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A generalized eigenfunction solves the eigenvalue equation for a linear operator in an enlarged space, often a space of distributions, although it may fail to belong to the underlying Hilbert space. A plane wave $e^{ikx}$ on the whole real line is a generalized momentum eigenstate, with momentum $\hslash k$, and is not square-integrable. This usage differs from a generalized eigenvector in a Jordan chain, where a higher power of $L-\lambda$ vanishes rather than $L-\lambda$ itself.

The Volterra integration operator is injective and has dense range. Its range of the Volterra integration operator is $\{f\in H^1(0,1):f(0)=0\}$, and its Moore–Penrose inverse of an operator is differentiation on this domain. Thus the error estimate for $K^{†}f$ implicitly requires $f(0)=0$, in addition to the printed $H^2$ assumption. For an arbitrary $H^2$ function the same calculation bounds error against $f^{\prime }$, but that function need not be in $\mathcal{D}(K^{†})$.

Put $a=(1-\alpha )/2$, $b=(1+\alpha )/2$ and $h=\alpha /2$. The three stencils are forward, central finite differences, and backward on $[0,a)$, $[a,b)$ and $[b,1]$, respectively. All shifted evaluation points lie in $[0,1]$. For $g\in L^2(0,1)$, apply $|A-B{|}^2\le 2(|A{|}^2+|B{|}^2)$ and change variables in each of the six evaluation terms. The resulting integration intervals arePair $[0,a]$ with $[a,1-\alpha ]$, $[\alpha ,b]$ with $[b,1]$, and the two middle intervals. Each pair has disjoint interiors, so almost every point is counted at most three times. This overlap multiplicity bound for piecewise difference operators provesFor the bias on the middle interval, write the central finite difference as an average of $f^{\prime }$. The fundamental theorem of calculus giveswhere $\Vert k_{\alpha }{\Vert }_1=\alpha /4$. Extend $f^{\prime \prime }$ by zero to the real line. Young's convolution inequality bounds the middle-interval $L^2$ error by $\alpha c/4$. Combining its squared error with the supplied outer-interval estimate gives the central-difference noise-bias boundThe triangle inequality and the operator norm estimate now giveFor $c>0$, balancing noise and approximation bias minimizes this upper bound atfor sufficiently small $\delta$ that $\alpha <1/2$. Capping the parameter at $1/4$ supplies an admissible rule for all positive noise levels. More generally, $\alpha \to 0$ and $\delta /\alpha \to 0$ suffice; $\alpha =\sqrt{\delta }$ avoids needing $c$ or dividing by $c=0$.

To prove convergence for every admissible datum, not just $H^2$ data, let $f=Ku$ with $u\in L^2(0,1)$ and extend $u$ by zero. Each stencil is a forward, centred or backward average of $u$. Translation continuity in Lp makes each averaged difference from $u$ tend to zero in $L^2(\mathbb{R})$. Restricting those three bounds to their respective intervals proves $R_{\alpha }Ku\to u$. The noise-bias decomposition for linear regularization, with $\delta /\alpha \to 0$, finishes the convergent regularization of an inverse problem.

The wavefunction is a plane wave and a simultaneous formal eigenfunction of the momentum operator and the time-translation energy operator:Its momentum and energy are therefore definite, while its probability density is spatially uniform:For a particle of mass $m$, the probability current is $j=(\hslash /m)\mathrm{Im}(\overline{\varphi }{\partial }_x\varphi )=\hslash k{A}^2/m$, directed according to the sign of $k$.

On the whole real line, a nonzero constant-amplitude plane wave is not square-integrable and cannot be normalised to total probability one. It represents an idealised momentum eigenstate, interpreted through generalized eigenfunctions or as a limit of wave packets. Also, arbitrary $k,E$ do not automatically solve the Schrodinger equation for a specified potential energy: for a free particle, $E={\hslash }^2{k}^2/(2m)$, or this value plus the constant potential energy for a constant-potential region. If $A=0$, the zero function is not a physical quantum state.

Let $X$ be compact and $q:X\to X/R$ the quotient map. If $(U_i{)}_{i\in I}$ is an open cover of $X/R$, then $(q^{-1}(U_i){)}_{i\in I}$ is an open cover of $X$. A finite subfamily covers $X$; because $q$ is surjective, the corresponding $U_i$ cover $X/R$. Thus every quotient of a compact space is compact.

The Hausdorff property need not survive. On the real line, defineThe quotient $\mathbb{R}/\mathbb{Q}$ has more than one point. If two nonempty open subsets of the quotient were disjoint, their inverse images would be disjoint nonempty open subsets of $\mathbb{R}$ invariant under rational translation. But any two nonempty open intervals acquire an intersection after one is translated by a suitably chosen rational number, so two such saturated open sets cannot be disjoint. Distinct quotient points cannot be separated, and the quotient is not Hausdorff. This is the non-Hausdorff quotient of the real line by rational translation.

Finally let $f:X\to Y$ be a continuous bijection, with $X$ compact and $Y$ Hausdorff. Every closed subset $C\subseteq X$ is compact. Its continuous image $f(C)$ is compact, and every compact subset of a Hausdorff space is closed. Hence $f$ is a closed map. For every closed $C\subseteq X$,is closed in $Y$, so $f^{-1}$ is continuous. Therefore $f$ is a homeomorphism. This is the compact-to-Hausdorff continuous bijection theorem.

The Schwartz space on the real line isThese seminorms define its Fréchet space topology. The space of tempered distributions is its continuous dual space,

With the angular-frequency convention, the Fourier transform of a Schwartz function isIt maps $\mathcal{S}$ continuously to itself. The transform of $u\in {\mathcal{S}}^{\prime }$ is defined through the dual pairing:up to the fixed reflection and $2\pi$ factor if the inverse-transform convention is used for the test function.

Translation shifts the argument without changing the shape of a function. On the real line, the displayed convention shifts a graph to the right by $h$. Lebesgue measure invariance preserves every finite Lp norm; its Fourier transform is multiplied by $e^{-ih\xi }$.

Suppose a contour follows the real axis, closes in the upper half-plane, and avoids each simple real pole $a_j$ by a small semicircle above it. Each indentation is clockwise and tends toIf the large arc vanishes and there are no enclosed poles away from the real axis, the Cauchy principal value is therefore