Set $a=\sqrt{m\omega /(2\hslash )}x+\sqrt{\hslash /(2m\omega )}d/dx$ and let $a^{†}$ have the opposite derivative sign. On the bound-state domain, $[a,a^{†}]=1$ and $H=\hslash \omega (a^{†}a+1/2)$. Thus every energy eigenvalue has $E/\hslash \omega -1/2=\nu \ge 0$.

The lowering operator maps an eigenstate of $a^{†}a$ of eigenvalue $\nu$ to one of eigenvalue $\nu -1$, with squared norm multiplied by $\nu$. Repeated lowering cannot pass below zero, so it must terminate at zero. Hence $\nu$ is a nonnegative integer. The equation $a{\psi }_0=0$ gives a Gaussian ground state, and repeated raising gives a nonzero square-integrable state for every integer $n\ge 0$. ThereforeThese are also the Gaussian-times-Hermite-polynomial solutions obtained by the suggested series method.

Put $\alpha =m\omega /\hslash$. Normalization gives ${\psi }_0=(\alpha /\pi {)}^{1/4}{e}^{-\alpha x^2/2}$, soThe expectation of $x$ itself is zero by parity; that cancellation does not apply to its absolute value.

At fractional position $s$ along a straight rigid rod, the velocity is $(1-s)u+sv$. Uniform mass per unit length givesThe string constraints are $x_1^2+y_1^2+(-l+z_1{)}^2=l^2$ and $x_2^2+y_2^2+(-2l+z_2{)}^2=4l^2$. Their first-order terms give $z_1=z_2=0$. Their second-order terms then giveThe rod constraint is $(l+x_2-x_1{)}^2+(y_2-y_1{)}^2+(-l+z_2-z_1{)}^2=2l^2$. Its first-order part is $2l(x_2-x_1)=0$, hence $x_2=x_1$ to first order.

The centre of mass rises by $(z_1+z_2)/2$. Substituting the quadratic string expansions and the linear relation between the horizontal displacements givesVertical velocities contribute only at fourth order to the energy. Substituting $u=({\dot{x}}_1,{\dot{y}}_1,0)$ and $v=({\dot{x}}_1,{\dot{y}}_2,0)$ into the rod energy givesThe decoupled $x_1$ equation is ${\ddot{x}}_1+3g{x}_1/(4l)=0$. For the other two coordinates put $\kappa =3g/(2l)$; their equations areA normal-mode ansatz gives $3{\omega }^4-6\kappa {\omega }^2+2{\kappa }^2=0$, whose roots are ${\omega }^2=g(3\pm \sqrt{3})/(2l)$. The corresponding displacement ratios are $y_2/y_1=1\mp \sqrt{3}$. Therefore the three periods are

For a variation $y+\epsilon \eta$ with $\eta (0)=\eta (1)=0$, stationarity givesThe boundary term vanishes. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equation $d{F}_{y^{\prime }}/dx-F_y=0$.

Differentiating the proposed first integral givesThus it is constant along an extremal when $F_x=0$, proving the Beltrami identity.

For the positive branch relevant to the endpoint data, $F=\sqrt{1+y^{\prime 2}}/y$. The identity becomesThe Euler-Lagrange equation itself gives $y{y}^{\prime \prime }=-(1+y^{\prime 2})$, so $(y^2{)}^{\prime \prime }=-2$. Integrating, or solving the first integral, yields a circle $y^2+(x-c{)}^2=R^2$. The equal endpoint values force $c=1/2$ and $R^2=5/4$. Hence the unique positive smooth extremal isIt stays positive on the entire interval and satisfies the second-order equation, including its turning point. Merely differentiating the first integral after dividing by $y^{\prime }$ would need extra care there. Geometrically this is an arc of a hyperbolic geodesic in the upper half-plane model.

For the quotient map $\pi$, a subset $U$ of the quotient is open exactly when ${\pi }^{-1}(U)$ is open in $X$. This definition immediately makes $\pi$ continuous. If $g$ is continuous then $g\circ \pi$ is continuous. Conversely, continuity of the composition gives, for every open $V\subseteq Z$, that ${\pi }^{-1}(g^{-1}(V))$ is open. By the defining criterion $g^{-1}(V)$ is open, so $g$ is continuous.

A map $h$ constant on each equivalence class defines a unique map $\bar{h}([x])=h(x)$. It is well-defined by the assumed class constancy, and its composition with $\pi$ is $h$. The preceding criterion proves that $\bar{h}$ is continuous.

For the integer lattice quotient, defineThis continuous map is onto the product of the two unit circles, and its fibres are exactly the lattice equivalence classes. It therefore induces a continuous bijection $\bar{F}:{\mathbb{R}}^2/{\mathbb{Z}}^2\to {\mathbb{T}}^2$. Every class has a representative in $[0,1{]}^2$, so the quotient is the continuous image of that compact square and is compact. The target is Hausdorff as a subspace of ${\mathbb{C}}^2$. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism, provingFinally, if $U\subseteq {\mathbb{R}}^2$ is open, then ${\pi }^{-1}(\pi (U))={\bigcup }_{m\in {\mathbb{Z}}^2}(U+m)$ is open, being a union of translates. The quotient criterion makes $\pi (U)$ open. Thus the quotient map is an open map.

The nested closed sets have the finite intersection property: the intersection of finitely many is the one with the largest index and is nonempty. Compactness gives $\boxed{Y_{\infty }\ne \varnothing }$.

For an open neighbourhood $W$ of the intersection, suppose no $Y_n$ were contained in $W$. Then $Z_n=Y_n\cap (X\setminus W)$ would be nested, nonempty and closed. The same argument gives a point in all $Z_n$, contradicting $Y_{\infty }\subseteq W$. Hence

For connectedness, suppose $Y_{\infty }=A\cup B$ were a separation into two nonempty relatively closed sets. Since $Y_{\infty }$ is compact, $A,B$ are disjoint compact subsets of $X$. The supplied separation fact gives disjoint open neighbourhoods $U_A,U_B$. Their union contains $Y_{\infty }$, so some $Y_n$ lies in that union. But $Y_n$ meets both neighbourhoods, since it contains $A$ and $B$. Its intersections with the two neighbourhoods are then a separation, contradicting its connectedness. Thus the nested intersection is connected.

Lebesgue measurability of $g$ means that $g^{-1}(U)$ is a Lebesgue measurable subset of $\mathbb{R}$ for every open subset $U\subseteq \mathbb{R}$. Equivalently one can check all sets $\{x:g(x)>t\}$ with real $t$.

There are two conventions for a step function on the whole real line. Under the bounded-support integration convention it is a finite linear combination of indicators of bounded intervals; under the unrestricted piecewise-constant convention nonzero constant tails are allowed. The conclusions below state both conventions explicitly, since only the first assertion depends on this choice.

The monotone convergence theorem says that if $0\le f_j\uparrow f$ almost everywhere on a measure space, then $\int f_j\uparrow \int f$, allowing infinity. The dominated convergence theorem says that if $g_j\to g$ almost everywhere and $|g_j|\le H$ for an integrable $H$, then $g$ is integrable and $\int g_j\to \int g$.

For $a\ne 0$, the truncated functions $f_a{1}_{[-j,j]}$ increase to $f_a$. Each has the elementary integral $2\arctan (|a|j)/|a|$. Thus monotone convergence givesFor fixed $x$, eventually $|x|\le n\pi /2$ and $n\sin (x/n)\to x$, so $g_n(x)\to 1/(1+x^2)$. Inside the support the supplied inequality yields $n^2{\sin }^2(x/n)\ge 4x^2/{\pi }^2$. Outside the support $g_n=0$. Consequently the global bound isand this majorant is integrable by the already proved formula with $a=2/\pi$. Dominated convergence now gives $\boxed{\int g_n\to \int (1+x^2{)}^{-1}dx=\pi }$. The majorant is independent of $n$, including on the moving support.