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Characteristics beginning at $\xi$ have $x=\xi +t{u}_I(\xi )$ and carry the constant value $u_I(\xi )$. In the sloping interval, $x=t+(1-t)\xi$. For $0\le t<1$ the resulting single-valued solution isIt is continuous and piecewise smooth, locally Lipschitz for $t<1$, but not a classical $C^1$ solution across the two derivative corners. Its weak derivatives satisfy the equation, since the function itself has no jumps there. The sloping gradient is $-1/(1-t)$ and blows up at $t=1$, when those characteristics coalesce.

Thus the continuous characteristic description has maximal interval $0\le t<1$. If “weak solution” is taken literally, it does not cease to exist then: the entropy continuation is a shock with left/right states 1 and 0 and Rankine-Hugoniot speed $1/2$, situated at $x=(t+1)/2$ for $t\ge 1$. It is a global entropy weak solution, with the step at $t=1$ as its limiting trace. This distinguishes breakdown of the characteristic parametrization from nonexistence of a weak continuation.

Set $p_{\ast }(x)=2^{1-n}{T}_n(x)$, a monic Chebyshev polynomial. Its norm on $[-1,1]$ is $2^{1-n}$. At the ordered points $x_j=\cos ((n-j)\pi /n)$, its values alternate between the two extreme values $\pm 2^{1-n}$.

Approximate $f(x)=x^n$ by polynomials of degree at most $n-1$. The error corresponding to $q_{\ast }=x^n-p_{\ast }$ is exactly $p_{\ast }$. The Chebyshev alternation theorem therefore makes $q_{\ast }$ the unique best approximant. For any monic $p$, the polynomial $q=x^n-p$ is an admissible approximant, soEquality holds precisely for $p=p_{\ast }$. Alternatively, a smaller norm would force $p-p_{\ast }$ to change sign between each consecutive pair of alternating extrema, giving at least $n$ zeros to a polynomial of degree at most $n-1$.

The Chebyshev alternation theorem says that for $f\in C([a,b])$ a polynomial $q$ of degree at most $n-1$ is a best uniform approximation precisely when there are $n+1$ ordered points $a\le x_0<\cdots <x_n\le b$ such thatThe best polynomial is unique. These equally large alternating extrema are the equal-ripple criterion; the zero-error case is interpreted in the evident way.

Take $R_t=0$ when no record has occurred below $t$, the needed convention for the maximum of an empty set. Assume $F(t)<1$ and write $S(t)=1-F(t)$. Given the latest record $v_k<t$, the chance that the next one exceeds $t$ is $S(t)/S(v_k)$. Multiplying this by the joint density from part (a) leaveson the ordered simplex $0<v_1<\cdots <v_k<t$. Its integral is $S(t)\Lambda (t{)}^k/k!$, where $\Lambda (t)={\int }_0^{t}f(s)/S(s)ds=-\log S(t)$. The $k=0$ probability is directly $S(t)=e^{-\Lambda (t)}$. HenceThis is Poisson record counts in cumulative hazard coordinates; the instantaneous rate is the original distribution's hazard function. If the distribution has a finite upper endpoint and $t$ reaches it, $F(t)=1$ and infinitely many record values lie below $t$ almost surely; there is then no finite-parameter Poisson law. The finite-hazard assumption is thus necessary for the literal assertion.

After a record of value $v$, the first subsequent observation exceeding it has the original distribution conditioned on $X>v$. Indeed, summing over the number of failed observations gives the densityIndependence makes this conditional transition depend only on the latest record. Multiplying these transitions, starting with the density $f(v_1)$, provesThe ordering indicator is part of the density; it is zero elsewhere. Continuity of the observation distribution avoids ties.

For the positively oriented simple closed boundary, the turning tangent theorem gives $\int k(s)ds=2\pi$. If $L$ is its length, continuity gives a point with $k(s_0)\le 2\pi /L$. The planar isoperimetric inequality gives $L^2\ge 4\pi A(K)$, henceFor the reversed orientation the curvature integral is $-2\pi$, so a point of nonpositive curvature already proves the stated positive upper bound. The region-bounding interpretation is the ordinary Jordan-curve one; self-intersections would require specifying a different notion of enclosed area.

At a fixed initial parameter choose an orthogonal map $Q$ sending $\tilde{T}$ to $T$, $\tilde{N}$ to $N$ and $\tilde{B}$ to $-B$. Its determinant is $-1$. The transformed curve $Q\tilde{\alpha }$ has Frenet frame $(Q\tilde{T},Q\tilde{N},-Q\tilde{B})$, so its curvature is $k$ and its torsion is $-\tilde{\tau }=\tau$.

Both its frame and the frame of $\alpha$ solve the same linear Frenet system with the same initial data. The uniqueness theorem for a linear ordinary differential system with continuous coefficients makes the frames identical on the whole interval. In particular $Q{\tilde{\alpha }}^{\prime }={\alpha }^{\prime }$, so their difference is a constant vector $v$. ThusThe orientation-reversing rigid motion accounts for the reversal of torsion.

For an arclength-parametrized curve define the Frenet frame by $T={\alpha }^{\prime }$, $k=|{\alpha }^{\prime \prime }|>0$, $N=T^{\prime }/k$ and $B=T\times N$. Differentiating the unit lengths gives diagonal frame-derivative coefficients zero. Also $T^{\prime }=kN$ implies $N^{\prime }\cdot T=-k$, while $B^{\prime }\cdot T=-B\cdot T^{\prime }=0$. Define the torsion of a space curve by $\tau =-B^{\prime }\cdot N$. Orthogonality then givesThese are the Frenet formulae. Differentiating $B=T\times N$ also verifies the chosen torsion sign; equivalently $\tau =\det ({\alpha }^{\prime },{\alpha }^{\prime \prime },{\alpha }^{\prime \prime \prime })/|{\alpha }^{\prime }\times {\alpha }^{\prime \prime }{|}^2$.

The monodromy theorem states that a germ admitting analytic continuation along every path in a simply connected domain extends to a single-valued holomorphic function there. Equivalently continuations along homotopic paths with fixed endpoints agree. For a biholomorphism between the two tori, lift its composition with $\mathbb{C}\to \mathbb{C}/{\Lambda }_1$ through the second covering. Simple connectivity, or monodromy applied to local inverse charts, yields a holomorphic lift $F:\mathbb{C}\to \mathbb{C}$.

Lift the inverse similarly. The compositions of the lifts differ from the identity by lattice translations; adjusting the lifts shows that $F$ is bijective. By the allowed classification, $F(z)=az+b$ with $a\ne 0$. For $\lambda \in {\Lambda }_1$, its two arguments $z,z+\lambda$ represent the same torus point, so $a\lambda =F(z+\lambda )-F(z)\in {\Lambda }_2$. Thus $a{\Lambda }_1\subseteq {\Lambda }_2$. Apply the inverse lift to obtain the reverse inclusion. ThereforeThe translation $b$ affects the chosen origin, not the period lattice.

An elliptic function for a lattice $\Lambda$ is a meromorphic function on $\mathbb{C}$ satisfying $f(z+\lambda )=f(z)$ for all $\lambda \in \Lambda$. Its order is the number of poles, counted with multiplicity, in a fundamental parallelogram, equivalently the degree of its map from the torus to the Riemann sphere. Differentiation preserves the periods. If the distinct poles have orders $r_1,\dots ,r_s$, the derivative has exactly the same pole locations with orders $r_j+1$. ThereforeThe bounds use $1\le s\le m$, valid for a nonconstant elliptic function.

For the Weierstrass elliptic function $\wp$, $f=\wp {\wp }^{\prime }$ has a single pole of order five, so its derivative has order six. For the second example choose $c$ not a critical value of $\wp$ and choose $a$ away from zero and the two points where $\wp =c$. Thenhas a double pole at $a$ and three simple poles: the quotient has residue $-2$ at zero and residue 1 at each of its two other poles. No locations overlap. Hence its order is five and its derivative has order $3+2+2+2=9$, as required.

Assume $K$ is nonempty; without that necessary hypothesis the assertion is false. Put $d={\inf }_{z\in K}\Vert x-z{\Vert }_p$ and choose $y_n\in K$ with $\Vert x-y_n{\Vert }_p\to d$. Convexity gives $(y_n+y_m)/2\in K$, so its distance to $x$ is at least $d$. Apply the supplied uniform-convexity inequality to $x-y_n$ and $x-y_m$:Thus the minimizing sequence is Cauchy. Completeness of the Banach space ${\ell }^p$ gives $y_n\to y$; closedness gives $y\in K$, and continuity of the norm gives $\Vert x-y{\Vert }_p=d$. The minimum distance is attained. This is the nearest point in a uniformly convex Banach space property. The same inequality applied to two minimizers shows uniqueness, though existence alone was requested.

Apply Hölder's inequality to the finite sum with the second sequence $|x_j+y_j{|}^{p-1}$:Divide by the last power if the sum is nonzero. The resulting bound is independent of $N$, so passing to the limit proves both membership in ${\ell }^p$ and the Minkowski inequality

Let $q=p/(p-1)$. The scalar Young inequality $ab\le a^p/p+b^q/q$ follows by maximizing $ab-a^p/p$ over $a\ge 0$, obtaining $b^q/q$. If both norms are nonzero, apply it to $|x_j|/\Vert x{\Vert }_p$ and $|y_j|/\Vert y{\Vert }_q$, and sum over a finite initial segment. Passing to the increasing limit givesZero norms give the trivial case. This proves Hölder's inequality without assuming convergence of the product series in advance.

Suppose $y^2+21=x^3$. Its positive left side gives $x>0$. Reduction modulo four shows that $x$ is odd and $y$ even. Moreover $3\mid y$ would make the valuation of $y^2+21$ at 3 equal to one, impossible for a cube. The same argument excludes $7\mid y$.

In $\mathbb{Z}[s]$, factor $(y+s)(y-s)=(x{)}^3$. Any common prime ideal divisor divides both $2y$ and $2s$, so must lie above 2, 3 or 7. The odd norm excludes 2, and the previous divisibility arguments exclude 3 and 7. The two ideals are therefore coprime. Unique factorization of ideals gives $(y+s)=I^3$. The class group has exponent two, so $[I{]}^3=1$ implies $[I]=1$. Thus $I=(a+bs)$, and the only units, $\pm 1$, can be absorbed into a cube. It follows thatComparing the coefficient of $s$ gives $1=3b(a^2-7b^2)$, an impossibility over the integers. Consequently there are no integer solutions: this is the no integer solutions of the Mordell equation with constant minus twenty-one result. The class-group argument is essential: element factorization was not assumed to be unique.

For the ideal class group of the imaginary quadratic field of discriminant minus eighty-four, write $s=\sqrt{-21}$. Here ${\mathcal{O}}_K=\mathbb{Z}[s]$, $D_K=-84$, and the Minkowski bound for ideal classes is $(2/\pi )\sqrt{84}<6$. Thus every class has a representative of norm at most five. The relevant prime ideals areThe first two are ramified: ${\mathfrak{p}}_2^2=(2)$ and ${\mathfrak{p}}_3^2=(3)$. Also ${\mathfrak{p}}_5^2=(2-s)$, as both sides have norm 25 and $s\equiv 2(\mathrm{mod}{\mathfrak{p}}_5)$. The element $3+s$ has norm 30 andThere are no elements of norms 2, 3, 5 or 6, since the element norm is $a^2+21b^2$. Hence the classes of ${\mathfrak{p}}_2$, ${\mathfrak{p}}_3$ and their product are all nontrivial and distinct. The relation makes the last product equal to the class of ${\mathfrak{p}}_5$. Ideals of norm four are principal, and the two primes above five have the same class of order two. The norm bound therefore gives no further classes. Thus

Choose a nonzero integral ideal $I$ in the inverse of any desired ideal class. For the nonzero element $x\in I$ supplied by the norm bound, $J=(x)I^{-1}$ is integral, belongs to the desired class and satisfiesThere are only finitely many integral ideals of bounded norm: identify ${\mathcal{O}}_K$ with ${\mathbb{Z}}^n$ as an additive group, so ideals of norm $m$ are among its subgroups of index $m$, of which there are finitely many. Since every ideal class has one of these bounded-norm representatives, the ideal class group is finite.

Use the given square-class assignments and ${\omega }^3=1$, $1+\omega +{\omega }^2=0$. The character values, in the specified class order, areFor example the fourth entry of ${\chi }_S$ is $((-{\omega }^2{)}^2-\omega )/2=0$, and that of ${\chi }_A$ is $\omega$. A class of centralizer order $c$ has size $24/c$, so a character norm is the sum of its squared moduli divided by these centralizer orders. ThereforeBoth are genuine characters by their symmetric/exterior-square construction. A character norm is the sum of squares of its irreducible multiplicities, so both characters are irreducible.

For a representation space $V$, let ${\chi }_S$ and ${\chi }_A$ be the characters on ${\mathrm{Sym}}^{2}V$ and ${\bigwedge }^{2}V$. If the eigenvalues of $\rho (g)$ are ${\lambda }_1,\dots ,{\lambda }_d$, the respective traces are ${\sum }_{i\le j}{\lambda }_i{\lambda }_j$ and ${\sum }_{i<j}{\lambda }_i{\lambda }_j$. Since $\chi (g)=\sum {\lambda }_i$ and $\chi (g^2)=\sum {\lambda }_i^2$, this givesFinite-order matrices over $\mathbb{C}$ are diagonalizable, so this eigenvalue argument applies to every group element. Equivalently the symmetric and antisymmetric projections on $V\otimes V$ are $(I\pm \text{swap})/2$.

If $\rho$ affords an irreducible character $\chi$, entrywise complex conjugation gives a representation with character $\overline{\chi }$. An invariant complex subspace for the conjugate representation conjugates back to one for $\rho$, so irreducibility is preserved. If representations afford $\chi ,\psi$, their tensor product has trace $\chi (g)\psi (g)$, proving that their product is a character.

With the usual character inner product,The character orthogonality and the first assertion therefore make this equal to one if $\chi =\overline{\psi }$ and zero otherwise. This also counts the invariant vectors in the tensor-product representation.

For $r\ge 3$, join a clique on $r-2$ vertices to an independent set on $n-r+2$ vertices. The resulting complete $(r-1)$-partite graph hasIts largest clique has size $r-1$. Every missing edge joins two vertices of the independent set; adding it produces a $K_r$ with the original $(r-2)$-clique. It is therefore maximal $K_r$-free. For $n>r$ its part sizes are unbalanced, so its edge count is strictly less than that of $T_{r-1}(n)$.

For $r=3$ the construction is a star with $n-1$ edges. Every maximal triangle-free graph on $n>3$ vertices is connected: an edge between distinct components would create no triangle. A connected graph has at least $n-1$ edges, so the minimum is $n-1$, attained by the star. Literally the requested construction is impossible for $r=2$, since $T_1(n)$ has zero edges. The nontrivial assertion thus requires the customary $r\ge 3$ hypothesis.