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Two differentiable functions can be linearly independent on a connected interval while their Wronskian vanishes everywhere. For example, let $f(x)=(\max \{-x,0\}{)}^2$ and $g(x)=(\max \{x,0\}{)}^2$. Both have a continuous first derivative, and their values and first derivatives vanish at zero. On either side of zero one of the functions is identically zero, so $f{g}^{\prime }-f^{\prime }g=0$ everywhere. Evaluating a constant linear combination $Af+Bg=0$ at $-1$ and $1$ forces $A=B=0$.

The obstruction is that the two halves supply independent information without simultaneous nonzero value-and-derivative data at any point. For solutions of a common regular second-order linear differential equation, the uniqueness theorem for ordinary differential equations removes this possibility; Abel's identity gives the zero-or-nowhere-zero alternative for their Wronskian.

The equation ${\int }_0^{\infty }{e}^{-|x-t|}f(t)dt=f(x)$ has this solution family in the ordinary bounded-growth transform class. Differentiating twice gives $f^{\prime \prime }=-f$, and the integral's right derivative at zero gives $f^{\prime }(0)=f(0)$. Alternatively its positive-support Fourier transform is $A(ik-1)/(k^2-1)$; inversion above both real poles gives the same answer.

Suppose an integer-valued random walk starts at $S_0=0$, has upward increments at most one, and satisfies $S_n/n\to \mu >0$ almost surely. Write $T_j$ for its first-passage time to $j\ge 1$. On the event of almost sure convergence, the walk tends to $+\infty$ and cannot jump over any positive integer, so every $T_j$ is finite. The bound $S_n\le n$ gives $T_j\ge j\to \infty$. Along this subsequence, $j/T_j=S_{T_j}/T_j\to \mu$ on each such sample path, proving the displayed reciprocal limit. No independence between the hitting times and the path is required. The strong law of large numbers supplies the hypothesis for integrable independent and identically distributed random variables as steps with positive mean.

Fix $n\ge 2$ bins. In repeated uniform balls-in-bins allocation, throw each remaining ball independently into those same $n$ bins, remove every ball that is alone in its bin, and repeat with the survivors. Let $B_j$ be the number remaining before round $j+1$.

Conditional on $B_j=b\ge 2$, a specified ball is alone with probability $(1-1/n{)}^{b-1}$. Its survival probability is therefore $1-(1-1/n{)}^{b-1}$, so linearity of expectation givesFor $b=0$ or $1$ the next count is zero. Here $n$ denotes the fixed bin count, not the current number of balls; confusing those two quantities changes the process.

The survival indicator random variables are generally dependent. For two initial balls and two bins, they survive together exactly when they choose the same bin, of probability $1/2$ each round. Hence $\mathbb{P}(B_j=2)=2^{-j}$ and $\mathbb{E}{B}_j=2^{1-j}$ for $B_0=2$. In particular, replacing the conditional formula by the same nonlinear function of $\mathbb{E}{B}_j$ would not be justified by linearity of expectation.

Let $B=\mathrm{diag}(b,a,a)$, with $a,b>0$, $a\ne b$, and $A=R\mathrm{diag}(a,b,a)$ be a twin-related gradient with $\cos \theta =2ab/(a^2+b^2)$, $\sin \theta =\kappa (b^2-a^2)/(a^2+b^2)$, $\kappa =\pm 1$. Their jump is $(b^2-a^2)(-b,\kappa a,0)\otimes (1,\kappa ,0)/(a^2+b^2)$.

A plane through the origin with normal $p=(p_1,p_2,0)$ is spanned by $e_3$ and $t=(-p_2,p_1,0)$. Both gradients map $e_3$ to the same direction, so the two image half-planes are coplanar exactly when $At$ and $Bt$ are parallel in their transverse section. Direct calculation givesIts nonzero prefactor makes planarity equivalent to $p\parallel (1,\kappa ,0)$, the twin normal. On that plane the two affine maps agree, proving sufficiency as well. The restriction on the third normal component matters; this statement is not a classification of every possible material plane.

Suppose a two-species reaction-diffusion instability has an unstable squared-wave-number band $(q_{-},q_{+})$. On a rectangle with Neumann boundary conditions, $q=(n_x\pi /L_x{)}^2+(n_y\pi /L_y{)}^2$, with both indices nonnegative. The displayed thin-width condition excludes every $n_y\ge 1$ mode but can leave $n_y=0$ modes in the band for sufficiently large $L_x$. Omitting zero indices would incorrectly remove the longitudinal pattern. Linear onset selects allowed directions; it does not alone establish the final nonlinear pattern.

For a reaction-stable two-species equilibrium with both Jacobian matrix diagonal entries negative and positive diagonal diffusivities, each spatial mode has more negative trace after diffusion is added. Its determinant is the positive reaction determinant plus positive coefficients of $k^2$ and $k^4$. Hence every mode stays stable. Unequal diffusion alone is insufficient for a Turing instability; the reaction feedback must also permit a destabilizing weighted diagonal sum.

If a one-dimensional reaction-diffusion model has a strict unstable squared-wavenumber band $(q_{-},q_{+})$, its nonzero Neumann modes on $[0,L]$ have $q_n=(n\pi /L{)}^2$. Increasing $L$ from zero first reaches the band with $n=1$ at the displayed threshold. Equality is neutral; the first mode is strictly unstable for $\pi /\sqrt{q_{+}}<L<\pi /\sqrt{q_{-}}$. Thus positive growth has an infimum in domain length rather than an attained smallest length. The zero mode must be checked separately through reaction stability.

For $f\in C^2[a,b]$ with $a<b$, the trapezoidal rule satisfies the displayed bound, and the constant $1/12$ is best possible. The polynomial interpolation error for the endpoint linear interpolant satisfies $|f(x)-p(x)|\le \frac{1}{2}M(x-a)(b-x)$, where $M=\max |f^{\prime \prime }|$. Integrating this inequality and using the triangle inequality for an integral proves the bound because ${\int }_a^b(x-a)(b-x)dx=(b-a{)}^3/6$. Equality holds for $f(x)=(x-a{)}^2$: the exact integral is $(b-a{)}^3/3$, its trapezoidal rule estimate is $(b-a{)}^3/2$, and $M=2$.

For real rectangular matrices, expanding $(A+H{)}^t(A+H)$ leaves the quadratic remainder $H^{t}H$, whose matrix norm is $O(\Vert H{\Vert }^2)$. Thus the displayed Fréchet derivative is a linear map in the increment $H$. At an orthogonal matrix, its kernel consists of increments $H=A\Omega$ with $\Omega$ skew-symmetric. This is the tangent-space condition obtained by differentiating $A^{t}A=I$.

The directed vertex-edge incidence matrix has one $+1$ and one $-1$ per nonloop column and a zero column for a loop. Every square subdeterminant belongs to $\{0,1,-1\}$. Induct on its size: a column with no nonzero entries makes the determinant zero; a column with one nonzero entry permits a cofactor expansion to a smaller determinant; and if every column has two nonzero entries, their opposite signs make the column sums zero, giving linearly dependent rows. Negating rows preserves absolute subdeterminants. Consequently the unsigned vertex-edge incidence matrix of a bipartite graph is totally unimodular: negate all rows in one part. The assignment problem is an instance, with row vertices and column vertices forming the two parts.

A matrix with entries in $\{0,1,-1\}$ is a totally unimodular matrix if every column has at most two nonzero entries, with opposite signs when both occur. To prove this, induct on the size of a square submatrix. A zero column gives determinant zero, a one-entry column reduces by cofactor expansion, and if every column has two entries then all column sums vanish, giving dependent rows. The total unimodularity of directed incidence matrices is a central instance.

The fourth derivative is the derivative of the third derivative. It vanishes for a polynomial of polynomial degree at most three. For a cubic spline, this statement applies inside each knot interval; a jump of the third derivative can prevent a classical fourth derivative from existing at a spline knot.

Suppose ${\sum }_n\Vert a_n{\Vert }_1<\infty$. By summable L1 norms imply almost-everywhere absolute convergence, let $A={\sum }_n{a}_n$ where the series converges absolutely, and choose zero on the exceptional null set. Then $A$ is a Lebesgue measurable function and $|A|\le S={\sum }_n|a_n|$ almost everywhere. Since $S$ is Lebesgue integrable, so is $A$. The signed partial sums are bounded in absolute value by $S$, so the dominated convergence theorem proves termwise integration. It also gives the quantitative L1 norm estimateThe numerical series of integrals is absolutely convergent because $|\int a_n|\le \int |a_n|$.

For example, on $(0,1)$ take $a_n(x)=(-1{)}^{n}n{1}_{(0,2^{-n})}(x)$. Its L1 norm is $n{2}^{-n}$, and these norms sum to two. Every positive $x$ lies in only finitely many supports, so the series is pointwise finite, although its terms are not uniformly small. Termwise integration gives ${\int }_0^1{\sum }_n{a}_n={\sum }_n(-1{)}^{n}n{2}^{-n}=-2/9$, by differentiating the geometric series. This illustrates why a small L1 norm can permit tall terms on short intervals.

Replacing a designated subformula by a logically equivalent formula preserves logical equivalence of the surrounding propositional formulas. For any Boolean valuation the two expressions at the replaced node have the same truth value. Each logical connective preserves this agreement when its other inputs are unchanged. Induction from that node to the root proves agreement of the whole formulas under every valuation.

Subformula replacement is the operation of replacing one specified subformula occurrence by another logical formula. Its preservation of well-formedness is the subformula replacement theorem.

For $dX=\mu (t)Xdt+\sigma (t)dB$ with deterministic coefficients, $X(t,x)=a(t)[x+{\int }_0^{t}a(s{)}^{-1}\sigma (s)d{B}_s]$, where $a(t)=\exp ({\int }_0^t\mu )$. The random part does not depend on the initial value. For $f\in C_b^1$, the mean value theorem bounds the initial-value difference quotient by $a(t)\Vert f^{\prime }{\Vert }_{\infty }$, and dominated convergence theorem gives the displayed derivative. The transition law is normal with mean $a(t)x$ and variance ${\int }_0^t\exp (2{\int }_s^t\mu )\sigma (s{)}^{2}ds$.

For a state on a C-star algebra $x$ on a unital algebra $A$, put $⟨[a],[b]⟩=x(b^{\ast }a)$ on $A/N_x$, with the inner product linear in its first argument. The Cauchy–Schwarz inequality for positive C-star functionals makes the null vectors pair to zero with every vector. Moreover $c^{\ast }c\le \Vert c{\Vert }^{2}1$ gives $x((ca{)}^{\ast }(ca))\le \Vert c{\Vert }^{2}x(a^{\ast }a)$, so $N_x$ is a left ideal and $[a]\mapsto [ca]$ is well-defined and bounded by $\Vert c\Vert$. Completion gives a Hilbert space $H_x$ and a star-representation ${\pi }_x(c)[a]=[ca]$. The adjoint identity follows from $x(b^{\ast }ca)=⟨[a],{\pi }_x(c^{\ast })[b]⟩$. The unit vector ${\xi }_x=[1]$ is cyclic and recovers the state as $x(a)=⟨{\pi }_x(a){\xi }_x,{\xi }_x⟩$.

For a unital represented C-star algebra $A\subseteq \mathcal{B}(H)$, the displayed functional is a state. It is linear under the first-argument-linear inner-product convention, has $x_{\eta }(1)=1$, and is positive because $x_{\eta }(a^{\ast }a)=\Vert a\eta {\Vert }^2\ge 0$. The Cauchy-Schwarz inequality gives $|x_{\eta }(a)|\le \Vert a\Vert$, while evaluation at the unit gives equality of its norm with one. Its GNS null left ideal consists exactly of the operators annihilating $\eta$.

For a nonzero unital C-star algebra, a positive functional on a C-star algebra obeys $\Vert x\Vert =x(1_A)$, and conversely this equality for a bounded functional implies positivity. Therefore its state on a C-star algebra functionals are exactly the maximizers of the real linear evaluation $x\mapsto \mathrm{Re}x(1_A)$ on the closed dual unit ball. If a convex combination reaches that maximum, each constituent must reach it: each real part is at most one, and their weighted mean is one. The modulus bound then forces evaluation to equal one, so each constituent is again a state. Hence the state space is an exposed face and a face of a convex set.