Use Picard iteration on $[a-h,a+h]$:By induction, every iterate stays in the prescribed rectangle: if $u_n$ does, then $|u_{n+1}(x)-b|\le M|x-a|\le Mh\le k$. Continuity makes each next integral well defined. The Lipschitz condition gives, for $n\ge 0$,For $n=0$ this is the previous bound; integration of the bound multiplied by $A$ proves the next step, on either side of $a$.

The sum of these bounds converges uniformly on the entire interval. Thus $u_n$ converges uniformly to a continuous $u$, whose graph remains in the closed rectangle. The Lipschitz condition also makes $f(t,u_n(t))$ converge uniformly to $f(t,u(t))$, so passage to the limit givesThe fundamental theorem of calculus makes $u$ a continuously differentiable solution with $u(a)=b$.

For uniqueness, let $u,v$ be two solutions and put $D_0=\sup |u-v|$ on the interval. Repeated use of the Lipschitz condition in their integral equations givesfor every $n$. Letting $n\to \infty$ gives $u=v$. The global bound $M$ prevents any competing solution whose graph lies in $D$ from exiting the rectangle through a horizontal boundary before the end of the interval. Therefore a unique solution exists on the full stated interval. No extra restriction $Ah<1$ is needed; the factorial estimates replace that restriction.

For a holomorphic function on $|z-a|<R$, the complex Taylor theorem stateswith uniform convergence on each smaller closed disc. Here is a proof. Choose $r<\rho <R$. For $|z-a|\le r$ and $|\zeta -a|=\rho$, the geometric seriesconverges uniformly. Substitute it into Cauchy integral formula and integrate term by term to getIf $M_{\rho }={\max }_{|\zeta -a|=\rho }|f(\zeta )|$, then $|c_n|\le M_{\rho }{\rho }^{-n}$. These bounds justify differentiating the power series any fixed number of times on a smaller closed disc, by comparison with convergent differentiated geometric series. Evaluation at $a$ gives $f^{(n)}(a)=n!c_n$. Since every $|z-a|<R$ belongs to some such smaller disc, the formula holds throughout the stated domain.

The identity theorem says that two holomorphic functions on a connected open subset of $\mathbb{C}$ are equal everywhere if their equality set has an accumulation point in that subset. Equivalently, a holomorphic function with an interior accumulation point of zeros is identically zero there.

The proposed Taylor coefficients at $i$, including the specified value for $n=0$, giveThe function $1/z$ is holomorphic on $\mathbb{C}\setminus \{0\}$ and has exactly the given derivatives, so it supplies existence. Any other admissible holomorphic function has the same Taylor series on $|z-i|<1$. The punctured plane is connected, so the identity theorem gives uniqueness:

Use the following boundary version of Cauchy's integral theorem: if a function is holomorphic on a neighborhood of a compact region bounded by finitely many piecewise smooth simple closed curves, its integral around the positively oriented boundary is zero. On an annulus this means that its integrals around the two circles, both taken counterclockwise, are equal.

For $0<\rho <r$, the function $f(z)/(z-w)$ is holomorphic on a neighborhood of $\rho \le |z-w|\le r$. The theorem therefore givesAs $\rho \downarrow 0$, continuity of $f$ makes the integrand converge uniformly to $f(w)$. The right side tends to $f(w)$, proving Cauchy integral formula without assuming that formula in an intermediate step.

Let $I$ denote the requested boundary integral. Parametrize the unit circle by $z=e^{it}$. Since $dz=i{e}^{it}dt$,Taking complex conjugates and using $\overline{e^{it}}=e^{-it}$ givesThis is the conjugated Cauchy integral on a circle identity. The function $h(z)=f(z)/(1-\overline{a}z)$ is holomorphic on a neighborhood of the closed unit disc, because $|a|<1$. Applying Cauchy integral formula at zero gives $\overline{I}=h(0)=f(0)$. ThusThe complex conjugates in these formulas are essential; the boundary integrand contains $\overline{f(z)}$.

A maximal ideal is a proper ideal $I$ of $R$ such that an ideal $J$ with $I\subseteq J\subseteq R$ is either $I$ or $R$. Suppose first that $I$ is a maximal ideal. If $a\notin I$, then $I+(a)=R$, so $1=i+ra$ for some $i\in I$ and $r\in R$. In the quotient ring, $(r+I)(a+I)=1+I$. Every nonzero element of $R/I$ therefore has a multiplicative inverse, so $R/I$ is a field. Conversely, if $R/I$ is a field and $J$ properly contains $I$, choose $a\in J\setminus I$. The element $a+I$ has a multiplicative inverse $r+I$, whence $1-ra\in I\subset J$, giving $1\in J$ and $J=R$. ThusThe squares in ${\mathbb{F}}_7$ are $0,1,2,4$, which do not include $-1=6$. Hence $X^2+1$ has no root in ${\mathbb{F}}_7$. A polynomial of degree two or three over a field is an irreducible polynomial exactly when it has no root: any proper polynomial factorization of such a degree has a factor of degree one. The polynomial ring over a field is a Euclidean domain, so the ideal generated by an irreducible polynomial is a maximal ideal. It follows thatis a field. Every coset has a unique representative $a+bX$, by polynomial division, so $|F|=7^2=49$.

Let $j$ be the coset of $X$ in this field. Since $j^2=-1$, the mapis a ring homomorphism. It is surjective by the representative description. Since $1,j$ are linearly independent over ${\mathbb{F}}_7$, its kernel of a ring homomorphism isThere is also a field with $343$ elements. The cubes in ${\mathbb{F}}_7$ are $0,1,-1$, so $X^3-2$ has no root and is an irreducible polynomial. Thereforeis a field whose unique representatives have degree below three. There are $7^3=343$ such representatives.

The adjoint on an inner product space ${\alpha }^{\ast }:V\to V$ is characterized byHere is a finite-dimensional construction that supplies every required basis fact. If $v_1,\dots ,v_n$ is any basis, its coordinate linear functionals form a basis of $V^{\ast }$: a linear functional is uniquely specified by its values on these $n$ vectors. Hence $\dim V^{\ast }=n$. The linear mapis injective, because $J(w)=0$ implies $⟨w,w⟩=0$, and positive definiteness implies $w=0$. Equal dimensions make $J$ an linear isomorphism. If ${\alpha }^{\prime }$ is the dual map, thenThis establishes existence and linearity. If two possible vectors $u,w$ represent the same linear functional through the inner product, then $J(u-w)=0$, so $u=w$. Applied to every argument, this proves uniqueness of the adjoint on an inner product space.

For linear maps $S,T$, two applications of the defining inner product identity giveUniqueness therefore gives $(ST{)}^{\ast }=T^{\ast }{S}^{\ast }$. Likewise $(S+T{)}^{\ast }=S^{\ast }+T^{\ast }$, $(cT{)}^{\ast }=c{T}^{\ast }$ for real $c$, and $I^{\ast }=I$. An induction gives $({\alpha }^n{)}^{\ast }=({\alpha }^{\ast }{)}^n$, and applying additivity to a polynomial with real coefficients givesSymmetry of the real inner product also gives $({\alpha }^{\ast }{)}^{\ast }=\alpha$. If $m$ is the minimal polynomial of $\alpha$, the displayed identity gives $m({\alpha }^{\ast })=0$, so the minimal polynomial of ${\alpha }^{\ast }$ divides $m$. Applying the same argument to ${\alpha }^{\ast }$ gives reverse divisibility. Both polynomials are monic, hence their minimal polynomials are equal.

The distinct eigenvalues are exactly the roots of the minimal polynomial in the chosen field. An eigenvector $v$ with eigenvalue $\lambda$ satisfies $m(\alpha )v=m(\lambda )v$, so $m(\lambda )=0$. Conversely, if $m(X)=(X-\lambda )q(X)$ and $\alpha -\lambda I$ were a linear isomorphism, then $m(\alpha )=0$ would imply $q(\alpha )=0$, contradicting minimal degree. Thus $\alpha -\lambda I$ is not injective, giving an eigenvector. It follows that $\alpha$ and ${\alpha }^{\ast }$ have the same distinct eigenvalues. This includes the complex eigenvalues after complexification of a real vector space; over the real field it states equality of their real eigenvalues.

Write $V^{\ast }=V^{\prime }$ for the dual space. Define $\theta (v)(f)=f(v)$. For fixed $v$, this is a linear functional on $V^{\ast }$; evaluation is also linear in $v$, so $\theta :V\to V^{\ast \ast }$ is a linear map. If $v\ne 0$, extend $v$ to a basis and take its first coordinate linear functional $f$; then $f(v)=1$. Thus $\theta$ is injective. A dual basis has the same number of elements as the original basis, so $\dim V=\dim V^{\ast }=\dim V^{\ast \ast }$. The rank-nullity theorem now makes $\theta$ an linear isomorphism.

The annihilator of a vector subspace of a linear subspace $U\le V$ isExtend a basis $v_1,\dots ,v_k$ of $U$ to a basis $v_1,\dots ,v_n$ of $V$, with dual basis $f_1,\dots ,f_n$. A linear functional belongs to $U^{\circ }$ exactly when its first $k$ coordinates in this dual basis vanish. Hence $U^{\circ }$ has basis $f_{k+1},\dots ,f_n$, provingThe dual map is ${\alpha }^{\ast }(h)=h\circ \alpha$; here this is the algebraic dual map, denoted ${\alpha }^{\prime }$ in the source. If $v\in \ker \alpha$, then $({\alpha }^{\prime }h)(v)=h(\alpha v)=0$. Consequently $\mathrm{im}{\alpha }^{\prime }\subseteq (\ker \alpha {)}^{\circ }$. Moreover,because coordinate linear functionals separate vectors. This proves $(\mathrm{im}{\alpha }^{\prime }{)}^{\circ }=\theta (\ker \alpha )$, including the requested second inclusion. Taking dimensions givesThe first inclusion has equal dimensions on both sides, soFor a matrix $A=(a_{ij})$ representing $\alpha$, the matrix of ${\alpha }^{\prime }$ in the corresponding dual bases is $A^{\mathsf{T}}$: its $ji$ entry is $h_i(\alpha v_j)=a_{ij}$. Thus the column rank of $A^{\mathsf{T}}$ is the row rank of $A$, while the column rank of $A$ is $\dim \mathrm{im}\alpha$. Applying the established equality to these two linear maps proves row rank equals column rank over every field.