If $(a^{(n)})$ is Cauchy in the maximum norm on ${\mathbb{C}}^m$, every coordinate is Cauchy in $\mathbb{C}$ and has a limit $a_j$. Given $\epsilon >0$, for sufficiently large $n,k$ every coordinate difference is at most $\epsilon$. Letting $k\to \infty$ gives ${\max }_j|a_j^{(n)}-a_j|\le \epsilon$. Thus ${\mathbb{C}}^m$ with its maximum norm is complete.

Let $F(a)=\sum a_j{z}_j$. The triangle inequality givesTake $\delta =\sum \Vert z_j\Vert$. This estimate also makes $a\mapsto \Vert F(a)\Vert$ continuous on the maximum-norm unit sphere, which is compact by coordinatewise Bolzano-Weierstrass. Linear independence makes this function strictly positive there, so its minimum $\gamma$ is positive. Homogeneity then givesA Cauchy sequence in $X$ consequently has Cauchy coordinate vectors in ${\mathbb{C}}^m$. Their coordinate limit maps by the upper estimate to a limit in $X$. Therefore every finite-dimensional normed complex space is a Banach space.

Anti-self-adjoint means $T^{\ast }=-T$. Unitary means $T^{\ast }T=T{T}^{\ast }=I$, so $T^{-1}=T^{\ast }$. For a unitary operator $\Vert Tu{\Vert }^2=⟨u,T^{\ast }Tu⟩=\Vert u{\Vert }^2$, and the space is nonzero. Thus $\Vert T\Vert =1$.

The series for the exponential converges absolutely in the Banach algebra $B(V)$, since ${\sum }_n\Vert T^n/n!\Vert \le e^{\Vert T\Vert }$. The adjoint is norm continuous and conjugate linear, and its action on these real scalar coefficients gives $(e^T{)}^{\ast }=e^{T^{\ast }}$. The absolutely convergent Cauchy product givesbecause its degree-$n$ coefficient for $n>0$ is $T^n{\sum }_{j=0}^n(-1{)}^j/[j!(n-j)!]=0$. If $T^{\ast }=-T$, these identities prove $e^T$ is unitary.

If $T$ is invertible, zero is outside the spectrum of $T^{-1}$. For $\mu \ne 0$ the factorizationshows that the left side is invertible exactly when the final factor is. HenceFor unitary $T$, both $T$ and $T^{-1}=T^{\ast }$ have norm one. The norm bound on each spectrum gives $|\lambda |\le 1$ and $|{\lambda }^{-1}|\le 1$, forcing $\sigma (T)$ to lie on the unit circle.

The converse fails: on ${\mathbb{C}}^2$ with its usual inner product takeIts spectrum is $\{1\}$, but $T^{\ast }T=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)\ne I$, so it is not unitary.

The classical Fourier partial sums areHere $a_0={\pi }^{-1}\int f$, and functions are extended periodically. One version of the Dini test for Fourier series states that if one-sided limits $L_{+},L_{-}$ at $t$ exist andthen $s_n^f(t)\to (L_{+}+L_{-})/2$. A step function on a finite interval is constant on each side of any given point for sufficiently small distance, so its displayed numerator is zero there. Dini test for Fourier series gives exactly the asserted midpoint of the two one-sided values, including at jumps.

Let $P_n=s_n$ on $L^2$. It is the orthogonal projection onto the span of the displayed trigonometric functions through frequency $n$, hence $\Vert P_n\Vert \le 1$. For a step-function approximation $\psi$,The convergence on a dense set under a uniform operator bound, here applied using the allowed convergence for $\psi$ and density of step functions, proves $P_{n}f\to f$ in $L^2$. Consequently the union of these finite trigonometric spans is dense, which is precisely completeness of the orthonormal system. It also proves Parseval's equality by passing to the limit in the Pythagorean projection identity.

For $f(t)=t^2$, oddness gives $b_j=0$, direct integration gives $a_0=2{\pi }^2/3$, and two integrations by parts give $a_j=4(-1{)}^j/j^2$. ThusDini test for Fourier series justifies this pointwise on the interval, or absolute coefficient summability gives its continuous extension after the $L^2$ identification. At zero it yields ${\sum }_{j\ge 1}(-1{)}^{j-1}/j^2={\pi }^2/12$. Parseval givesTherefore ${\sum }_{j\ge 1}{j}^{-4}={\pi }^4/90$.

The spectrum of a bounded operator consists of $\lambda$ for which $T-\lambda I$ has no bounded two-sided inverse. The point spectrum consists of eigenvalues, so $\ker (T-\lambda I)\ne 0$. The approximate point spectrum consists of $\lambda$ admitting unit vectors $x_n$ with $\Vert (T-\lambda I)x_n\Vert \to 0$; equivalently $T-\lambda I$ is not bounded below.

If $T-\lambda I$ is invertible, its inverse bounds it below and its Banach-space adjoint is invertible. Thus neither alternative on the right of the claimed union occurs. Conversely suppose $\lambda$ belongs to neither term and write $S=T-\lambda I$. There is $c>0$ with $\Vert Sx\Vert \ge c\Vert x\Vert$, so $S$ is injective and has closed range: convergence of $S{x}_n$ makes $x_n$ Cauchy, and completeness supplies a preimage of its limit. Also $\ker S^{\prime }=0$. The allowed annihilator identity makes the range dense, hence equal to $X$. Its inverse has norm at most $1/c$. This proves the spectrum decomposition through the Banach-space adjoint:For the right shift on ${\ell }^1$, $\Vert T\Vert =1$, so the allowed norm bound puts its spectrum in the closed unit disc. Under the linear pairing with ${\ell }^{\infty }$, its Banach-space adjoint is the left shift $T^{\prime }(b_1,b_2,\dots )=(b_2,b_3,\dots )$. For each $|\lambda |\le 1$ the bounded nonzero sequence $(1,\lambda ,{\lambda }^2,\dots )$ is an eigenvector of $T^{\prime }$ with eigenvalue $\lambda$. The union identity gives the reverse inclusion. Hence $\sigma (T)=\{\lambda :|\lambda |\le 1\}$. This Banach-space adjoint uses the complex-linear dual, so its eigenvalue is $\lambda$ itself.

The complex Hahn-Banach theorem extends every bounded complex-linear functional on a subspace to the whole normed space with exactly the same norm. Let $d=\mathrm{dist}(x_0,M)>0$ and define on $M+\mathbb{C}{x}_0$The representation is unique. The distance inequality $\Vert y+\lambda x_0\Vert \ge |\lambda |d$ gives $\Vert g_0\Vert \le 1$. Taking $y=-m$ with $\Vert x_0-m\Vert \downarrow d$ proves the reverse norm bound. Extend by the Hahn-Banach theorem to $g\in X^{\prime }$; then $g{|}_M=0$, $\Vert g\Vert =1$, and $g(x_0)=d\ne 0$. The labels (a)-(c) specify these simultaneous properties of this one functional.

Let $Y=\overline{\mathrm{span}F}$. Every functional vanishing on $F$ vanishes on $Y$ by linearity and continuity, so $Y\subseteq (F^{\circ }{)}_{\circ }$. If $x\notin Y$, the separation just proved gives a bounded functional zero on $Y$ and nonzero at $x$. It belongs to $F^{\circ }$ and excludes $x$ from its preannihilator. ThereforeFor the dual-side counterexample take $X={\ell }^1$, $X^{\prime }={\ell }^{\infty }$, and $G=c_0$. It is norm closed, since a uniform limit of sequences tending to zero still tends to zero. Every coordinate unit vector belongs to $G$, so the only $x\in {\ell }^1$ annihilated by every element of $G$ has all coordinates zero. Thus $G_{\circ }=\{0\}$ andThe dual-side double annihilator corresponds to weak-star closure, which need not equal norm closure.

The continuous dual space $X^{\prime }$ consists of all bounded complex-linear functionals $f:X\to \mathbb{C}$, with pointwise vector operations and norm $\Vert f\Vert ={\sup }_{\Vert x\Vert \le 1}|f(x)|$. This is finite by boundedness. The scalar norm identity and triangle inequality follow under the supremum; if the norm is zero, scaling every nonzero vector into the unit ball shows that $f$ vanishes. Hence $X^{\prime }$ is a normed complex vector space.

For the dual of the convergent sequence space, take $\beta \in {\ell }^1$. The displayed formula is a bounded linear functional, since the series is absolutely convergent andIt is linear in $\beta$, so $\Vert \varphi (\beta )\Vert \le \Vert \beta {\Vert }_1$. For equality fix $N$, choose ${\alpha }_j$ for $j\le N$ of modulus at most one so that ${\alpha }_j{\beta }_{j+1}=|{\beta }_{j+1}|$, and make all later terms equal a constant $c$ of modulus at most one with $c{\beta }_1=|{\beta }_1|$. The choices for a zero coefficient can be arbitrary. This sequence lies in the unit ball of $c$ and converges to $c$. The reverse triangle inequality givesLetting $N\to \infty$ proves $\Vert \varphi (\beta )\Vert =\Vert \beta {\Vert }_1$, including injectivity.

For surjectivity let $F\in c^{\prime }$. By the permitted description of $c_0^{\prime }$, there is $\gamma \in {\ell }^1$ with $F(z)={\sum }_j{\gamma }_j{z}_j$ on $c_0$. Every $\alpha \in c$ decomposes as $\alpha -{\alpha }_{\infty }1\in c_0$ plus ${\alpha }_{\infty }1$. ConsequentlyTake ${\beta }_{j+1}={\gamma }_j$ and ${\beta }_1=F(1)-{\sum }_j{\gamma }_j$. This is an ${\ell }^1$ sequence and $F=\varphi (\beta )$, proving the stated isometric linear isomorphism onto $c^{\prime }$. No complex conjugation is inserted, since these are complex-linear dual functionals.

A norm on a complex vector space is a map $\Vert \cdot \Vert :X\to [0,\infty )$ that vanishes only at zero, satisfies $\Vert \lambda x\Vert =|\lambda |\Vert x\Vert$, and obeys the triangle inequality. The space is Banach when every Cauchy sequence in the resulting metric converges in $X$.

If $X$ is Banach and ${\sum }_j\Vert x_j\Vert <\infty$, the estimate $\Vert {\sum }_{j=m}^n{x}_j\Vert \le {\sum }_{j=m}^n\Vert x_j\Vert$ shows that its partial sums are Cauchy, hence converge. Conversely let every such series converge and take a Cauchy sequence $(y_n)$. Choose a subsequence with $\Vert y_{n_{k+1}}-y_{n_k}\Vert <2^{-k}$. The series consisting of $y_{n_1}$ followed by these successive differences is absolutely summable in norm, so converges by hypothesis; its partial sums are exactly this subsequence. A Cauchy sequence with a convergent subsequence converges to the same limit, proving the absolute-series criterion for completeness.

On the absolutely summable sequence space ${\ell }^1$, coordinatewise addition and scalar multiplication preserve membership, and summing the scalar triangle inequality verifies its norm axioms. To prove completeness using the criterion, let $x^{(k)}=(x_j^{(k)})\in {\ell }^1$ with ${\sum }_k\Vert x^{(k)}{\Vert }_1<\infty$. For each $j$ the sum $x_j={\sum }_k{x}_j^{(k)}$ converges absolutely, and the permitted interchange of nonnegative double sums givesThus $x\in {\ell }^1$. MoreoverEvery absolutely norm-summable series therefore converges in ${\ell }^1$, proving ${\ell }^1$ is a Banach space.

For any affine variety and point $p$, with ${\mathfrak{m}}_p=\{f\in A(X):f(p)=0\}$, the standard localization description isAlso ${\mathcal{O}}_X(D(g))\cong A(X{)}_g$ for a principal open subset. One way to see the latter is to realize $D(g)$ as the affine variety $\{(x,z):g(x)z=1\}$; its coordinate ring is $A(X)[z]/(gz-1)=A(X{)}_g$ and part (d) applies.

For this cusp use $x=t^3,y=t^2$, identifying $A(X)$ with $k[t^2,t^3]$ inside $k(t)$. The parametrization gives exactly the relation $x^2=y^3$; its kernel is generated by this binomial. For example reduce a polynomial modulo $x^2-y^3$ to $a(y)+xb(y)$; substitution has even powers from $a(t^2)$ and odd powers at least three from $t^{3}b(t^2)$, so it can vanish only when both polynomials are zero.

Write $g=x-y=t^2(t-1)$. The open set corresponds to $t\ne 0,1$, andTo simplify this ring without assuming the normalization coordinate is present, observe that in $A(X)$Thus $y$ becomes a unit after inverting $g$, with inverse $(y^2-2x+y)/g^2$. Consequently $t=x/y$ belongs to the localization. Since $g=t^2(t-1)$ is a unit, both $t$ and $t-1$ become units. Conversely every original generator lies in the corresponding two-point localization of $k[t]$. HenceFinally $p=(1,1)$ corresponds to $t=1$. It lies outside $U$, so compute its local ring from $X$, at $\mathfrak{m}=(x-1,y-1)$. The element $y$ is a unit there, hence $t=x/y$ belongs to $A(X{)}_{\mathfrak{m}}$. Every polynomial in $t$ nonzero at one is then a unit. Conversely a denominator from $A(X)\setminus \mathfrak{m}$ remains nonzero at $t=1$. These inclusions proveThis smooth point has the ordinary one-variable local ring, even though the global curve has a cusp at its origin.

The localization calculation describes regular functions on a cusp with two points removed.

Polynomial functions are regular, giving an injective map $A(X)\hookrightarrow {\mathcal{O}}_X(X)$; injectivity follows from quotienting by the full vanishing ideal. To prove surjectivity, let $f$ be globally regular and consider the denominator idealIt is an ideal because sums and polynomial multiples preserve this property.

Fix $p\in X$. Locally $f=a/b$ on a neighbourhood $V$ of $p$, with $b$ nonzero there. Principal affine opens form a basis, so choose $h\in A(X)$ with $h(p)\ne 0$ and $D(h)\subseteq V$. Thenas functions on all of $X$: on $D(h)$ it follows from the local fraction, and off $D(h)$ both sides vanish because $h=0$. Hence $hb\in J$ and $(hb)(p)\ne 0$.

Thus $J$ has no common zero on $X$. By the weak Nullstellensatz for $A(X)$, every proper ideal lies in a maximal ideal belonging to a point of $X$; therefore $J$ cannot be proper. We get $1\in J$, so $f$ itself is polynomial. This provesThe argument works for reducible affine varieties as well: it uses functions and denominator ideals rather than assuming a single fraction field.

This is the denominator ideal proof for global regular functions.

A regular function on $U$ is a function $f:U\to k$ locally representable, in an affine chart of the ambient projective space, as a quotient $a/b$ of polynomial functions with denominator nonzero on that neighbourhood. Equivalently it is locally a quotient of homogeneous forms of the same degree. The equal-degree condition makes the quotient independent of the chosen projective representative.

The ring ${\mathcal{O}}_{X,p}$ is the ring of germs of such functions near $p$. Two representatives $(U,f)$ and $(V,g)$ define the same germ when they agree on some smaller neighbourhood of $p$. Addition and multiplication use a common neighbourhood. It is a local ring: its unique maximal ideal consists of germs vanishing at $p$, and any germ nonzero at $p$ has a regular reciprocal on a smaller neighbourhood.

For a multiplicatively closed set $S$ containing one, the localization of a ring $S^{-1}R$ consists of fractions $r/s$ with $r\in R,s\in S$, under the equivalenceThe factor $u$ is needed when the ring has zero divisors. Addition and multiplication areThe canonical map sends $r$ to $r/1$ and makes every element of $S$ a unit. It is universal with this property: a ring homomorphism $\varphi :R\to T$ taking $S$ into units extends uniquely by $r/s\mapsto \varphi (r)\varphi (s{)}^{-1}$. Usually zero is excluded from $S$; if it is included, this construction is the zero ring.

A quasi-projective variety is a locally closed subset of projective space: equivalently an open subset of some projective variety, with the induced algebraic variety structure. Affine varieties are examples, via the standard affine chart in projective space.

It is irreducible if it is nonempty and cannot be written as the union of two proper closed subsets. Equivalently, every two nonempty open subsets meet, or every nonempty open subset is dense. Closed and open are interpreted in the induced Zariski topology of $X$.

The homogeneous coordinate ring of the Segre image is the subalgebra $k[x_i{y}_j]\subseteq k[x_0,\dots ,x_n,y_0,\dots ,y_m]$, grading each product $x_i{y}_j$ as degree one. Its degree-$t$ part is precisely the space of polynomials bihomogeneous of bidegree $(t,t)$.

Indeed, every product of $t$ generators has those two degrees. Conversely, any monomial of bidegree $(t,t)$ can pair its $t$ individual $x$ factors with its $t$ individual $y$ factors, expressing it as a product of generators. The distinct resulting monomials are linearly independent. Counting them givesIn fact this agrees with the Hilbert function for every integer $t\ge 0$. Its leading coefficient is $1/(n!m!)$, so the two degree quantities areThus the Segre variety has dimension $n+m$ and embedded projective degree $\left(\genfrac{}{}{0ex}{}{n+m}{n}\right)$.

This gives the Segre Hilbert polynomial and degree directly from its graded monomials.