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At every regular point of the Riemann sphere, there is a neighborhood containing no poles; at each pole, there is a neighborhood containing no other poles. These neighborhoods cover the sphere. Its compactness gives a finite subcover, each member containing at most one pole. Hence the total number of poles, including a possible pole at infinity, is finite. Equivalently, an infinite pole set would have an accumulation point on the compact sphere, contrary to local regularity or isolation there.

Use the local coordinate $w=1/z$ around infinity. A pole at $w=0$ is isolated: for some $\epsilon >0$, $g(w)=f(1/w)$ is holomorphic at every point $0<|w|<\epsilon$. Taking $r=1/\epsilon$, the change of variable maps this punctured disk to $r<|z|<\infty$. Therefore there are no finite poles outside that disk. This is a property of the local meromorphic chart at infinity, not an assertion that infinity itself is regular.

Continuity on the compact product of unit spheres bounds the bilinear map: there is $C$ such that $\Vert f(h,k){\Vert }_3\le C\Vert h{\Vert }_1\Vert k{\Vert }_2$. This follows by scaling each nonzero vector to unit norm; if one factor space is zero, the map and the relevant assertion are trivial.

Bilinearity gives the exact expansionThe proposed derivative $L(h,k)=f(h,b)+f(a,k)$ is linear and bounded, with $\Vert L(h,k){\Vert }_3\le C(\Vert b{\Vert }_2\Vert h{\Vert }_1+\Vert a{\Vert }_1\Vert k{\Vert }_2)$. Set $\rho =(\Vert h{\Vert }_1^2+\Vert k{\Vert }_2^2{)}^{1/2}$. Its remainder obeysThus the derivative of a continuous bilinear map is the Fréchet derivative

By equivalent norms, choose $c,C>0$ with $c\Vert h{\Vert }_1\le \Vert h{\Vert }_2\le C\Vert h{\Vert }_1$. Differentiability in the first norm means that there is a bounded linear map $L:V\to \mathbb{R}$ with $f(a+h)-f(a)-Lh=r(h)$ and $|r(h)|/\Vert h{\Vert }_1\to 0$. The same $L$ is bounded in the second norm, since $|Lh|\le K\Vert h{\Vert }_1\le K{c}^{-1}\Vert h{\Vert }_2$. Moreover,The two norms define the same approach $h\to 0$, so this proves differentiability with the same Fréchet derivative in the second norm. Reversing the norms proves the converse.

For completeness, the derivative is unique: if $L_1,L_2$ both work, fix $v$ and take $h=tv$. The two remainder estimates imply $|(L_1-L_2)v|=o(1)$ as $t\to 0$, so $(L_1-L_2)v=0$ for every $v$. Thus differentiability and its derivative are independent of the chosen norm. This is differentiability under equivalent norms, not merely invariance of continuity.

An enclosing loop has no spanning disk wholly inside the fluid, so initial zero vorticity does not force its circulation to vanish. Kelvin's circulation theorem says only that the initial circulation is retained:For example $\mathbf{u}={\Gamma }_0{\mathbf{e}}_{\theta }/(2\pi r)$ has zero vorticity outside the cylinder but circulation ${\Gamma }_0$ around a once-enclosing positively oriented loop. Any two such loops have equal circulation, since their difference bounds an annulus in the fluid and Stokes theorem gives zero over it. More generally a loop with winding number $m$ has circulation $m{\Gamma }_0$. Thus irrotational flow need not have a single-valued velocity potential on the whole exterior domain.

A simple closed loop that does not surround the cylinder bounds a surface wholly in the fluid. At time zero, Stokes theorem gives zero circulation because its vorticity is zero. The smooth impermeable exterior flow transports this loop without changing its winding about the obstacle. Pulling a loop at time $t$ back to time zero and using Kelvin's circulation theorem therefore givesThis is the contractible-loop case of irrotational circulation in an exterior domain. Equivalently, the initially vanishing vorticity remains zero under the inviscid barotropic vorticity equation.

True. Choose a basis $u_1,\dots ,u_r$ of $U$ and extend it to a basis $u_1,\dots ,u_r,w_1,\dots ,w_s$ of the finite-dimensional vector space $V$. Put $W=\mathrm{span}(w_1,\dots ,w_s)$. Every basis vector belongs to $U+W$, so $U+W=V$. Linear independence of the complete basis makes a vector belonging to both spans necessarily zero. ThereforeThis constructs a direct-sum complement; it need not be unique.

False. In ${\mathbb{R}}^2$ let $U=\mathrm{span}(1,1)$, $W=\mathrm{span}(1,0)$ and $X=\mathrm{span}(0,1)$. Then $W+X={\mathbb{R}}^2$, so $U\cap (W+X)=U$, but $U\cap W=U\cap X=\{0\}$ and their sum is zero. Thus intersection need not distribute over the sum of vector subspaces. The right side is always contained in the left; this example makes the inclusion strict.

False. Distinct vector subspaces may be nested: in $V={\mathbb{R}}^2$, take $U=\{0\}$ and $W=\mathrm{span}(1,0)$. Their union is $W$, a subspace. More precisely, $U\cup W$ is a subspace exactly when one is contained in the other. If neither contains the other, choose $u\in U\setminus W$ and $w\in W\setminus U$. Were $u+w$ in the union, subtracting $u$ or $w$ in the appropriate subspace would contradict those choices.

True. Both vector subspaces contain zero, and if $u,v\in U\cap W$, every linear combination $\alpha u+\beta v$ belongs to each of $U$ and $W$. Thus it belongs to their intersection. This proves the intersection of vector subspaces is a vector subspace.

The map $h:X\to X\times Y$, $h(x)=(x,f(x))$, is continuous: the inverse image of a basic open set $U\times V$ is $U\cap f^{-1}(V)$. Its image is exactly the graph of a function $G_f$. The continuous image of a compact space is compact, soThe compact graph of a continuous map conclusion does not need $Y$ to be Hausdorff. In fact projection onto $X$ is a continuous inverse of $h:X\to G_f$, so the graph is homeomorphic to $X$.

Take $(x_0,y_0)$ outside the graph of a function, so $y_0\ne f(x_0)$. Since $Y$ is a Hausdorff space, choose disjoint open neighborhoods $V$ of $y_0$ and $W$ of $f(x_0)$. Continuity makes $U=f^{-1}(W)$ an open neighborhood of $x_0$. The product neighborhood $U\times V$ contains $(x_0,y_0)$ and cannot meet the graph, because every $f(x)$ with $x\in U$ belongs to $W$. The complement is therefore open in the product topology, provingThis is the closed graph of a map into a Hausdorff space property; no separation assumption on $X$ is required.

Here $H_2=\left(\begin{array}{cc}0& i\\ -i& 0\end{array}\right)$ and $H_2^2=I$. Its time evolution for an additional interval $\tau$ isAt $\tau =T=\pi \hslash /2$, $U_2(T)=-i{H}_2$. Acting on the state from the previous part givesIt differs from the initial state only by a global phase. For every observable or projection the phase cancels between bra and ket, so the final state is observationally indistinguishable from the original isolated state. A relative phase between two components or between coherently controlled alternatives would be observable; this overall phase of the entire state is not.

In the ordered eigenbasis, $H_1=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, a Pauli matrix. The coefficient equations are $i\hslash {\dot{a}}_0=a_1$ and $i\hslash {\dot{a}}_1=a_0$, with $a_0(0)=1$, $a_1(0)=0$. Differentiating the first gives ${\ddot{a}}_0=-a_0/{\hslash }^2$, and its initial derivative vanishes. ThusThe population of ${\chi }_1$ is ${\sin }^2(t/\hslash )$. To be an $H_3$ eigenstate of eigenvalue $+1$, the coefficient of ${\chi }_0$ must vanish. Its first positive zero isThe global phase does not change its eigenstate or measurement probabilities.

The Time-dependent Schrödinger equation is $i\hslash {\partial }_t\Psi =H\Psi$. The normalized eigenstates of the Hermitian Hamiltonian $H_3$ with distinct eigenvalues are orthogonal, so they form an orthonormal basis of this two-state system. Substitute the expansion of $\Psi$ and compare the two coefficients:Hence $a_0(t)=a_0(0)e^{it/\hslash }$ and $a_1(t)=a_1(0)e^{-it/\hslash }$. For the normalized test state $\varphi =\alpha {\chi }_0+\beta {\chi }_1$, the Born rule givesThe complex conjugates are essential; relative phases can affect this probability even though the energy-state populations remain constant under $H_3$.

Write $A=D+L+U$, with $D$ diagonal, and use only old iterates on the right side. The Jacobi method isStrict row diagonal dominance ensures every $a_{ii}\ne 0$. The exact solution satisfies the same coordinate relation. For $e^k=x^k-x^{\ast }$, subtraction givesTaking the maximum and iterating provesThis direct contraction proof requires no eigenvector assumption and applies to every strictly diagonally dominant matrix with the stated uniform $\gamma <1$.

Since every eigenvalue has modulus less than one, $1$ is not an eigenvalue and $I-H$ is invertible. Hence the iteration has a unique fixed point $\hat{x}=(I-H{)}^{-1}v$. The stated consistency implication gives $A\hat{x}=b$, and nonsingularity of $A$ identifies $\hat{x}=x^{\ast }$. This justifies using the fixed-point relation even though consistency was stated in only one direction.

Expand the initial error in the eigenvector basis, $x^0-x^{\ast }={\sum }_i{c}_i{w}_i$. Subtract the fixed-point equation from the stationary iterative method for a linear system and iterate:With $q={\max }_i|{\lambda }_i|<1$, any norm obeys $\Vert x^k-x^{\ast }\Vert \le q^k{\sum }_i|c_i|\Vert w_i\Vert \to 0$. ThusThe argument uses the given diagonalizable eigenbasis; the more general convergence criterion is spectral radius less than one.

For nonnegative integers $m,n$, put $B(z)=z^m[(z-a)/(1-\bar{a}z){]}^n$. Since $|a|<1$, the denominator has no zero on or inside the unit circle. On $|z|=1$,so $|B(z)|=1$. Consequently $|-a|<|B(z)|$, and Rouché's theorem gives the same interior zero count for $B-a$ as for $B$. The latter has $m$ zeros at zero and $n$ at $a$, with the multiplicities combining when $a=0$. ThereforeThis is the interior zero count for a finite Blaschke product minus a constant. If $m+n=0$, the nonzero constant $1-a$ has zero zeros, so the same conclusion includes that endpoint case.

The function is holomorphic on a neighborhood of the closed unit disk. On $|z|=1$, the mapping assumption gives $|g(z)|<1=|z|$. Apply Rouché's theorem to $z$ and $-g(z)$. Their sum $z-g(z)$ has the same number of interior zeros as $z$, namely one counting multiplicity. ThusThe zero is simple since its total multiplicity is one. No boundary fixed point is possible because $g$ maps the boundary strictly inside.

For $0\le t\le 1$, put $F_t=f+tg$. On the curve, $|F_t|\ge |f|-t|g|>0$, so no member of this homotopy has a boundary zero. The argument principle counts the interior zeros, with multiplicity, by its winding number $N(t)=\mathrm{wind}(F_t\circ \gamma ,0)$. For a piecewise smooth curve this isThere are no poles, and the null-homotopy of the Jordan curve in the domain permits this interior count. The loops $F_t\circ \gamma$ form a continuous homotopy avoiding zero, so their integer winding number stays constant. Equivalently, for a piecewise smooth curve the integral varies continuously with $t$ and is integer-valued. In particularThis proves Rouché's theorem. The counting is with multiplicity and assumes the usual positive orientation; reversing orientation reverses both winding counts without changing the conclusion.