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Take a variation $x+\epsilon \xi$ with $\xi (a)=\xi (b)=0$. Expanding the action givesAn integration by parts and the fixed endpoint conditions give the first variationThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equationand the second variation is

Linearizing the equation of motion about $x$ givesso the Jacobi equation isWhen $u$ has no zero on $[a,b]$,The total derivative integrates to zero because $\xi (a)=\xi (b)=0$, hence

For the simple harmonic oscillator, the Jacobi equation is $\ddot{u}+{\omega }^{2}u=0$. Put $t_0=(a+b)/2$ and chooseIf $b-a<\pi /\omega$, then $|\omega (t-t_0)|<\pi /2$ throughout $[a,b]$, so $u$ is positive there. The preceding square identity proves that the classical path is a local minimum of the action whenever the elapsed time is less than half an oscillation period.

The Poincaré half-plane model iswith the orientation inherited from the complex plane. Forone hasThese identities show directly that ${\gamma }^{\ast }d{s}^2=d{s}^2$. The map is holomorphic with nonzero derivative, so it preserves orientation. The matrices $I$ and $-I$ induce the same map, giving the action of PSL2(R).

Conversely, let $F$ be an orientation-preserving isometry. The PSL2(R) action is transitive on $\mathbb{H}$, so compose $F$ with an element taking $F(i)$ back to $i$. The resulting isometry fixes $i$ and acts on $T_i\mathbb{H}$ by an orientation-preserving orthogonal map, hence a rotation. The stabilizer of $i$ in PSL2(R),realizes every such tangent rotation. An isometry is determined by its value and differential at one point because it preserves geodesics and the exponential map. Thus the composed isometry belongs to PSL2(R), and so does $F$.

The map $\tau (z)=-\overline{z}$ is an orientation-reversing isometry. Composing any orientation-reversing isometry with $\tau$ gives an orientation-preserving one, so

A hyperbolic line is a vertical Euclidean line or a semicircle orthogonal to the real axis. Its hyperbolic reflection ${\sigma }_{\ell }$ is the unique orientation-reversing isometry that fixes every point of $\ell$. If $\ell ,{\ell }^{\prime }$ meet at angle $\theta$, thenis the hyperbolic rotation about $A$ through angle $2\theta$. The generators satisfyIf $\theta /\pi =p/q$ in lowest terms, then $\rho$ has order $q$ and the generated group is the finite dihedral group of order $2q$. If $\theta /\pi$ is irrational, $\rho$ has infinite order and the generated group is the infinite dihedral group. In the degenerate case $\ell ={\ell }^{\prime }$, the group has order two.

Let $C\subset \mathbb{R}$ be compact and let $M=\Vert f{\Vert }_{\infty }$. Since every $k_n$ is supported in $[-R,R]$, the function $f$ is uniformly continuous on the compact setGiven $\epsilon >0$, choose $\delta >0$ so thatwhenever $x\in C$ and $|t|<\delta$. Since $\int k_n=1$ and $k_n\ge 0$,Property 3 makes the final term smaller than $\epsilon /2$ for all sufficiently large $n$, uniformly in $x\in C$. Hence $f_n\to f$ uniformly on every compact set. The sequence $(k_n)$ is an approximate identity.

For the second part, extend $g$ to a continuous function $\tilde{g}$ on $\mathbb{R}$ by setting it equal to zero outside $[0,1]$; continuity at the endpoints uses $g(0)=g(1)=0$. DefineThese nonnegative kernels have integral one. For every $\delta >0$, their mass outside $[-\delta ,\delta ]$ tends to zero exponentially relative to the mass near zero, so they satisfy property 3.

For $x\in [0,1]$, the convolution isBecause $|x-y|\le 1$ on the square $[0,1{]}^2$, no cutoff remains in this formula. Expanding the $n$th power shows that $p_n$ is a polynomial in $x$. The first part, applied to the compact interval $[0,1]$, givesThis proves the Weierstrass approximation theorem for functions with the stated endpoint values.

Suppose first that every ideal of $R$ is finitely generated. For an ascending chainthe union $I={\bigcup }_n{I}_n$ is an ideal. Its finite generating set lies in one $I_N$, so $I_n=I_N$ for all $n\ge N$. Thus $R$ satisfies the ascending chain condition and is a Noetherian ring.

Conversely, if an ideal $I$ is not finitely generated, choose $a_1\in I$, and after choosing $a_1,\dots ,a_n$, chooseThis creates a strictly ascending chain of ideals, contradicting Noetherianity. Hence every ideal is finitely generated.

If $\phi :R\to S$ is surjective and $J⊲S$, then ${\phi }^{-1}(J)$ is an ideal of $R$. If it is generated by $a_1,\dots ,a_m$, then $J$ is generated by $\phi (a_1),\dots ,\phi (a_m)$. Thus every ideal of $S$ is finitely generated, so $S$ is Noetherian.

The Hilbert basis theorem states that if $R$ is a commutative Noetherian ring, then $R[X]$ is Noetherian. To prove it, let $I⊲R[X]$. The leading coefficients of polynomials in $I$ generate an ideal of $R$; choose generators that occur as leading coefficients of $f_1,\dots ,f_s\in I$, and let $N={\max }_i\mathrm{deg}{f}_i$. For each degree below $N$, the leading coefficients of members of $I$ of at most that degree likewise form a finitely generated ideal; choose finitely many corresponding polynomials. Any $f\in I$ of degree at least $N$ can have its leading term cancelled by a linear combination of monomial multiples of the $f_i$. Repeating lowers its degree below $N$, where the second finite list completes the reduction. These finitely many selected polynomials generate $I$, proving the theorem.

For every $\alpha \in \mathbb{C}$, the ring $\mathbb{Z}[\alpha ]$ is Noetherian. If $\alpha$ is transcendental, evaluation identifies it with $\mathbb{Z}[X]$. If $\alpha$ is algebraic, it is a quotient of $\mathbb{Z}[X]$. The integers are Noetherian, the Hilbert basis theorem handles $\mathbb{Z}[X]$, and quotients preserve Noetherianity.

For a unique factorization domain that is not Noetherian, takeEvery polynomial and every factorization uses only finitely many variables, so existence and uniqueness of factorization reduce to a finite-variable polynomial ring, which is a UFD. Butis a strictly ascending chain, so the ring is not Noetherian.

Finally, the ring $C(\mathbb{R},\mathbb{R})$ is not Noetherian. For $n\ge 1$, letThese are ideals and $I_n⊊I_{n+1}$: the distance function to $[-1/(n+1),1/(n+1)]$ belongs to $I_{n+1}$ but not to $I_n$. This strict ascending chain proves the claim.