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An algebra norm is a norm satisfying $\Vert ab\Vert \le \Vert a\Vert \Vert b\Vert$. Its completion is a Banach algebra in which the original algebra embeds densely.

Every nonzero complex unital Banach algebra in which every nonzero element is invertible is isomorphic as a complex algebra to $\mathbb{C}$. Indeed, the nonemptiness of the Banach-algebra spectrum gives, for any $a$, some $\lambda \in {\sigma }_A(a)$. The element $a-\lambda 1$ is noninvertible, so the hypothesis forces $a=\lambda 1$. Thus $\lambda \mapsto \lambda 1$ is an algebra isomorphism onto $A$.^[ref]

The isometry conclusion additionally uses $\Vert 1\Vert =1$, since $\Vert \lambda 1\Vert =|\lambda |\Vert 1\Vert$. Without that convention, $\mathbb{C}$ equipped with $\Vert z\Vert =2|z|$ is still a Banach algebra with every nonzero element invertible, but its unital complex algebra isomorphism from the usual $\mathbb{C}$ is not isometric. The complex scalar field is also essential: $\mathbb{R}$ itself gives a real Banach algebra satisfying the invertibility condition.

The canonical image of the closed unit ball of a Banach space $X$ is weak-star dense in the closed unit ball of $X^{\ast \ast }$.

Let $n\in {\mathbb{R}}^3$ be a unit vector, let $P=n{n}^T$ be the orthogonal projection onto its axis, put $Q=I-P$, and let the cross-product matrix $J$ satisfy $Jx=n\times x$. A real axis-angle decomposition has the form $A=aP+bQ+cJ$. It multiplies the axial component by $a$ and acts on the perpendicular plane as the complex number $b+ic$. It is an invertible matrix exactly when $a\ne 0$ and $b^2+c^2>0$, and thenTo check the formula, use $P^2=P$, $Q^2=Q$, $PQ=PJ=JP=0$ and $J^2=-Q$, the last identity following from the vector triple product. Thus $(bQ+cJ)(bQ-cJ)=(b^2+c^2)Q$. The axial and perpendicular direct summands can be inverted independently. For a pure rotation through an angle $\theta$, $a=1$, $b=\cos \theta$, $c=\sin \theta$, and the inverse reverses that angle.

Let $M$ be a transitive class. If $M$ is closed under the ambient subsets defined by ${\phi }^M$, then $M$ satisfies the $\phi$-instance of separation, because

The principle of epsilon induction says that every progressive class is universal: ifthen $\phi (x)$ holds for every set $x$. Over the other axioms of ZF, it is equivalent to the Axiom of foundation.

For any sets $X,Y$, either $X$ injects into $Y$ or $Y$ injects into $X$. Applying this to $X$ and its Hartogs ordinal proves the well-ordering theorem, so over ZF cardinal comparability is equivalent to choice.

Every set admits a well-order. In ZF this statement is equivalent to the axiom of choice.

A choice function on a family $\mathcal{A}$ of nonempty sets is a function $f$ with $f(A)\in A$ for every $A\in \mathcal{A}$.

Forthe zero-input stable states are $0$ and $(k+\sqrt{k^2-4})/2$. The switching threshold is the saddle-node bifurcation determined by $f(g_c,s_c)={\partial }_{g}f(g_c,s_c)=0$ on the low-concentration branch.

Termwise asymptotic equivalence of two asymptotic sequences does not ensure that an asymptotic expansion in one has an expansion in the other. As $x\to +\infty$, takeBoth are asymptotic sequences and ${\psi }_n/{\varphi }_n\to 1$ for every fixed $n$. The constant function $f=1$ has the exact expansion $\sum a_n{\varphi }_n$ with $a_0=1$ and all later coefficients zero.

An expansion in the $\psi$ scale must also start with coefficient one, since ${\psi }_0\to 1$. But after that first term its remainder is $-x^{-1/2}$. No constant coefficient $b_1$ makesThe ratio to $x^{-1}$ diverges. The perturbation of the leading scale is small relative to that scale but too large relative to the next one. This is why replacing scales needs remainder control at all subsequent orders.

Let $({\varphi }_n)$ be a positive asymptotic sequence. Define ${\chi }_0={\varphi }_0$ and ${\chi }_n=\sqrt{{\varphi }_{n-1}{\varphi }_n}$ for $n\ge 1$, using the positive geometric mean. ThenThe latter limit follows by multiplying the two adjacent ratios of the original asymptotic sequence. However,Thus these are new intermediate scales, not termwise equivalent replacements. For ${\varphi }_n=x^{-n}$ as $x\to \infty$, the new scales are $1,x^{-1/2},x^{-3/2},x^{-5/2},\dots$. Positivity is needed for this real square-root construction.

For a positive asymptotic sequence $({\varphi }_n)$, let ${\psi }_0={\varphi }_0$ andThis uses half the usual two-variable harmonic mean. Since ${\varphi }_n/{\varphi }_{n-1}\to 0$,Thus each refined scale is asymptotically equivalent to its original scale. Alsoso $({\psi }_n)$ is again an asymptotic sequence. For ${\varphi }_n=x^{-n}$ as $x\to \infty$, one gets ${\psi }_n=x^{-n}/(1+x^{-1})$ for $n\ge 1$. Termwise equivalence preserves the ordering of scales; it does not automatically preserve all coefficients of a represented asymptotic expansion.

For an independent sample from $\mathrm{Uniform}[-\theta ,\theta ]$, the maximum likelihood estimator is $M_n={\max }_i|X_i|$ andThe parameter-dependent support makes the convergence rate $n$ rather than $\sqrt{n}$, so the regular maximum-likelihood central limit theorem does not apply.

If $\sqrt{n}(T_n-\theta )$ is asymptotically normal, differentiability gives $\sqrt{n}(g(T_n)-g(\theta ))$ the variance multiplied by $g^{\prime }(\theta {)}^2$.

Suppose an estimating equation $n^{-1}{\sum }_i\psi (X_i,\hat{\theta })=0$ has the expansionwhere $A=\mathbb{E}[\partial \psi (X,{\theta }_0)/\partial {\theta }^T]$ is nonsingular and the centered estimating function has covariance matrix $B$. A central limit theorem gives asymptotic covariance $A^{-1}B{A}^{-T}$ for the rescaled error, or approximately $n^{-1}{A}^{-1}B{A}^{-T}$ for $\hat{\theta }$.

The transpose is essential when $A$ is not symmetric; the simplified form $A^{-1}B{A}^{-1}$ applies only in the symmetric case. Plugging in consistent estimates gives a sandwich covariance estimate. Regression heteroskedasticity-consistent standard errors and cluster-level generalized estimating equations use instances of this construction.

If $X_n\stackrel{d}{\to }X$ and $Y_n\stackrel{\mathbb{P}}{\to }c$ for a deterministic constant $c$, then $(X_n,Y_n)\stackrel{d}{\to }(X,c)$. The continuous mapping theorem therefore givesand $X_n/Y_n\stackrel{d}{\to }X/c$ when $c\ne 0$. No independence between the two sequences is required. The deterministic limit matters: convergence of two separate nondegenerate marginal distributions does not determine their joint limit.

For example, if $\sqrt{n}(\hat{\theta }-\theta )\stackrel{d}{\to }N(0,v)$ with $v>0$ and $\hat{v}\stackrel{\mathbb{P}}{\to }v$, then $\sqrt{n}(\hat{\theta }-\theta )/\sqrt{\hat{v}}\stackrel{d}{\to }N(0,1)$ wherever the denominator is defined. This justifies substituting a consistent estimator of asymptotic variance when constructing a Wald confidence interval.

Forrepeated integration by parts gives

For fixed $a$ and real $y\to +\infty$, it is the upper incomplete gamma function that has the endpoint-tail expansionThe empty product is one. Repeated integration by parts gives each coefficient and an integral remainder. For a positive integer $a$, the product eventually vanishes and the expansion terminates. The lower function instead satisfies $\gamma (a,y)=\Gamma (a)-\Gamma (a,y)$ for $\mathrm{Re}a>0$; confusing the two functions would incorrectly make the lower function decay to zero.^[ref]

If an algebra is defined over a parameter ring $R$, a ring homomorphism $R\to k$ produces a specialized $k$-algebra by extension of scalars. Defining relations specialize by replacing every parameter by its image in $k$.