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Use the Fourier transform convention $\tilde{F}(\omega )={\int }_{\mathbb{R}}F(x)e^{-i\omega x}dx$. Reindexing the convergent periodization of an integrable function gives $g(\tau +2\pi )=g(\tau )$; absolute integrability also ensures that the defining sum is absolutely convergent almost everywhere in a period. To calculate its Fourier series coefficients, uniform convergence permits passing the limit of the partial sums through the finite-interval integral. Changing variables $x=\tau +2\pi k$ and using $e^{2\pi ink}=1$ givesThe sum of these integrals is absolutely convergent, bounded by ${\int }_{\mathbb{R}}|F|$, so the last rearrangement is justified. This proves the proposed Fourier expansion at the level of coefficients.

There is a genuine missing hypothesis in the pointwise assertion as printed. Take $F(x)=1_{\{0\}}(x)$. It is absolutely integrable with integral zero, and its periodization has only finitely many nonzero terms on $[0,2\pi ]$, so the series converges uniformly there. Yet $g(0)=1$ while $\tilde{F}(n)=0$ for every $n$. Thus the displayed pointwise identity does not follow from the stated assumptions. Additional regularity ensuring Fourier convergence to the prescribed values of $g$, for example a continuous piecewise continuously differentiable periodization, makes the coefficient identity into the intended Poisson summation formula.

The particular function $F(x)=e^{-|x|}$ satisfies that stronger condition. Direct integration yieldsOn $0\le \tau \le 2\pi$, summing the two geometric tails gives $g(\tau )=\cosh (\tau -\pi )/\sinh \pi$, a continuous periodic function which is piecewise smooth. Its Fourier series has absolutely summable coefficients and converges to $g$. At $\tau =0$,Consequently

Choose the principal branch of the complex logarithm on $H$, with $-\pi <\arg z<\pi$. ThenThus $-\pi <\mathrm{Re}w<\pi$ and the imaginary part ranges over all real numbers. Conversely, for $w=x+iy$ in the strip, $z=e^{iw}=e^{-y}{e}^{ix}$ is in $H$ and its principal logarithm is $iw$. Therefore $\boxed{w=-i\log z:H⟶S}$ is an analytic bijection, with analytic inverse $z=e^{iw}$. Every other continuous branch on the connected domain is $\log z+2\pi ik$ for some fixed integer $k$, so its image is the translated strip $\boxed{S+2\pi k}$.

For $G$, introduce the Möbius transformation $\zeta =(z-1)/(z+1)$, whose inverse is $z=(1+\zeta )/(1-\zeta )$. Writing $\zeta =a+ib$, its defining circle inequalities transform as follows:andHence $G$ maps bijectively onto the wedge $\pi /2<\arg \zeta <3\pi /4$. The principal branch of the complex logarithm is analytic throughout this wedge. Multiplying its logarithm by $-8i$ expands the interval of real parts to $(4\pi ,6\pi )$, and subtracting $5\pi$ centers it at zero. The required conformal map is thereforeIts real part is $8\arg \zeta -5\pi \in (-\pi ,\pi )$, and its imaginary part $-8\log |\zeta |$ takes every real value. More explicitly, its inverse on $S$ is obtained from $\zeta =\exp (i(W+5\pi )/8)$ and $z=(1+\zeta )/(1-\zeta )$, which verifies both surjectivity and injectivity.

A basis of a topology is a family $\mathcal{B}$ of open sets such that every open set is a union of members of $\mathcal{B}$. Equivalently, whenever $x\in U$ and $U$ is open, there is $B\in \mathcal{B}$ with $x\in B\subseteq U$. For a metric space, all open balls form a basis. For the product topology on $X\times Y$, the rectangles $U\times V$ with $U,V$ open in their respective factors form a basis; rectangles made from factor bases also suffice.

If $C\subseteq X$ and $D\subseteq Y$ are countable dense subsets, then $C\times D$ is countable. Every nonempty basic rectangle meets it, since $U\cap C$ and $V\cap D$ are nonempty. Every nonempty open set contains such a rectangle, proving that a product of two separable spaces is separable.

Suppose a metric space has a countable dense set $C$. The balls $B(c,r)$ with $c\in C$ and positive rational $r$ form a countable basis. Indeed, given $x\in U$ open, choose $ϵ>0$ with $B(x,ϵ)\subseteq U$, then choose $c\in C$ with $d(c,x)<ϵ/3$ and rational $r$ satisfyingThe triangle inequality gives $x\in B(c,r)\subseteq B(x,ϵ)\subseteq U$. Conversely, from any countable basis select one point of each nonempty member. These points form a countable dense subset, because every nonempty open set contains a basis member. This direction holds for arbitrary topological spaces. Thus a metric space is separable exactly when it is a second-countable space. Intersecting a countable basis with a subspace gives a countable basis for its subspace topology; selecting points as above shows that every subspace of a separable metric space is separable.

For the Sorgenfrey line, $\mathbb{Q}$ meets every nonempty interval $[a,b)$, so it is a countable dense subset. In the Sorgenfrey plane, however, for any $x\in \mathbb{R}$ and $ϵ>0$,Indeed, $(t,-t)$ in that rectangle requires both $t\ge x$ and $t\le x$. Every singleton is therefore open in the subspace topology on $Y$, which is the discrete topology. A dense subset of a discrete space must contain every point. Since $Y$ is in bijection with the uncountable set $\mathbb{R}$, it has no countable dense subset. Hence $X$ is separable and $Y$ is not separable, illustrating why the metric assumption in the inheritance statement matters.

Two norms $\Vert \cdot {\Vert }_a,\Vert \cdot {\Vert }_b$ are Lipschitz equivalent norms when constants $c,C>0$ satisfy $c\Vert x{\Vert }_a\le \Vert x{\Vert }_b\le C\Vert x{\Vert }_a$ for every vector $x$. Applying these inequalities to $x_j-x_k$ shows that a sequence is a Cauchy sequence for one norm exactly when it is so for the other; applying them to $x_j-x$ likewise equates convergence. If one norm is complete, a Cauchy sequence for the other converges using the first norm and hence also the second. Interchanging the norms proves completeness is preserved in both directions.

Write $x={\sum }_i{x}_i{e}_i$ and set $C={\sum }_i\Vert e_i\Vert$. If $\Vert x{\Vert }_2=1$, then $|x_i|\le 1$, and the triangle inequality givesScaling a nonzero $x$ gives $\Vert x\Vert \le C\Vert x{\Vert }_2$ for every $x$. The reverse triangle inequality now impliesso $x\mapsto \Vert x\Vert$ is continuous, indeed Lipschitz, in the Euclidean norm.

Suppose there were no positive lower bound on the Euclidean unit sphere. We could choose $x_j$ with $\Vert x_j{\Vert }_2=1$ but $\Vert x_j\Vert <1/j$. The Bolzano-Weierstrass theorem supplies a subsequence converging in the Euclidean norm to $x$. Continuity of both norms gives $\Vert x{\Vert }_2=1$ and $\Vert x\Vert =0$, contradicting definiteness. Therefore some $m>0$ satisfies $\Vert x\Vert \ge m$ on the unit sphere. Scaling again proves the finite-dimensional equivalence of norms:

An R-module $M$ is a free module if it has a basis $(e_i{)}_{i\in I}$: every $m\in M$ has a unique expression as a finite sum ${\sum }_i{r}_i{e}_i$, with $r_i\in R$. Equivalently, it is a direct sum of copies of $R$. Let $q:M\to M/N$ be the quotient map. When $M/N$ is free, choose a basis $({\overline{f}}_j)$ and lifts $f_j\in M$. The universal property of a free module defines an $R$-linear map $s:M/N\to M$ by $s({\sum }_j{r}_j{\overline{f}}_j)={\sum }_j{r}_j{f}_j$. It satisfies $q\circ s=\mathrm{id}$, so every $m\in M$ decomposes uniquely asThis gives a split short exact sequence and $M\cong N\oplus (M/N)$. If $N$ is also free, the union of its basis and the lifted quotient basis is a basis of $M$: spanning follows from the decomposition, and applying $q$ to a putative relation kills all quotient coefficients before independence in $N$ kills the remaining ones. Thus free submodule and free quotient imply a free module.

For the first claim, take $R=\mathbb{Z}$ andHere $N\cong N^{\prime }\cong \mathbb{Z}$ are free modules and $M/N\cong M^{\prime }/N^{\prime }\cong \mathbb{Z}/2\mathbb{Z}$. But $M$ is a torsion-free module, whereas $M^{\prime }$ contains the nonzero torsion element $(0,\overline{1})$. Therefore claim (1) is false.

For the second claim, the same lifting construction works without assuming that $N$ is free. Since both isomorphic quotient modules are free, it yieldsThus claim (2) is true. Equivalently, this uses the fact that free modules are projective.

A Hermitian matrix satisfies $A=A^{†}$, where $†$ means conjugate transpose. If $Av=\lambda v$ with $v\ne 0$, then $v^{†}Av=\lambda \Vert v{\Vert }^2$ is real, since it equals its complex conjugate $v^{†}{A}^{†}v$. Thus every eigenvalue is real. To prove the finite-dimensional spectral theorem, choose a unit eigenvector $v$; one exists because the complex characteristic polynomial has a root. Its orthogonal complement is invariant, since for $w\perp v$,The restriction to that complement is again Hermitian. Induction on dimension supplies an orthonormal basis of eigenvectors, so $A=U\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_n)U^{†}$ for a unitary matrix $U$.

In this basis, $x^{†}Ax={\sum }_i{\lambda }_i|(U^{†}x{)}_i{|}^2$. Therefore $\boxed{A>0⟺{\lambda }_i>0\text{ for every }i}$. In that case the square root of a matrixis Hermitian, positive definite and satisfies $C^2=A$. For uniqueness, any positive-definite Hermitian $D$ satisfying $D^2=A$ commutes with $A$ and hence preserves each eigenspace of $A$. Its restriction to the $\lambda$-eigenspace is Hermitian, and its positive eigenvalues all solve ${\mu }^2=\lambda$. Thus that restriction is $\sqrt{\lambda }I$, forcing $D=C$. This is the principal square root of a positive semidefinite matrix, here strictly positive definite.

Now let $D=\sqrt{B}$ and $X=C-D$. Expanding without assuming that $C$ and $D$ commute givesThe Hermitian matrix $X$ has $X^2\ge 0$ because $v^{†}{X}^{2}v=\Vert Xv{\Vert }^2$, whereas $A-B>0$. Hence $CX+XC>0$. If $Xv=\mu v$, $v\ne 0$, thenSince $C>0$, every real eigenvalue $\mu$ of $X$ is positive. The positive-eigenvalue criterion now gives the strict finite-dimensional operator monotonicity of the square root:

A state of a Markov chain is a recurrent state if, starting there, the first strictly positive return time is finite with probability one. A recurrent Markov chain has every state recurrent; in an irreducible Markov chain recurrence is a class property. Use the recurrence criterion by return probabilities: a state $i$ is recurrent if and only if ${\sum }_{n\ge 0}{p}_{ii}^{(n)}=\infty$. To see the criterion, this sum is the expected number of visits including the initial one. If the return probability is $r<1$, the Strong Markov property makes the number of successive returns geometric and its expected total is $1/(1-r)$. If $r=1$, every fixed number of returns occurs with probability one, so the expectation is infinite.

For the symmetric simple random walk on the integers, odd-time return probabilities vanish, whileBy Stirling's formula, $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)\sim 4^n/\sqrt{\pi n}$. Hence $p_{00}^{(2n)}\sim 1/\sqrt{\pi n}$, whose sum diverges. The return-probability criterion proves recurrence of zero. Translation invariance gives the same result at every integer, or one can use irreducibility. ThusThis proof establishes almost-sure return, not a finite expected return time; every state of the walk is in fact a null recurrent state.

A 95 percent confidence set is a measurable data-dependent set $C(X_1,\dots ,X_n)$ whose repeated-sampling coverage satisfies ${\mathbb{P}}_{\theta }\{\theta \in C(X_1,\dots ,X_n)\}\ge 0.95$ for every parameter value, with equality for an exact 95 percent procedure. The parameter is fixed; the set is random. This is not a posterior probability statement about the parameter after the data have been observed.

For independent normal random variables of known variance, $\overline{X}\sim N(\mu ,{\sigma }^2/n)$, so $(\overline{X}-\mu )/(\sigma /\sqrt{n})$ has the standard normal distribution. The supplied quantile gives the exact confidence intervalIts deterministic length is $3.920\sigma /\sqrt{n}$. Requiring this to be at most $\epsilon$ givesAmong fixed-length intervals based equivariantly on the sample mean, the centred interval has greatest coverage: an interval of a given length captures most mass of a symmetric unimodal normal distribution when centred on its mean. Thus the width calculation is sharp for the usual fixed-length normal-mean construction.

In the position representation, the orbital angular momentum operator isOn a common smooth domain use $[x_a,p_b]=i\hslash {\delta }_{ab}$, $[x_a,x_b]=[p_a,p_b]=0$. Expanding the product commutator givesContracting with the two alternating tensors gives the orbital angular momentum commutation relations:For $L_{\pm }=L_1\pm i{L}_2$, these imply $[L_3,L_{\pm }]=\pm \hslash L_{\pm }$. Therefore $L_3\psi =0$ yieldsThusThe nonzero qualification is necessary: a rotationally invariant wavefunction is annihilated by all three components, and the zero vector is not an eigenvector.

Assume real $k\ne 0$, and put $a=|k|$. Away from the source point the Green function solves $G^{\prime \prime }-a^{2}G=0$. The two decay conditions select $G=A{e}^{a(x-\xi )}$ on $x<\xi$ and $G=B{e}^{-a(x-\xi )}$ on $x>\xi$. Continuity at $\xi$ requires $A=B$: a jump in $G$ would create a derivative of the Dirac delta that is absent from the equation. Integrating the equation across the source gives $G^{\prime }(\xi +)-G^{\prime }(\xi -)=1$, so $-2aA=1$. Thus the negative of the one-dimensional modified Helmholtz Green function isIts distributional derivative verifies the delta equation, including its sign.

For a source for which the integral converges and the resulting function has the required decay, defineApplying the differential operator under the integral, or in the sense of distributions, gives $\mathcal{L}y=\int \delta (x-\xi )S(\xi )d\xi =S(x)$. A continuous integrable source, for example, makes this convolution vanish at both infinities and gives a classical solution. If there were two solutions, their difference would be $A{e}^{ax}+B{e}^{-ax}$; decay at positive infinity forces $A=0$, and at negative infinity forces $B=0$. Thus the decaying solution is unique whenever it exists.

The assumptions needed for existence should not be dropped: an arbitrary source need not admit a decaying solution, as $S=1$ demonstrates. Also the displayed formula presupposes $k\ne 0$; for $k=0$ no Green function for a point source can decay at both infinities, since its derivative must jump by one while both exterior linear pieces would have to be zero.

Differentiability means existence of a linear map $A:{\mathbb{R}}^2\to \mathbb{R}$ such thatThis is the Fréchet derivative. Substituting increments on the coordinate axes shows $A(h,k)=f_x(x_0,y_0)h+f_y(x_0,y_0)k$. For the unit direction $v=(\cos \alpha ,\sin \alpha )$, the remainder is $r(tv)=o(|t|)$. Dividing the increment by $t$ proves differentiability of the restricted function and gives its directional derivative:For the supplied rational function, both coordinate-axis restrictions are zero, so $f_x(0,0)=f_y(0,0)=0$. Any Fréchet derivative would therefore be the zero map. But $f(t,t)=t$, and hence $|f(t,t)|/\sqrt{t^2+t^2}=1/\sqrt{2}$ for nonzero $t$. This contradicts the required vanishing remainder. ThusIn fact every directional derivative exists: along $v$ it equals ${\cos }^2\alpha \sin \alpha +\cos \alpha {\sin }^2\alpha$. These values do not depend linearly on the direction, illustrating why existence of directional derivatives is weaker than differentiability.

Set $D=1+u^2+v^2$. The two tangent vectors to the Enneper surface areTheir cross product is $D(-2u,2v,1-u^2-v^2)$. Since $4u^2+4v^2+(1-u^2-v^2{)}^2=D^2$, its length is $D^2>0$, proving regularity and giving the unit normalFor the first fundamental form, direct dot products give $E=G=D^2$ and $F=0$. The second derivatives are ${\sigma }_{uu}=(-2u,2v,2)$, ${\sigma }_{uv}=(2v,2u,0)$, and ${\sigma }_{vv}=(2u,-2v,-2)$. Taking their dot products with $\mathbf{n}$ gives the second fundamental form coefficients $e=2$, $f=0$, $g=-2$. Therefore the Gaussian curvature isHere curvature means Gaussian curvature; the opposite signs of the second-form diagonal entries also give zero mean curvature, as expected for this minimal surface.

Set $y=x_3+5$, so $y\ge 0$ and $y\le 10$. Introduce nonnegative slack and surplus variables $s_1,s_2,s_3$:The objective is $z=x_1+2x_2-6y+30$. A feasible simplex basis is $(x_2,s_1,s_3)$ with nonbasic variables $(x_1,s_2,y)=0$; its basic values are $(3,9,10)$. The initial simplex dictionary isIn the simplex algorithm, choose $s_2$ as the entering variable since it has positive objective coefficient. With the other nonbasic variables zero, only $s_1$ decreases; the simplex ratio test gives $s_2\le 18$. Thus $s_1$ leaves. Pivoting its row givesAll nonbasic variables are nonnegative, and all objective coefficients in the final dictionary are nonpositive, proving optimality. Setting them to zero gives $x_1=0$, $y=0$, $x_2=12$, henceThe surplus value is $s_2=18$, so the lower constraint is indeed satisfied, not mistakenly treated as equality at the optimum. The strict negative objective coefficients also show uniqueness of the optimizer.

Use additive notation for this finitely generated abelian group. The change of generators $A=a+3d$, $e=c-b$, together with $b,d$, is invertible over the integers: $a=A-3d$ and $c=b+e$. The second relation gives $2e=0$. Subtracting it from the first gives $4A=0$, while the third becomes $4b+2e=0$, hence $4b=0$. The last relation becomes $8A+4b+4e=0$ and is consequently redundant. Conversely $4A=4b=2e=0$ imply every original relation, so no further relation has been introduced or lost. Thus the cyclic group decomposition isThe infinite cyclic group factor contributes only zero to an element killed by two. Each finite factor has two such elements, so the kernel of multiplication by two has $2^3=8$ elements. Removing the identity leavesThe invertible generator change also supplies a direct presentation proof of the decomposition, without relying solely on a Smith normal form computation.

Start with Taylor's theorem with integral remainder, written on the whole interval using truncated powers:For the Peano kernel theorem, require that $L$ be a linear error functional, annihilate all polynomials of degree at most $k$, and permit interchange with this integral. These conditions hold, in particular, for finite linear combinations of point evaluations of $f$ and its derivatives through order $k$, with the truncated-power derivatives interpreted almost everywhere in $\theta$; dominated integration justifies the interchange. Another sufficient setting is a bounded functional on a differentiability space where the truncated-power integral is convergent in its norm. Polynomial exactness alone, without linearity and an integration-interchange justification, is not sufficient.

Applying $L$ annihilates the Taylor polynomial and givesThis is the unnormalized Peano kernel convention used by the question; including $1/k!$ within the kernel instead removes the factor outside the integral.

For the stated derivative error, constants and linear and quadratic polynomials are annihilated, so $k=2$. Acting on $(x-\theta {)}_{+}^2$ givesThus the sharp central-secant derivative error satisfiesHence $c=1/6$ works. It is the minimum: for $f(x)=x^3$, $L(f)=3-4=-1$ and $\Vert f^{\prime \prime \prime }{\Vert }_{\infty }=6$, so no smaller constant can satisfy the inequality. ThereforeThe sign-definite kernel explains why a constant third derivative attains the bound exactly.