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On a suitable dense domain of smooth rapidly decreasing wave functions, the canonical commutator is $[x,p]=i\hslash$. Expanding the two ladder operators givesTheir sum and difference giveDirect multiplication, retaining the operator order, givesUsing ${\beta }^2=m\omega /\hslash$ identifies the Hamiltonian operator:For any admissible wave function, adjointness givesFor the conventional normalized state $\Vert \Psi \Vert =1$, this is the stated lower bound $E\ge \hslash \omega /2$. For an unnormalized function the norm factor must be retained; rescaling a wave function cannot leave its unnormalized energy integral bounded below by a fixed positive number.

Equality holds exactly when $a{\Psi }_0=0$. In position representation this equation is $(\beta x+{\beta }^{-1}{\partial }_x){\Psi }_0=0$, whose normalized square-integrable solution isThis construction establishes existence of a state attaining the bound, and hence proves that it is the ground state.

The commutator $[a,a^{†}]=1$ implies $[H,a^{†}]=\hslash \omega a^{†}$. Repeatedly commuting through the product givesThese vectors are nonzero: the identity $a(a^{†}{)}^n{\Psi }_0=n(a^{†}{)}^{n-1}{\Psi }_0$ gives their norm squared $n!$ by induction. ThereforeFor time-dependent stationary states multiply each spatial eigenfunction by $e^{-i{E}_{n}t/\hslash }$.

Use first-kind Chebyshev polynomials $T_j$, characterized by $T_j(\cos \theta )=\cos (j\theta )$. Substituting $x=\cos \theta$ converts the weighted inner product into ${\int }_0^{\pi }g(\cos \theta )h(\cos \theta )d\theta$. Hence these polynomials are orthogonal, with squared norms $\pi$ for $j=0$ and $\pi /2$ for $j\ge 1$.

For the specified function, $f(\cos \theta )=\sin \theta$. Its constant projection coefficient is ${\pi }^{-1}{\int }_0^{\pi }\sin \theta d\theta =2/\pi$. For $j\ge 1$ the coefficient is $(2/\pi ){\int }_0^{\pi }\sin \theta \cos (j\theta )d\theta$. Symmetry about $\pi /2$ makes it zero for odd $j$. For $j=2m$, the product-to-sum identity givesTherefore the Chebyshev projection of a semicircle isIn particular the best degree-at-most-$n$ polynomial is even, so for odd $n$ its actual degree is at most $n-1$.

A function $f:I\to \mathbb{R}$ is uniformly continuous ifThe same $\delta$ must work at every point of $I$.

Every uniformly continuous function on $(0,1)$ is bounded. Choose the tolerance $\delta$ for $\epsilon =1$ and an integer $N$ with $1/N<\delta$. The finitely many centers $c_j=(j+1/2)/N$, $0\le j<N$, cover the interval by cells of radius $1/(2N)$. For each $x$, one center satisfies $|f(x)-f(c_j)|<1$. Thus $|f(x)|\le 1+{\max }_j|f(c_j)|$. This is the uniformly continuous function on a totally bounded set is bounded argument; closedness of the interval is unnecessary.

The reciprocal function is not uniformly continuous on $(0,1)$. It is unbounded, already contradicting the preceding result. Directly, $x_n=1/(n+1)$ and $y_n=1/(n+2)$ have distance tending to zero while $|1/x_n-1/y_n|=1$.

The oscillating function $\sin (1/x)$ is not uniformly continuous either, despite being bounded. TakeTheir distance tends to zero, but their function values are $1$ and $-1$. The fixed output gap of two contradicts the uniform-continuity condition, for example with $\epsilon =1$.

Identify the circle group with $[0,1)$ modulo its endpoints, and let $\lambda$ be normalized Lebesgue measure. We first check invariance. On the $r$th inverse branch, $x=(y+r)/m$ for $0\le r<m$. For an integrable function $g$, substitution givesIn particular, testing indicator functions shows that the integer multiplication map on the circle preserves $\lambda$.

To prove ergodicity, let $B$ be a measurable set with $\lambda (E_m^{-1}B△B)=0$ and put $h=1_B$. Then $h(E_{m}x)=h(x)$ almost everywhere. Its Fourier coefficients areUsing invariance of $h$ and substituting separately on the same $m$ branches givesIn the last line, the finite geometric series of $m$th roots of unity is zero unless $m$ divides $k$, when it equals $m$. For any nonzero integer $k$, divide repeatedly by $m$ until the resulting integer is no longer divisible by $m$. The recurrence then implies $c_k=0$.

Since the Fourier basis is a complete orthonormal basis of $L^2([0,1),\lambda )$, the absence of every nonconstant Fourier coefficient implies $h=c_0=\lambda (B)$ almost everywhere. An indicator function takes only the values zero and one, so $\lambda (B)$ must be zero or one. This proves

For the indicated cyclic Galois resolvent, reindexing its terms and using $g^m=1$ givesThere exists $w$ with $R(w)\ne 0$ because distinct field automorphisms are linearly independent as functions. A short proof is as follows. Take a nonzero relation with the fewest nonzero terms, normalize one coefficient, evaluate it at $uv$ and subtract a scalar multiple of the relation at $v$. Choose $u$ on which two participating automorphisms differ. This produces a nonzero relation with fewer terms, a contradiction.

Set $\alpha =R(w)\ne 0$. Then $g^j\alpha ={\zeta }^j\alpha$ are $m$ distinct conjugates. Also $g({\alpha }^m)={\alpha }^m$, so $a={\alpha }^m\in K$. The orbit has size $m$, giving $[K(\alpha ):K]=m$ and hence $K(\alpha )=M$. All the roots ${\zeta }^j\alpha$ belong to $M$, soThis also treats $m=1$ with a nonzero $w\in K$.

The irreducible characters are an orthonormal basis of the space of complex class functions for the inner productThus the row orthogonality relation and completeness statement areIn particular the number of irreducible characters equals the number of conjugacy classes. In class-representative notation, the orthogonality relation is ${\sum }_j\overline{{\chi }_i(g_j)}{\chi }_k(g_j)/|Z(g_j)|={\delta }_{ik}$.

A function element is a pair $(f,D)$ where $D$ is a nonempty disk and $f$ is holomorphic there; the germ at its center captures its local information. Elements continue one another when their functions agree on an overlapping connected neighborhood. A complete analytic function is a maximal collection reachable from one element by chains of analytic continuation, including all further continuations of its members.

If $f^n=p$ in the initial disk, the identity theorem propagates this identity across every overlapping disk in a continuation chain. Induction along the chain proves every element of the complete analytic function also satisfies ${\tilde{f}}^n=p$.

Put $F(z,w)=w^n-p(z)$. Its gradient is $(-p^{\prime }(z),n{w}^{n-1})$. At a point of the curve with $w\ne 0$, the second entry is nonzero; at $w=0$, the root of $p$ is simple and the first entry is nonzero. Thus the holomorphic implicit function theorem supplies charts. Its precise local statement is that a holomorphic equation with nonzero derivative in one variable can locally be solved uniquely and holomorphically for that variable as a function of the other. For $w\ne 0$, use $z$ as coordinate and write $w=w(z)$; at a simple root use $w$ as coordinate and write $z=z(w)$. The transition maps are holomorphic with nonzero derivatives, giving a Riemann surface. The two projections are holomorphic because they are holomorphic expressions in either kind of chart.

Away from the zeros of $p$, the projection onto $z$ is an unramified $n$-sheeted covering. A germ of a root corresponds to $(z,f(z))$ and its analytic continuations follow this covering. Near a simple root, $z-z_0$ is a nonzero analytic factor times $w^n$; a circuit about $z_0$ cyclically permutes all $n$ branches. For nonconstant square-free $p$, these circuits connect all sheets. Hence the germ surface $S(F)$ is the curve with the points over the roots removed, and $C$ is its branched completion obtained by adding those ramification points. An ordinary function element in the variable $z$ cannot include such a point: the zero order of a holomorphic $n$th power is divisible by $n$, whereas $p$ has zero order one. If a convention already includes these branch points in the Riemann surface of a multivalued function, that completed surface is $C$ itself. For a nonzero constant $p$, the curve has $n$ separate components, and the chosen element determines just one of them. The identically zero polynomial is excluded by the non-singularity premise.

A fundamental solution of a linear differential operator $P$ is a distribution $G$ satisfying $PG={\delta }_0$. The locally integrable function $G=e^{-|x|}/2$ defines a distribution by integration against compactly supported smooth test functions. Away from zero it satisfies $-G^{\prime \prime }+G=0$. It is continuous at zero, while $G^{\prime }(0+)=-1/2$ and $G^{\prime }(0-)=1/2$. Consequently its distributional second derivative is its ordinary derivative on the two sides plus the derivative jump times ${\delta }_0$:This derivative-jump formula follows by integration by parts on each half-line; the boundary terms equal $-\varphi (0)$ in $G^{\prime \prime }$.

Convolution with the indicator forcing givesThe two values and their first derivatives agree at $\pm 1$, so no delta terms arise there. This is a decaying weak solution, with the equation holding classically away from the forcing jumps.

For real Schwartz $\varphi$, expand the energy:The mixed term is zero by the weak equation $-u_0^{\prime \prime }+u_0=V$, or integration by parts using the continuous first derivative and decay. The remaining integral is strictly positive whenever $\varphi$ is not identically zero. HenceThe real-valued function space is implicit in the stated ordered quadratic functional; for complex functions one instead uses absolute squares and the corresponding real part of the forcing term.

Let ${\Phi }_t$ be the local flow of a vector field $V$. The Lie derivative of a differential form is ${\mathcal{L}}_V\eta ={\frac{d}{dt}|}_{t=0}{\Phi }_t^{\ast }\eta$. We prove Cartan's magic formulaBoth sides are degree-zero derivations of an algebra on the exterior algebra of differential forms. They agree on a function $f$, since ${\iota }_{V}df=V(f)$. They also agree on $df$: pullback commutes with the exterior derivative, so ${\mathcal{L}}_{V}df=d(Vf)$, while $(d{\iota }_V+{\iota }_{V}d)df=d(Vf)$. Locally every differential form is a sum of products of functions and coordinate differentials, proving the identity in every degree.

For a compactly supported Hamiltonian vector field $V_H$, take the convention ${\iota }_{V_H}\omega =-dH$. Since $d\omega =0$, Cartan's magic formula gives ${\mathcal{L}}_{V_H}\omega =0$, so its complete flow of a vector field consists of symplectomorphisms. Given $p$ and $v\in T_{p}M$, a bump function lets us choose a compactly supported $H$ with $d{H}_p=-{\iota }_v{\omega }_p$, and hence $V_H(p)=v$. Choose such fields for a basis of $T_{p}M$. Their successive small-time flows give a map with invertible derivative at the origin. The inverse function theorem makes the orbit of $p$ open. Every orbit is open; because $M$ is connected, there is only one orbit. Thus the compactly supported symplectomorphisms act transitively on points, by Hamiltonian transitivity on a connected symplectic manifold.

On the unit sphere, the standard symplectic area is $4\pi$. The equator has complementary disks of areas $2\pi ,2\pi$, whereas the latitude at height $1/2$ has complementary disks of areas $\pi ,3\pi$: the cap above height $h$ has area $2\pi (1-h)$. A symplectomorphism must preserve the unordered pair of complementary areas. Therefore no such symplectomorphism exists, by complementary areas obstruct symplectic equivalence of separating curves.

For an infinite cardinal $\rho$ put ${\beth }_0(\rho )=\rho$ and ${\beth }_{n+1}(\rho )=2^{{\beth }_n(\rho )}$. The Erdős-Rado theorem for finite arities isThe partition relation requires a homogeneous subset of order type ${\rho }^{+}$ for every coloring of $(n+1)$-element subsets by at most $\rho$ colors. In particular it gives uncountable homogeneous sets. The case $n=0$ is the infinite pigeonhole principle: a union of $\rho$ sets each of size at most $\rho$ cannot have size ${\rho }^{+}$.

We give the induction step by a closed elementary-submodel construction of an end-homogeneous sequence. Let $\mu$ be infinite, $\theta =(2^{\mu }{)}^{+}$, and $c:[\theta {]}^{r+1}\to \rho$ with $\rho \le \mu$. Choose a sufficiently large regular $\chi$ and an elementary submodel $M\prec H_{\chi }$ with $c,\theta \in M$, all ordinals below ${\mu }^{+}$ in $M$, $|M|=2^{\mu }$, and closure under externally given sequences of length at most $\mu$. To construct it, take ${\mu }^{+}$ successive elementary Skolem-hull closures, at each step including all such sequences from the preceding stage. Cardinality stays $2^{\mu }$ because $(2^{\mu }{)}^{\mu }=2^{\mu }$. In the final union every sequence of length at most $\mu$ has all its entries in some earlier stage, by regularity of ${\mu }^{+}$, and so was included at the next stage.

Put $\beta =\sup (M\cap \theta )$. Regularity of $\theta$ gives $\beta <\theta$, and $\beta \notin M$ since otherwise its successor would contradict the definition of the supremum. Recursively, for $\alpha <{\mu }^{+}$ choose $x_{\alpha }\in M\cap \theta$ above all earlier $x_{\xi }$ and satisfyingThere are at most $\mu$ constraints. Their parameters and their colors belong to $M$, and closure puts their complete code in $M$. The point $\beta$ witnesses their simultaneous satisfiability in $H_{\chi }$; elementarity supplies a witness in $M$. This gives an increasing end-homogeneous sequence of length ${\mu }^{+}$: on it, the color of an $(r+1)$-tuple depends only on the first $r$ entries.

For the induction from arity $n$ to $n+1$, take $\mu ={\beth }_{n-1}(\rho )$ and $r=n$. Color the $n$-tuples of the resulting sequence by their common color with a later point, equivalently by their color with $\beta$. Its index order has type ${\mu }^{+}={\beth }_{n-1}(\rho {)}^{+}$, so the induction hypothesis supplies a homogeneous index set of type ${\rho }^{+}$. The original $(n+1)$-tuples on those indices have that same color. This completes the proof without substituting the theorem's name for its induction.

A useful ordinal partition-bound function isIt is total: choose an infinite cardinal $\rho \ge |\alpha |$ with ${\rho }^{+}>\alpha$; the pairs case supplies $(2^{\rho }{)}^{+}$ with the required homogeneous order type for all these color numbers. It is nondecreasing and satisfies $f(\alpha )\ge \alpha$. Start iterating $f$ and take a supremum $\kappa$ at a limit stage. If the increasing iterates do not attain that supremum, then for every $\alpha <\kappa$ some iterate $a_{\xi }$ satisfies $\alpha \le a_{\xi }<\kappa$, andIf the iterates have already stabilized, their supremum is a fixed point directly. Thus the relevant nontrivial case is an uncountable cardinal $\kappa$ closed below itself under $f$.

Such closure implies that $\kappa$ is a strong limit. Suppose $\lambda <\kappa$ is infinite and $\kappa \le 2^{\lambda }$. Take distinct binary strings of length $\lambda$, indexed in order by $\kappa$, and color a pair by its first differing coordinate. No three strings are homogeneous: three bits cannot be pairwise different at the same coordinate. Closure would give $f(\lambda +1)<\kappa$, and its restriction would have a homogeneous subset of order type $\lambda +1$ in $\lambda$ colors, a contradiction. Hence $2^{\lambda }<\kappa$; finite exponents are harmless for an uncountable $\kappa$.

Use the usual tree property: every set-theoretic tree of height $\kappa$ with nonempty levels of cardinality below $\kappa$ has a cofinal branch. For an uncountable cardinal satisfying this all-set-theoretic trees convention, $\kappa$ must be regular. If it were singular with cofinality $\tau <\kappa$, attach to one root $\tau$ chains whose lengths increase cofinally to $\kappa$. Every level has at most $\tau$ nodes, the height is $\kappa$, and no branch is cofinal. This contradiction proves regularity. Combined with the preceding closure argument it proves that the cardinal under discussion is strongly inaccessible.

For any $c:[\kappa {]}^2\to \nu$, $0<\nu <\kappa$, make a separated set-theoretic tree of functions. Its level $\alpha$ consists of the functions $t:\alpha \to \nu$ realized as $t(\xi )=c(\{\xi ,\beta \})$ for some $\beta \ge \alpha$, with $\beta <\kappa$. Order these functions by restriction. Levels are nonempty and have size at most ${\nu }^{|\alpha |}<\kappa$, using the strong-limit property. The tree property gives a branch $b:\kappa \to \nu$. Recursively select $x_{\eta }$ for $\eta <\kappa$, larger than all earlier selections, whose coloring function agrees with $b$ through a level above those earlier selections; regularity allows that level to remain below $\kappa$, and its node has a realizing point. Then $c(\{x_{\xi },x_{\eta }\})=b(x_{\xi })$ for $\xi <\eta$. One of the fewer-than-$\kappa$ colors occurs on $\kappa$ selected points, by regularity. These points are homogeneous, proving $\kappa \to (\kappa {)}_{\nu }^2$ for every $\nu <\kappa$. ConsequentlyA strictly increasing countable iteration has a supremum of countable cofinality, so it cannot meet this regular set-theoretic tree-property condition; no continuity of $f$ has been assumed.

The final printed assertion needs its convention specified. The usual tree property alone does not imply strong inaccessibility; it can consistently hold at successor cardinals. In the argument just given, strong inaccessibility follows from function closure together with that set-theoretic tree property. There is also a stronger branching form of the tree property for which the assertion holds literally: require a cofinal branch in every separated rooted height-$\kappa$ set-theoretic tree whose nodes each have fewer than $\kappa$ immediate successors, without restricting level sizes. The singular-chain example proves regularity under this convention too. If $2^{\lambda }\ge \kappa$ for $\lambda <\kappa$, take the full binary set-theoretic tree through level $\lambda$, choose $\kappa$ distinct nodes on that level, and attach to its $\xi$th chosen node a chain of length $\xi$, for $\xi <\kappa$. It has height $\kappa$ and at most two successors per node, but every branch has length below $\kappa$. Thus this branching property implies the strong-limit condition and hence strong inaccessibility. At a strongly inaccessible cardinal it agrees, for separated set-theoretic trees, with the usual level-size convention: regularity bounds the union of fewer than $\kappa$ successor sets, and strong-limit arithmetic bounds the possible predecessor sequences at limit levels.