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A radial profile is a one-variable function $\bar{u}$ representing a radial function $u$ by $u(x)=\bar{u}(|x|)$. On an annulus $a<|x|<b$, with $a>0$, the spherical coordinate system identifies its Lp norm with the weighted profile norm through $\int |u{|}^p=|S^{n-1}|{\int }_a^b|\bar{u}(r){|}^p{r}^{n-1}dr$. Thus equality of two radial functions almost everywhere identifies their profiles almost everywhere in the ordinary one-dimensional sense.

Let $0<a<b<\infty$ and $1\le p<\infty$. A radial function in $W^{1,p}(\{a<|x|<b\})$ has a radial profile in $W^{1,p}(a,b)$; its first weak derivative is the profile of the radial derivative. To see the identity, sum the weak derivative tests against the components of $\varphi (r)x/r$ to obtain $\int g\varphi r^{n-1}=-\int \bar{u}(\varphi r^{n-1}{)}^{\prime }$. Substitute $\varphi =\zeta /r^{n-1}$ for arbitrary $\zeta \in C_c^{\infty }(a,b)$. The weighted Lp norms and $a>0$ give the displayed bound. Radiality of the radial derivative follows by applying a rotation to the weak derivative identity: if $u(Qx)=u(x)$, then $\nabla u(Qx)=Q\nabla u(x)$ almost everywhere, so the radial scalar product is invariant.

For $p>1$, the one-dimensional Sobolev representative obeys $|\bar{u}(r)-\bar{u}(s)|\le |r-s{|}^{1-1/p}\Vert {\bar{u}}^{\prime }{\Vert }_p$. The reverse triangle inequality transfers this to $u$, giving a continuous embedding of this radial Sobolev space into $C^{0,1-1/p}$ on the annulus. The two Hölder seminorms are equal: one inequality follows from $||x|-|y||\le |x-y|$, and the reverse follows by putting both points on the same ray. The restriction $a>0$ is essential for the unweighted profile bound.

The reachable set ${\mathcal{R}}_T(x_0)$ collects all endpoints attainable from $x_0$ at exactly $T$ by admissible control inputs. If arrival any time up to $T$ is allowed, use ${\mathcal{R}}_{\le T}(x_0)={\bigcup }_{0\le t\le T}{\mathcal{R}}_t(x_0)$, with ${\mathcal{R}}_0(x_0)=\{x_0\}$. These definitions differ: a general dynamical system may drift away after early arrival, so waiting need not preserve the target. The reachable set is always relative to a declared input class and any constraints on the entire path.

For the double integrator starting from $(0,0)$, now require measurable functions as control inputs with $|u(t)|\le 1$ almost everywhere. Write $(p,v)$ for its endpoint at $T>0$. Two integrals determine the endpoint:Our mathematical example has the exact reachable setTo see the upper boundary, fix $v$: put $u=1$ first and $u=-1$ last, switching at $s=(T+v)/2$. This maximizes the weighted integral for position because early commands have the largest weight $T-t$; the required net velocity fixes $s$. Formally, $(T-t)-(T-s)=s-t$ has the same sign as this maximizing command, so changing to any other admissible command with the same integral cannot increase $p$. Reversing the order gives the lower boundary. Pointwise convex combinations of these two commands preserve $v$ and attain every intermediate $p$, so the formula describes the whole region, not merely a bound. The extreme commands are bang-bang control.

In particular, the bounded double integrator can move from rest to position $d$ and rest again exactly when $|d|\le T^2/4$. Its minimum duration is therefore $2\sqrt{|d|}$ in these normalized units. With freely chosen time every rest-to-rest distance is attainable; with a fixed deadline many are excluded. Changing the amplitude bound to $|u|\le a$ with $a>0$ scales this condition to $|d|\le a{T}^2/4$. This shows why specifying the controls matters even when the Kalman rank criterion gives full rank of a matrix.

In closed quantum control, a density operator reachable set lies inside its unitary orbit of a density operator, since coherent evolution preserves the spectrum. That orbit need not be fully reachable under the available control inputs.

For probability distributions $P,Q$, relative entropy satisfies $D_{\mathrm{KL}}(P\Vert Q)\ge 0$, with equality exactly when $P=Q$. Thus truthful prediction uniquely minimizes expected log-loss whenever the comparison is well defined, explaining the strictly proper scoring rule property of the logarithmic scoring rule.

For a short proof with probability density functions $f,g$ and $g>0$, set $r=f/g$. Their integrals both equal one, givingThe strictly convex function $r\log r-r+1$ is minimized only at $r=1$, so equality means $f=g$ almost everywhere. Its value at zero is one by $0\log 0=0$. Where $g=0<f$, relative entropy is infinite; where both vanish, the contribution is zero. These conventions extend the argument beyond positive reference probability density functions.

For nonnegative integrable functions converging almost everywhere to an integrable $f$, convergence of their integrals implies L1 convergence. Indeed $\min (f_n,f)\to f$ almost everywhere and is dominated by $f$, so the dominated convergence theorem gives convergence of its integral. Use $|f_n-f|=f_n+f-2\min (f_n,f)$. This applies to probability densities, giving convergence in total variation distance.

For a probability measure-preserving system and a countable measurable partition $\xi$ of finite entropy, $-n^{-1}\log \mu ({\xi }_0^{n-1}(x))$ converges almost everywhere and in $L^1$ to an invariant function whose integral is $h_{\mu }(T,\xi )$. For an ergodic transformation, the limit is the constant $h_{\mu }(T,\xi )$. In general it is $\mathbb{E}[I_{\mu }(\xi \mid {\bigvee }_{j=1}^{\infty }{T}^{-j}\xi )\mid \mathcal{I}]$. The martingale convergence theorem, maximal inequality for conditional information functions, and triangular ergodic averaging lemma prove this directly, without invertibility.

For any Lebesgue measurable set $A\subseteq \Omega$, the singular support of its indicator function, viewed as a regular distribution, is the intersection of the two essential closures in the displayed formula. Outside this intersection the indicator function is constant almost everywhere on a ball and hence represents a smooth function. At a point in the intersection, each sufficiently small ball has positive Lebesgue measure in both sets. Any continuous representative on that ball must take only the values zero and one everywhere: every other value would persist in a smaller ball and violate equality almost everywhere. By connectedness of the ball it would then be constant, a contradiction.

If $A$ is open, its essential closure is its relative closure. Its singular support can still be smaller than its topological boundary: $A={\mathbb{R}}^n\setminus \{0\}$ represents the constant one and has empty singular support.

Target: Support and singular support rules for convolution. Parent: Convolution of distributions with a compactly supported factor. Intended file: fourier-analysis.bigb.

If $u\in W_{\mathrm{loc}}^{1,2}$ and $\Phi$ is continuously differentiable with bounded derivative on the range in use, then $D(\Phi (u))={\Phi }^{\prime }(u)Du$ almost everywhere. One proves this first for smooth functions and then uses density of smooth functions in a Sobolev space and boundedness of ${\Phi }^{\prime }$. On compact sets one may truncate $\Phi$ away from the range of a bounded $u$.

For a Sobolev space function, almost every line parallel to a coordinate axis gives an absolutely continuous restriction. An absolutely continuous function has derivative zero almost everywhere on each of its level sets. Slicing and Fubini's theorem give the assertion in higher dimension. If $w\in W^{2,p}$, apply it to $w$ and then to each component of its gradient to get $D^{2}w=0$ almost everywhere on $\{w=0\}$. This supplies the contact differentiation used in noncontact of ROF level boundaries.

For a compact Hausdorff space $K$, $C(K)$ consists of continuous scalar-valued functions with the supremum norm. It is a Banach space and, with complex scalars, pointwise multiplication and conjugation make it a commutative unital C-star algebra.

Why does completeness hold? A Cauchy sequence in the supremum norm is Cauchy at every point, hence has a pointwise limit. The uniform Cauchy estimate passes to that limit, giving uniform convergence; the uniform limit theorem makes the limit continuous. Thus the norm and the chosen regularity of the functions cooperate to keep every limit inside $C(K)$.

Changing to the L2 norm on $C([0,1])$ loses this property. For integers $n\ge 2$, the continuous functions $h_n(x)=\min (1,\max (0,n(x-1/2)))$ approximate the step function $h(x)=0$ for $x\le 1/2$ and $h(x)=1$ for $x>1/2$. Direct integration givesThe triangle inequality makes $(h_n)$ a Cauchy sequence, but no continuous function agrees with $h$ almost everywhere: continuity would force value zero on the whole left open interval and one on the whole right open interval. Hence this normed space is incomplete. The missing limit exists in L2 space, its completion of a normed space, showing exactly what the larger space adds.

Let $X\ne \{0\}$ be a Banach space over the complex numbers, and let $T:X\to X$ be a bounded linear operator defined on all of $X$. Its spectrum consists of those $\lambda \in \mathbb{C}$ for which $T-\lambda I$ has no everywhere-defined bounded inverse. Its complement is the resolvent set of an operator $\rho (T)$. By the bounded inverse theorem, a bijective $T-\lambda I$ automatically has a bounded inverse. The spectrum is a nonempty compact subset of the complex plane, withThe bound uses the operator norm, and nonemptiness uses the complex field.^[ref] For a real linear operator, complex spectral values are understood by extending it to a complex vector space; restricting $\lambda$ to the real field can give an empty set.

In a finite-dimensional vector space, failure of inversion forces a nonzero kernel, hence an eigenvector. An infinite-dimensional vector space allows an injective shifted linear operator that cannot solve every equation $(T-\lambda I)u=g$. The spectrum therefore records both exact eigenvectors and failures of solvability or stable inversion. Searching only for eigenvalues can miss every spectral value of a perfectly simple bounded linear operator.

Consider the multiplication operator $M$ on the complex L2 space $L^2([0,1],dx)$, with Lebesgue measure:Its operator norm is one. The inequality $\Vert Mf{\Vert }_2\le \Vert f{\Vert }_2$ gives the upper bound; unit vectors concentrated near $x=1$ approach it. The result isThis is the interval example of the general connection between multiplication operators and the essential range of their multiplier.^[ref] Indeed, an eigenfunction would satisfy $(x-\lambda )f(x)=0$ almost everywhere. It vanishes except possibly on the singleton $\{\lambda \}$, a null set; such a function is the zero vector in L2 space. Concentrating at one point cannot produce a nonzero L2 function.

For $\lambda \notin [0,1]$, the inverse is completely explicit:The distance to a set is positive, so the inverse multiplier is bounded. Its essential supremum gives the displayed operator norm; arbitrarily small intervals near a closest point approach the bound. Our resolvent of an operator uses $(\lambda I-M{)}^{-1}$, the negative of this inverse, with the same operator norm. For $\lambda =-1/4$, $5/4$, and $1/2+i/10$, the inverse norms are respectively $4$, $4$, and $10$. The last parameter lies directly above the middle of the spectrum in the complex plane.

This calculation gives the spectrum a practical meaning: it marks where solving a shifted equation loses uniform control. At $\lambda =1/2+i\eta$, the largest amplification of an input error is $1/|\eta |$. Changing $\eta$ from $0.1$ to $0.01$ increases that factor from ten to one hundred. The equation remains uniquely solvable for each nonzero $\eta$, but its sensitivity grows as the parameter approaches the spectrum.

To see directly why every $\lambda \in [0,1]$ belongs to the spectrum, choose $0<\epsilon <1$ and define a clipped interval and a normalized indicator function:Its L2 norm is exactly one, including at both endpoints. On its support, $|x-\lambda |\le \epsilon$, givingThus any choice ${\epsilon }_n\to 0$ produces approximate eigenvectors. This specializes the localized-indicator construction in Nonnenmacher's spectral notes.^[ref] A hypothetical bounded inverse $B$ would imply $1=\Vert u_{\lambda ,\epsilon }{\Vert }_2\le \Vert B\Vert \epsilon$, impossible for arbitrarily small $\epsilon$.

For an interior point with $\epsilon \le \min (\lambda ,1-\lambda )$, $\ell =2\epsilon$ and the residual is exactly $\epsilon /\sqrt{3}$. At $\lambda =0$ or $1$, $\ell =\epsilon$ and the same residual formula holds, but the amplitude is ${\epsilon }^{-1/2}$ rather than $(2\epsilon {)}^{-1/2}$. At $\lambda =1/2$, taking $\epsilon =0.2,0.1,0.05$ gives amplitudes approximately $1.5811,2.2361,3.1623$ and residuals $0.11547,0.057735,0.028868$. The narrowing support and increasing amplitude keep total squared mass equal to one.

For every $\lambda \in [0,1]$, the continuous spectrum description is also concrete. Any L2 function $g$ can be approximated in L2 norm by its truncations to $|x-\lambda |\ge \delta$. Each truncation lies in the range of $M-\lambda I$, since division there is bounded. The range is therefore dense. However, the constant function $g=1$ has no L2 function preimage: $1/(x-\lambda )$ is not square-integrable, including at the endpoints.

These approximate eigenvectors do not approach an eigenvector in L2 norm. Any possible limit would have L2 norm one and satisfy $(M-\lambda I)u=0$ by continuity, contradicting the empty point spectrum. This is the striking feature: the linear operator acts increasingly like scalar multiplication on suitable unit vectors, although no direction realizes that scaling exactly.

A complementary example is the diagonal operator on sequence space $(Dx{)}_n=x_n/n$ on complex l2 sequence space, with $n=1,2,\dots$. The coordinate unit vectors satisfy $D{e}_n=(1/n)e_n$, soZero is not an eigenvalue, since $Dx=0$ forces every coordinate to vanish. Nevertheless $\Vert D{e}_n\Vert =1/n\to 0$: these are approximate eigenvectors at zero. More concretely, $g=(1/n{)}_{n\ge 1}$ belongs to l2 sequence space, but $Dx=g$ would require $x_n=1$ for all $n$, which does not. Outside the displayed closed set, division by $1/n-\lambda$ is uniformly bounded, proving that no further spectral values occur. Thus the spectrum contains the missing limit of genuine eigenvalues, even when that limit has no eigenvector.

A strictly proper scoring rule is a proper scoring rule for which the true probability distribution is the unique optimizer of the expected value. Equivalently, any different admissible probability distribution gives strictly larger expected loss. For probability density functions, two densities that agree almost everywhere represent the same probability distribution. Ordinary propriety permits ties: the constant reward $S(P,x)=0$ is a proper scoring rule, because every report ties, but it is not a strictly proper scoring rule and elicits no information.

Strict propriety concerns expectation, not guaranteed success on each observation or each realized dataset. With the stipulated Bernoulli distribution $q=0.7$, suppose the outcome is $Y=0$. The honest report $p=0.7$ then has Brier score loss $0.49$, while $p=0.2$ has loss $0.04$. The logarithmic loss similarly gives losses $-\ln (0.3)\approx 1.203973$ and $-\ln (0.8)\approx 0.223144$. The honest forecast loses this comparison on that draw, which occurs with probability $0.3$ in the stipulated statistical model.

The $Y=0$ loss curves favor $p=0$, and the $Y=1$ loss curves favor $p=1$. Their weighted average, with weights $0.3$ and $0.7$, favors $p=0.7$. Choosing a report after observing $Y$ would use information unavailable at forecast time. Even an average loss on a finite dataset can rank an honest forecast below a competing forecast through outcome variation; strict propriety guarantees the ranking in expected value under the specified information and true predictive distribution.

A nonnegative function-valued operator $S$ is sublinear if $S(f+g)\le S(f)+S(g)$ and $S(cf)=|c|S(f)$ almost everywhere. This pointwise domination notion includes maximal one-sided interval averaging operators and differs from requiring an ordinary linear map.

Suppose measurable $a_n$ on a Lebesgue measurable set $E$ satisfy ${\sum }_n{\int }_E|a_n|<\infty$. The nonnegative partial sums $S_N={\sum }_{n\le N}|a_n|$ increase to a measurable extended-valued $S$. The monotone convergence theorem gives ${\int }_{E}S={\sum }_n{\int }_E|a_n|<\infty$, forcing $S<\infty$ almost everywhere. Each $a_n$ is finite almost everywhere too; discard the union of these countably many null sets when defining signed partial sums. The series ${\sum }_n{a}_n$ then converges absolutely outside a null set.

Suppose $f_k\to f$ almost everywhere and ${\sup }_k|f_k|\in L^1$ in a probability measure-preserving system. Thenalmost everywhere and in $L^1$, where $\mathcal{I}$ is the invariant sigma-algebra. For the almost-everywhere assertion, bound the terms with index at least $M$ by ${\sup }_{k\ge M}|f_k-f|$, apply the Birkhoff ergodic theorem, and then let $M\to \infty$. The finitely many remaining end terms vanish by the linear growth bound for integrable observables.

On $(0,1)$, for $k\ge 0$ and $0\le j<2^k$, the displayed indicator functions form a sequence that converges to zero in every finite Lp norm while failing to converge pointwise anywhere. Indeed, for $1\le p<\infty$,Every point belongs to exactly one interval in each dyadic block, and for $k\ge 1$ it misses at least one other interval in that block. Thus every point sees infinitely many ones and infinitely many zeros. Half-open intervals specify the behaviour at dyadic endpoints. This separates norm convergence in an Lp space from pointwise convergence and convergence almost everywhere.

For nonnegative Lebesgue measurable $f$ on $(0,1)$, the displayed Volterra power-kernel integral is defined for every $0<x<1$ as a nonnegative extended Lebesgue integral. The factor is finite and continuous on the open integration interval $(0,x)$, and an endpoint value can be assigned arbitrarily. The result may initially be infinite; for example $f(y)=1/y$ makes every value infinite. At $\beta =1$, it reduces to the ordinary Volterra operator.

For joint integration, use the integral kernel $(x-y{)}^{\beta -1}$ on $0<y<x<1$ and zero elsewhere. This piecewise prescription avoids the singular diagonal and never raises a negative number to a noninteger power. The L1 bound for a Volterra power-kernel integral gives an integrable output, finite almost everywhere, whenever $f$ is integrable.

For a normed space $E$ and $0<q<\infty$, an operator $T:E\to L^0$ is of weak type $(E,q)$ when the displayed distribution bound holds uniformly over $f\in E$ and $a>0$. No completeness of $E$ is needed. A sublinear operator has pointwise triangle domination and absolute homogeneity; its outputs may be understood modulo almost everywhere equality.