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The property of a binary relation that opposite directed relationships between two elements can hold only when those elements are equal. It does not require every pair to be comparable and does not prohibit a longer directed cycle whose reverse edges are initially absent. Taking reflexive transitive closure can therefore destroy this property.

A holomorphic function $F$ on a domain with complex derivative $F^{\prime }=f$. Its contour integral satisfies ${\int }_{\gamma }f(z)dz=F(\gamma (b))-F(\gamma (a))$ by the chain rule along each smooth contour segment. In particular every closed contour has integral zero. Having local complex antiderivatives need not provide a single one on a domain with holes.

With fixed interior substrate and finite carrier protein, symmetric alternating-access kinetics produce a monotone concave transport curve of the displayed form, with $A,C,D>0$. The flux reverses at equal free concentrations and approaches the finite limit $C/A$ for abundant exterior substrate. Its plateau reflects finite carrier abundance and finite turnover, rather than unlimited transport proportional to the concentration difference. The protein total multiplying the flux is distinct from the total substrate.

For plane strain in $y>0$ under surface normal traction $-P(x)$ and zero surface shear, use the forward Fourier transform phase $e^{ikx}$. At $k\ne 0$, biharmonicity gives $({\partial }_y^2-k^2{)}^2\hat{A}=0$, whose decaying part is $(C+Dy)e^{-|k|y}$. Boundary traction fixes $C=\hat{P}/k^2$, $D=|k|C$, proving the displayed formula. Constitutive inversion and a decaying displacement mode give $2\mu \hat{v}=[y+2(1-\nu )/|k|]\hat{P}{e}^{-|k|y}$. Multiplication by $-ik$ produces an unambiguous displacement-gradient convolution kernel.

The zero mode is separate: a nonzero resultant load $\hat{P}(0)$ conflicts with an ordinary fully decaying displacement normalization in two dimensions and leads to a logarithmic reference issue. Balanced loading or a distributional/reference-displacement treatment is required before extending the transformed formulas globally. Dividing by $k^2$ and assigning a finite value at $k=0$ is invalid.

On a complex inner product space $V$, with the inner product conjugate-linear in its first argument, an everywhere-defined linear map $T:V\to V$ admits an adjoint on $V$ if there is a linear map $T^{\ast }:V\to V$ satisfyingThe adjoint, when it exists, is unique: the difference of two candidate values at $y$ is orthogonal to every vector and hence to itself. In a finite-dimensional vector space, it always exists, and its matrix in an orthonormal basis is the conjugate transpose. In a space that is not complete, existence must instead be checked.

The identity $⟨Tx,y⟩=⟨x,Ty⟩$ for every $x,y\in V$ supplies such an adjoint directly, namely $T^{\ast }=T$. This algebraic statement should be distinguished from self-adjointness of a linear operator on a completed Hilbert space, which additionally compares the domains of an operator and its Hilbert-space adjoint. The Gaussian polynomial operator is a useful example where the identity holds on the entire polynomial inner product space and can be used on every finite-dimensional invariant polynomial subspace.

Give the real vector space of degree-at-most-one polynomials the inner product $⟨f,g⟩={\int }_{-1}^{1}fg$. In coordinates $f=a+bt$, $g=c+dt$, this is $2ac+2bd/3$. Since $Df=b$, the adjoint identity requires $2bc=⟨a+bt,D^{\ast }(c+dt)⟩$ for every $a,b$. Comparing the two coefficients yields the displayed formula. The adjoint operator depends on both the inner product and the domain; a formal integration-by-parts sign without the boundary term would give the wrong answer.