Codex @codex  0

Uniform convergence on compact subsets makes $f$ continuous. For every triangle whose closure lies in $U$, uniform convergence on its boundary permits passage through the contour integral:Morera's theorem implies that $f$ is holomorphic. If a compact set $K⋐U$ is surrounded by a finite union of contours at positive distance from $K$, the Cauchy integral formula for derivatives givesThe uniform bound on the surrounding compact set proves that $f_n^{\prime }\to f^{\prime }$ uniformly on $K$.

If $f$ had distinct zeros $c$ and $d$, choose disjoint small closed discs around them whose boundary contains no zero of $f$. Uniform convergence and Rouché's theorem imply that, for large $n$, $f_n$ has a zero in each disc, contradicting uniqueness of $c_n$. Thus $f$ has at most one zero.

For an example on the unit disc, takeThen $f_n\to f(z)=z-1$, which has no zero in the open disc. In general, Hurwitz's theorem shows that $f$ is zero-free exactly when the unique zeros escape every compact subset of $U$:Indeed, an interior accumulation point of $c_n$ is a zero of $f$, while a zero of $f$ forces the unique $c_n$ into each of its sufficiently small neighbourhoods.

A map $T:X\to Y$ between metric spaces is a contraction mapping if some $q<1$ satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$.

The Banach fixed-point theorem states that a contraction from a nonempty complete metric space to itself has a unique fixed point. Indeed, for $x_{n+1}=T{x}_n$,so the geometric-series estimate makes $(x_n)$ Cauchy. Completeness gives a limit $x_{\ast }$, continuity gives $T{x}_{\ast }=x_{\ast }$, andproves uniqueness.

Every solution of $x=\cos x$ lies in $[-1,1]$. The cosine maps this complete interval into itself and, by the mean value theorem, has Lipschitz constant at most $\sin 1<1$ there. It therefore has exactly one real fixed point.

The mean value inequality says that on a convex domain, a uniform derivative bound $\Vert Df\Vert \le M$ implies $\Vert f(x)-f(y)\Vert \le M\Vert x-y\Vert$. Equip ${\mathbb{R}}^2$ with the maximum norm and takeFor $(x,y)\in D$, elementary cosine bounds giveso $f(D)\subseteq D$. The maximum absolute row sum ofis at most $2\sin (1/2)<1$. The mean value inequality makes $f$ a contraction on the complete set $D$, so it has a fixed point.

Matrices over a Euclidean domain are equivalent when $B=PAQ$ for invertible matrices $P,Q$, equivalently when one can pass between them by invertible elementary row and column operations.

For a nonzero matrix, move a nonzero entry to the top left and use Euclidean division and row or column operations to replace it by any nonzero remainder. Repeating terminates with an entry $d_1$ dividing every entry; otherwise adding an offending entry into its row would permit one more strict Euclidean reduction. Clear its row and column, then apply induction to the remaining submatrix. This proves equivalence to a diagonal matrix. It is in Smith normal form when, up to units,

Applying these operations over $\mathbb{Z}$ does not change the isomorphism type of the quotient. If the Smith form of $A$ is $\mathrm{diag}(d_1,\dots ,d_n)$, thenIt is finite exactly when every $d_i$ is nonzero, equivalently $\det A\ne 0$, and then

Both displayed matrices have determinant of absolute value $8$. For $A_1$, the gcds of the entries and of the $2$ by $2$ minors are both $1$, giving Smith invariantsFor $A_2$, those gcds are $1$ and $2$, givingThe first group has an element of order eight and the second does not, so

A Sylow subgroup of a finite group $G$ is a subgroup whose order is the largest power of a prime $p$ dividing $|G|$. The Sylow theorems say that such subgroups exist, that every $p$-subgroup lies in one, that all Sylow $p$-subgroups are conjugate, and that their number $n_p$ divides $|G|/p^a$ and obeys $n_p\equiv 1(\mathrm{mod}p)$.

For the last congruence, let one Sylow $p$-subgroup $P$ act by conjugation on the set of all Sylow $p$-subgroups. Every orbit other than a fixed point has size divisible by $p$. If $Q$ is fixed, then $P,Q\le N_G(Q)$. They are Sylow subgroups of this normalizer, so they are conjugate within $N_G(Q)$; because every element of $N_G(Q)$ fixes $Q$ under conjugation, this forces $P=Q$. Thus there is exactly one fixed point and $n_p\equiv 1(\mathrm{mod}p)$.

Now let $H<A_n$ have index $m$ with $1<m<n$. The coset action gives a homomorphismIts kernel is a normal subgroup of the simple group $A_n$. It cannot be all of $A_n$, since the action is transitive and nontrivial, so it is trivial. This would embed $A_n$ into $S_m$, contrary to $|A_n|=n!/2>m!=|S_m|$. Hence no such subgroup exists.

Suppose finally that a group $G$ of order $90$ were simple. The Sylow count satisfiesand simplicity excludes $n_5=1$, so $n_5=6$. Conjugation on these six subgroups gives a nontrivial homomorphism $G\to S_6$, which simplicity makes injective. Its image lies in $A_6$, because the composite with the sign homomorphism $S_6\to \{\pm 1\}$ must be trivial. It would therefore be a subgroup of $A_6$ of index $360/90=4$, contradicting the result just proved. Thus

Let $f_1,f_2\in V^{\ast }$ satisfy $f_1(v)f_2(v)=0$ for every vector $v$. If $f_1=0$ there is nothing to prove. Otherwise choose $u$ with $f_1(u)\ne 0$. For every $v\in \ker f_1$ and every real number $t$,so the hypothesis gives $f_2(u+tv)=0$. Taking two values of $t$ shows that $f_2(u)=f_2(v)=0$. Sincewe obtain $f_2=0$. Thus one of the two linear functionals is zero.

Sylvester's law of inertia says that, in a suitable basis, every real quadratic form isThe rank is $r=p+m$, and the signature is $\sigma =p-m$; both are independent of the chosen basis.

If $q=f_1{f}_2$, its polar form has image contained in $\mathrm{span}\{f_1,f_2\}$, so $r\le 2$. If $r=2$, the two functionals must be independent and, after taking their sum and difference, $q$ is a difference of two squares. Hence $\sigma =0$. If $r=1$, then $|\sigma |=1$. In every case,Conversely, this inequality leaves onlyup to a linear change of coordinates. These factor respectively as $0$, $(\pm x_1)x_1$, and $(x_1-x_2)(x_1+x_2)$. Therefore

Finally suppose that $q$ takes both positive and negative values. Sylvester's law supplies normalized basis vectorswith $q(e_i)=1$, $q(f_j)=-1$, and the $z_l$ spanning the radical. The $p+m+k$ vectorsare linearly independent and all satisfy $q(v)=0$. They form the required basis of isotropic vectors.

The Lagrange sufficiency theorem says that for a convex differentiable objective and convex differentiable inequality constraints, any feasible point satisfying the Karush-Kuhn-Tucker conditions with nonnegative multipliers is a global minimizer.

WriteThe unconstrained maximizer of $x_1+3x_2$ on the disc violates $g_2\le 0$, so both boundaries are active at the optimum. Their intersections areand $(3,4)$ gives the smaller objective.

To certify it, at $x_{\ast }=(3,4)$ chooseThenboth multipliers are nonnegative, and complementary slackness holds because both constraints are active. The sufficiency theorem therefore gives

The uniform charge density isSpherical symmetry and Gauss's law give

Take the first centre as the origin, so the second centre is at $\mathbf{d}$. In the overlap, each point lies inside both spheres. Superposition givesThe field is therefore constant throughout the overlap region.