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For the anticlockwise boundary of the region, the inward normal is the left normal, so the signed curvature obeys the turning tangent theorem ${\int }_0^{L}k(s)ds=2\pi$. Therefore some $s_0$ has $k(s_0)\le 2\pi /L$. The enclosed area satisfies $A\ge d_1{d}_2$, and the planar isoperimetric inequality gives $L\ge 2\sqrt{\pi A}$. ConsequentlyThis uses the interpretation of the curve as a simple boundary traversed once; the inward normal and enclosed region in the question have that meaning.

The Fenchel theorem states that every closed regular space curve has total curvature at least $2\pi$:Equality holds precisely for a convex plane curve, traversed once.

For a regular simple closed plane curve of length $L$ enclosing area $A$, the planar isoperimetric inequality is $L^2\ge 4\pi A$, with equality precisely for a circle.

The periodic Wirtinger inequality states that an absolutely continuous $L$-periodic function $h$ of mean zero with square-integrable derivative satisfieswith equality only for a linear combination of $\sin (2\pi s/L)$ and $\cos (2\pi s/L)$. Parametrize the curve by arc length and translate its position vector $r=(x,y)$ so that both coordinate means vanish. Then $\int |r{|}^2\le L^3/(4{\pi }^2)$, since $|r^{\prime }|=1$. Green's theorem and the Cauchy-Schwarz inequality giveEquality requires a first-harmonic parametrization and equality in Cauchy-Schwarz; its radius vector is a constant multiple of the rotated unit tangent. It is therefore a circle. Conversely, a circle attains equality.

Euler's criterion gives $\chi (-1)=(-1{)}^{(p-1)/2}=1$ in this case. There are equally many nonzero quadratic residues and quadratic nonresidues, so ${\sum }_{n=1}^{p-1}\chi (n)=0$. Replacing $n$ by $p-n$ and using the multiplicativity of the Legendre symbol givesConsequently $\boxed{S=0}$.

Put $h=(p-1)/2$. The nonzero quadratic residues in the finite field ${\mathbb{F}}_p$ form a subgroup of size $h$: the group homomorphism $x\mapsto x^2$ has kernel $\{1,-1\}$. Each square satisfies $u^h=1$ by Fermat's little theorem. Since a nonzero polynomial of degree $h$ has at most $h$ roots of a polynomial, these are all the roots of $X^h-1$. For any other nonzero $u$, Fermat's theorem gives $(u^h{)}^2=1$, so $u^h=-1$. Thus Euler's criterion follows, and multiplication givesBoth sides are $1$ or $-1$, which are distinct modulo the odd prime number $p$, so $\chi (mn)=\chi (m)\chi (n)$. Taking $n=-1$ in Euler's criterion gives $\chi (-1)=(-1{)}^{(p-1)/2}$, which is $1$ for $p\equiv 1(\mathrm{mod}4)$ and $-1$ for $p\equiv 3(\mathrm{mod}4)$.

Each $H$-conjugacy class represented by $x_i$ has $|H|/|C_H(x_i)|$ elements. For every element of that class, the number of conjugators from $g$ to it is $|C_G(g)|$. Grouping the induced character formula by these classes therefore gives

In the example, $a$ has order four, $b$ has order two, and $bab=a^{-1}$, so $H\cong D_8$. Its five conjugacy classes are $1$, $a^2$, $\{a,a^3\}$, $\{b,a^{2}b\}$ and $\{ab,a^{3}b\}$, of sizes $1,1,2,2,2$. The four one-dimensional irreducible characters are determined by $\psi (a)=\epsilon$ and $\psi (b)=\eta$, each sign independently chosen. The remaining two-dimensional irreducible character comes from the plane action of the square:Their squared degrees sum to eight, so this is the full character table.

In $S_4$, use class representatives $1,(12),(12)(34),(123),(1234)$. Their intersections with $H$ are respectively $\{1\}$, $\{b,a^{2}b\}$, $\{a^2\}\cup \{ab,a^{3}b\}$, the empty set, and $\{a,a^3\}$. The corresponding $S_4$ centralizer orders are $24,4,8,3,4$, while the noncentral $H$ centralizers have order four. The induced values are consequentlyIn the same $S_4$ class order, all five induced irreducible characters are

In the induced character formula above, there are no $x\in G$ with $x^{-1}gx\in H$ when the conjugacy class of $g$ misses $H$. The sum is empty, so $({\mathrm{Ind}}_H^G\psi )(g)=0$.

No. In the six-cycle with cyclic vertices $1,2,\dots ,6$, every edge belongs to one of the two alternating perfect matchings. However, the disjoint edges $12$ and $45$ cannot belong to a common perfect matching: the remaining vertices $3,6$ are not adjacent. Taking $X=\{1,3,5\}$ and $Y=\{2,4,6\}$ gives the required bipartite graph counterexample.

Take a disjoint union of $n$ single edges, each joining one vertex of $X$ to one of $Y$. Its unique matching in a graph covering $X$ contains every edge, while any singleton $A\subset X$ satisfies $|\Gamma (A)|=|A|=1$. For $n\ge 2$ this $A$ is nonempty and proper. This illustrates why connectedness was needed in part (ii).

Suppose a nonempty proper $A$ is tight. In every matching in a graph covering $X$, the vertices of $A$ use every vertex of $\Gamma (A)$. Hence no edge from $X\setminus A$ to $\Gamma (A)$ could occur in such a matching. The assumption that every edge occurs therefore rules out all such edges. By definition there are also no edges from $A$ to $Y\setminus \Gamma (A)$. This separates $A\cup \Gamma (A)$ from the rest, which contains the nonempty set $X\setminus A$, contradicting connectedness. Hall's inequality must therefore be strict on every nonempty proper $A$.

Fix any edge $xy$. After deleting $x$ and $y$, every nonempty subset $A$ of $X\setminus \{x\}$ satisfies $|\Gamma (A)\setminus \{y\}|\ge |A|$, by the strict inequality in the question. The Hall marriage theorem gives a matching in a graph for the remainder; adding $xy$ gives a matching covering $X$ and containing the prescribed edge. The case $|X|=1$ is immediate.

No such map exists. If $H$ fixed the boundary, it would satisfy the hypothesis of part (iii) for every $\epsilon >0$. Choose $0<\epsilon <1-|q|$. Then part (iii) would give $q\in H(B)$, contradicting the stipulated omission of $q$. Equivalently, radial projection from the omitted point would turn $H$ into a forbidden retraction.

Fix $z$ with $|z|<1-\epsilon$, and let $\Pi (w)=w/\max (1,|w|)$ be radial projection to the closed disc. The continuous map $p\mapsto \Pi (p-G(p)+z)$ has a fixed point $p$ by the Brouwer fixed-point theorem. On the boundary, $|p-G(p)+z|<\epsilon +|z|<1$, so its image under $\Pi$ is interior and cannot equal the boundary point $p$. Therefore $p\in B$. An interior fixed point of this projection must satisfy $p-G(p)+z=p$, because points outside the disc project to its boundary. Thus $G(p)=z$. Every $z\in B_{1-\epsilon }(0)$ belongs to $G(B)$.

Suppose first that a retraction $R:D\to C$ exists. The continuous map $p\mapsto -R(p)$ takes $D$ to itself, so the Brouwer fixed-point theorem gives $p=-R(p)$. Such $p$ lies on $C$, where $R(p)=p$, contradicting $p=-p$.

Conversely, suppose a continuous map $T:D\to D$ has no fixed point. For each $p$, follow the ray from $T(p)$ through $p$ to its last intersection with $C$. This defines a retraction. More explicitly, put $a=p-T(p)\ne 0$ and setThe positive-root formula is continuous, $t(p)\ge 1$, and $R(p)$ is on the unit circle. If $p\in C$, the last intersection is $p$ itself, so $R(p)=p$. Thus the nonexistence of a retraction rules out a fixed-point-free $T$, proving the equivalence.

The planar Brouwer fixed-point theorem states that every continuous map from a closed disc to itself has a fixed point.

Write $A=D+L+U$, with $D$ diagonal and $L,U$ the strict triangular parts. The Jacobi method updates all coordinates using the preceding iterate:For a symmetric positive definite matrix, $D$ has positive diagonal entries. In its splitting take $M=D$, $N=D-A$. The theorem requires $M^{\ast }+N=2D-A$ to be positive definite, which need not hold for every positive definite matrix.

For this tridiagonal matrix, let $S=\mathrm{diag}((-1{)}^1,\dots ,(-1{)}^n)$. Neighboring indices have opposite signs, so $SAS=2D-A$: the diagonal stays fixed and every nonzero off-diagonal entry changes sign. ThusThe Householder-John theorem now applies and proves Jacobi convergence for every initial vector. This is precisely why the tridiagonal structure matters.

The Householder-John theorem says that, for a splitting $A=M-N$ with $A$ Hermitian positive definite, positivity of $H=M^{\ast }+N=M^{\ast }+M-A$ implies $M$ is invertible and $\rho (M^{-1}N)<1$. Invertibility follows since $Mx=0$ for $x\ne 0$ would give $x^{\ast }Hx=-x^{\ast }Ax<0$.

For clarity, the spectral-radius conclusion follows from a useful energy identity. Put $T=M^{-1}N=I-M^{-1}A$. Direct multiplication givesIf $Tv=\lambda v$, thenThe eigenvalue one is impossible because $A$ is invertible. Both quadratic forms are positive, so $|\lambda |<1$. The splitting iteration $M{x}^{(k+1)}=N{x}^{(k)}+b$ propagates its error by $T$; spectral radius less than one therefore gives convergence from every initial vector.

The baffles create an induced pressure gradient even though no external gradient is applied. Away from them the same Couette-Poiseuille flow has flux per unit widthZero flux therefore gives $G=-6U/a^2$ andThe positive pressure gradient drives the return flow near the stationary lower plate, balancing the forward flow near the moving plate. The rectilinear approximation excludes the short end regions next to the baffles.

The steady rectilinear equation is $u^{\prime \prime }(y)=-G$, with no-slip conditions $u(0)=0$ and $u(a)=U$. Two integrations give the Couette-Poiseuille flowThe first term is driven by the moving wall and the second by the specified pressure gradient.

The substitution $u=e^{-x}$ changes the beta function toFor fixed $t>0$, its factor near zero expands asAway from zero the factor is bounded, so the extension from the finite-interval Watson lemma to this exponentially damped tail is immediate. ThereforeThe change $u\mapsto 1-u$ gives $B(s,t)=B(t,s)$, so, with $s>0$ fixed,These expansions hold with the other parameter fixed; the first does not claim uniformity when $t$ grows with $s$. The printed gamma-integral hint has inconsistent variable labels: the integration differential is $dx$ and the positive parameter is $y$.