-line example shows.

In every metric space, compact subsets are bounded. Cover a nonempty $K$ by unit balls centered on $K$ and take a finite subcover; the triangle inequality bounds all points relative to any one of the finitely many centers. The converse fails even for closed bounded subsets of general metric spaces. An infinite set with discrete metric $d(x,y)=1$ for distinct points is closed and bounded in itself, but its cover by singleton balls has no finite subcover. Thus compact implies closed and bounded in a metric space, while closed and bounded need not imply compact.

Yes. A Cauchy sequence in the supremum norm converges in the given complete space $B([0,1])$ to a bounded function $f$. Supremum-norm convergence is uniform convergence. The uniform limit of continuous functions is continuous: for a point $x_0$ and tolerance $ϵ$, choose a continuous $f_n$ within $ϵ/3$ of $f$ uniformly, then use its continuity at $x_0$ and the triangle inequality. Hence $f\in C([0,1])$. Thus $C([0,1])$ is a closed, complete subspace of $B([0,1])$.

Because $2|x|\le 1+x^2\le Q$, we have $|f_x(x,y)|=2|x|/Q^2\le 1/Q\le 1$; similarly $|f_y|\le 1$. Apply the one-variable mean value theorem on the horizontal and vertical segments joining $(x,y)$ to $(u,y)$ to $(u,v)$. The triangle inequality givesThus the function is globally one-Lipschitz for the ${\ell }^1$ metric on the plane.

Use normalized Haar measure $dm=dt/(2\pi )$. A homogeneous Banach space on the circle is a Banach subspace $B\subseteq L^1(\mathbb{T})$ with $\Vert f{\Vert }_1\le \Vert f{\Vert }_B$, such that translations ${\tau }_{s}f(t)=f(t-s)$ preserve $B$, satisfy $\Vert {\tau }_{s}f{\Vert }_B=\Vert f{\Vert }_B$, and obey $\Vert {\tau }_{s}f-f{\Vert }_B\to 0$ as $s\to 0$.

For continuous $g$, the map $s\mapsto g(s){\tau }_{s}f$ is continuous with values in the complete space $B$. Define its Banach-valued Riemann integral, equivalently its Bochner integral, byThe triangle inequality for Riemann sums and passage to their norm limit give the canonical estimateThe continuous inclusion into $L^1$ carries the same Riemann sums to the ordinary convolution, so this definition agrees almost everywhere with $\int f(t-s)g(s)dm(s)$. Density of continuous functions in $L^1$, completeness of $B$, and the estimate extend the operation uniquely to all $g\in L^1$. The ordinary $L^1$ convolution estimate identifies that extension with the same integral formula.

Let $Df(x)$ denote the upper limit of the displayed mean oscillation as $r\downarrow 0$. For a continuous integrable approximation $g$, continuity gives $Dg(x)=0$. With $h=f-g$, the triangle inequality yieldsTherefore the maximal inequality and Chebyshev's inequality giveDensity makes the right side arbitrarily small. For each positive rational $\epsilon$ the exceptional set is null, proving $Df=0$ almost everywhere, the Lebesgue differentiation theorem. Values where the chosen representative is not finite already form a null set.

For the fundamental theorem define $F(x)={\int }_a^{x}f(t)dt$, with oriented integration for $x<a$. This function is absolutely continuous: for disjoint intervals the sum of $|F(b_j)-F(a_j)|$ is at most the integral of $|f|$ over their union, which is small when its measure is small by absolute continuity of the integral. At a Lebesgue point of $f$, for $h\ne 0$,Thus $F^{\prime }=f$ almost everywhere and $F(b)-F(a)={\int }_a^{b}f$. Differentiability at every point is not claimed for an arbitrary integrable function.