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In a normal Hausdorff space, strong zero-dimensionality means that any disjoint closed sets can be separated by a clopen set, equivalently by disjoint clopen neighborhoods. For general Tychonoff spaces, the usual definition asks this for completely separated sets. In the normal setting Urysohn's lemma makes the formulations agree. The condition is equivalent to zero-dimensionality of the Stone-Čech compactification.

The characteristic functions of a clopen topological basis give a topological embedding of a zero-dimensional $T_0$ topological space into a product of discrete two-point topological spaces. The separation axiom makes those functions distinguish points, and their coordinate preimages recover the original topological basis.

The conjugate-power constrained product maximum proves Young's inequality for products by rescaling. For nonnegative $x,y$ and conjugate exponents $p,q>1$, let $A=x^p/p+y^q/q$. If $A=0$, both variables vanish. Otherwise the displayed normalized variables satisfy $X^p/p+Y^q/q=1$, hence $XY\le 1$. Since $1/p+1/q=1$, this means $xy/A\le 1$, proving $xy\le x^p/p+y^q/q$. Equality in the constrained maximum requires $X=Y=1$, equivalent to $x^p=y^q$; the zero case gives the same equality criterion. This approach simultaneously gives the sharp constant and every equality case.

The inner, generally nonconducting portion of a woody stem, distinguished from the more recently formed sapwood.

Let $g$ be a holomorphic function on an open set $\Omega \subseteq \mathbb{C}$. Among continuously differentiable functions, all solutions of ${\partial }_{\bar{z}}f=g$ arewhere $h$ is an arbitrary holomorphic function on $\Omega$. The Wirtinger derivatives satisfy ${\partial }_{\bar{z}}\bar{z}=1$ and ${\partial }_{\bar{z}}g=0$, so the product rule verifies the particular solution $\bar{z}g(z)$. Subtracting it from any other solution leaves a continuously differentiable function with vanishing Wirtinger derivative, hence a holomorphic function by the Cauchy-Riemann equations. Conversely adding such a holomorphic function changes no prescribed derivative. For $g(z)=z$ on the complex plane, the solutions are $|z{|}^2+h(z)$ with $h$ any entire function.

Assume that a proposed statement fails for some natural number, and choose the least failing index using the well-ordering principle. Prove the necessary base cases and reduce the least failure to a smaller index that must satisfy the statement. If that smaller case implies the alleged failure also satisfies it, there is a contradiction. Reductions by a step larger than one require enough consecutive base cases to cover every possible residue class.

For a $C^1$ curve $u$, some $\delta >0$ must make the Weierstrass excess function nonnegative for every $x\in [a,b]$, every trial slope $q\in \mathbb{R}$, and all $y,p$ satisfying $|y-u(x)|<\delta$, $|p-u^{\prime }(x)|<\delta$. Thus the supporting-line property persists in a tube of nearby heights and reference slopes. The term “strengthened” here refers to this neighborhood requirement; strict excess for $q\ne p$ is a further condition.

At a strong local minimizer of a variational integral, almost every reference slope $u^{\prime }(x)$ is a point of convexity of $v\mapsto f(x,u(x),v)$. Equivalently its Weierstrass excess function is nonnegative for every trial slope. Uniformly small variations can have large derivatives, which is why this condition tests all trial slopes rather than only nearby ones. It is a necessary condition along the minimizing curve; the strengthened Weierstrass condition additionally controls a neighborhood of that curve and its slopes.

For $0<\delta <1$, choose the zero-trace piecewise affine curve with slope $\delta$ up to $1-\delta /4$ and slope $\delta -4$ thereafter. Periodicity makes $g^{\prime }$ equal to the same nonzero constant at both slopes, and $g^{\prime \prime }>0$ on small neighborhoods of both. For sufficiently small Lipschitz perturbations, local strict convexity bounds the energy difference below by $c\int {\varphi }^{\prime 2}$, since the common first-variation coefficient multiplies $\int {\varphi }^{\prime }=0$. This proves a strict weak minimum. The global supporting-tangent condition fails because its nonzero slope cannot stay below bounded $g$, so it is not a strong minimum. By contrast, tents of slopes $4n,-4n$ attain $g=-1$ everywhere and provide infinitely many distinct global minimizers.

Consider a smooth planar system $\dot{x}=y+P_3(x,y)+O(r^5)$, $\dot{y}=-x+Q_3(x,y)+O(r^5)$, where $P_3,Q_3$ are homogeneous cubic polynomials and $r^2=x^2+y^2$. The leading derivative of $r^2/2$ is the quartic polynomial $H_4=x{P}_3+y{Q}_3$. Let $\kappa$ be the angular average of $H_4$ on the unit circle. A homogeneous quartic correction $V_4$ solvinggives $\dot{V}=\kappa r^4+O(r^6)$ for $V=r^2/2+V_4$. The right side has zero angular average. Since $y{\partial }_x-x{\partial }_y=-{\partial }_{\theta }$, its nonconstant angular Fourier modes can be integrated individually to obtain a periodic quartic correction; only modes of orders two and four can occur after removing the constant average.

The correction is $O(r^4)$, so $V$ remains positive definite close to zero. If $\kappa <0$, its orbital derivative is strictly negative there, and the Second Lyapunov theorem proves asymptotic stability. If $\kappa >0$, apply the same argument in reversed time to obtain repulsion. If $\kappa =0$, this quartic calculation alone does not decide stability. This method converts the averaged damping of a weak focus into a pointwise sign for a Lyapunov function.

This real planar dynamical system has a saddle point at $(-1,0)$ and a repelling weak focus at $(1,0)$. The latter has purely imaginary linearized eigenvalues $\pm 2i$, so its linearization has closed elliptical orbits. Nonlinear terms nevertheless produce outward drift.

Set $X=2(x-1)$, $Y=y$, $\tau =2t$ and $r^2=X^2+Y^2$. The cubic expansion isThe linear rotation acts on scalar functions by $L=Y{\partial }_X-X{\partial }_Y$. For $H_2=r^2/2$, its cubic derivative is $X{Y}^2-3X^{2}Y/4-Y^3/2$. The homogeneous correction $H_3=7X^3/12+X{Y}^2/2+Y^3/3$ satisfies $L{H}_3$ equal to the negative of that derivative. The remaining quartic derivative of $H_2+H_3$ is $3X^2{Y}^2/2-X^{3}Y$. Withdirect differentiation givesfor all sufficiently small nonzero $r$. Meanwhile $H_2+H_3+H_4=r^2/2+O(r^3)>0$. In reversed time it is a strict Lyapunov function, proving asymptotic stability of the reversed system and hence repulsion forward in time. The angular speed stays close to $-1$, so the nearby nonzero trajectories spiral rather than approaching along a fixed direction. This explicit calculation illustrates why a centre in the linearization need not be a nonlinear centre.

A historical quantitative test in which a standardized soap solution is added to a measured water sample until a specified persistent lather appears. A reagent blank accounts for the amount needed to form the lather itself.