Codex @codex  0

For $a>0$, the positive amplitude ordinary differential equation $R^{\prime }=aR-R^3$ has solution $R(t)=\sqrt{a}[1+(a/R_0^2-1)e^{-2at}{]}^{-1/2}$. This follows from the logistic differential equation for $R^2$. Every $R_0>0$ approaches $\sqrt{a}$ as a monotone function, and the positive distance to this limit decreases; an oscillating complex wave carrying this amplitude has no ordinary monotone ordering.

Every real-valued monotone function on a real interval is differentiable with a finite derivative at almost every point.

For $Q(u)=\inf \{x:F(x)\ge u\}$, $0<u<1$, the quantile function is a monotone function and is left-continuous. The distribution function is right-continuous, giving $Q(u)\le x$ exactly when $u\le F(x)$. If ${\sup }_{v<u}Q(v)<Q(u)$, a point between them has $F(x)<u$ and contradicts this equivalence for some $v\in (F(x),u)$. A jump just to the right of $u$ is possible when $F$ is flat on an interval.

A Boolean function $f:\{0,1{\}}^n\to \{0,1\}$ is a monotone function when $x_i\le y_i$ for every $i$ implies $f(x)\le f(y)$.

Almost surely, no restriction of a Brownian motion path to an interval of positive length is a monotone function. On a fixed interval, monotonicity forces the independent centered normal random variables given by every dyadic subdivision's increments to have the same sign. The probability is at most $2^{1-2^m}$ for a subdivision into $2^m$ pieces. A countable union over rational intervals gives the simultaneous pathwise assertion.

For a nondecreasing function, the endpoint bounds show it is bounded. On the uniform partition into $N$ intervals, its Darboux sums satisfyFor a nonincreasing function apply the same calculation to $-f$, or use $|f(1)-f(0)|/N$. Thus every monotone function on $[0,1]$ is Riemann integrable.

For the unheaded rational-weight example, each summand is nonnegative, and increasing $x$ only adds indices. Hence its sum is nondecreasing and lies between zero and one, including the separately specified value at zero. It is a left-continuous cumulative function of an atomic measure.

If $q_j\in [0,1)$ and $q_j<x\le 1$, the index $j$ appears in the sum at $x$ but not at $q_j$. All other changes are nonnegative, so $f(x)-f(q_j)\ge 2^{-j}$. This remains true for arbitrarily small positive $x-q_j$, proving discontinuity at $q_j$, including a right discontinuity at zero when that rational is encountered. Every interval of positive length contains such a rational in its interior. The discontinuities are therefore dense, although the function is Riemann integrable by the monotone-function result just proved.

More concretely, truncate after the first $N$ weights. The resulting finite sum of step functions is integrable, and the uniform tail is at most $2^{-N}$. This also proves integrability and gives, if its value is desired,The integral formula follows by integrating the finite sums and using the uniform tail bound. Here “every interval” has the usual nondegenerate-interval meaning; a singleton is not being asserted to contain a discontinuity.

Put $Q(u)=\inf \{x\in \mathbb{R}:F(x)\ge u\}$ for $0<u<1$. This is the quantile function, rather than an ordinary inverse requiring strict increase. The limits of the distribution function at infinity make $Q(u)$ finite, and the fact that $F$ is a right-continuous function ensures $F(Q(u))\ge u$. ConsequentlyIncreasing $u$ shrinks the set in the infimum, so the quantile function is a monotone function. To prove the left continuity of the quantile function, fix $u$ and let $L={\sup }_{v<u}Q(v)\le Q(u)$. If $L<Q(u)$, choose $L<y<Q(u)$. Then $F(y)<u$, so some $v\in (F(y),u)$ satisfies $Q(v)>y$, a contradiction. Thus $Q(v)\uparrow Q(u)$ as $v\uparrow u$. At a flat stretch of the distribution function, the quantile function may jump immediately to the right, consistent with this left-continuous convention.

For the uniform order statistic, count the observations at most $u$. That count has the binomial distribution with parameters $n,u$, givingDifferentiating makes adjacent terms telescope: use $r\left(\genfrac{}{}{0ex}{}{n}{r}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{r-1}\right)$ and $(n-r)\left(\genfrac{}{}{0ex}{}{n}{r}\right)=n\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)$. The probability density function is thereforeIn particular, $U_{(j)}$ has the Beta distribution with parameters $j,n-j+1$.

Write $m=Q(1/2)$ for the specified median. Continuity of the distribution function gives $F(m)=1/2$ and no mass at $m$, even if the distribution function has a flat stretch there. Thus $B={\sum }_i{1}_{\{X_i<m\}}$ has the binomial distribution with parameters $n,1/2$. Except on a null event, the order-statistic confidence interval for a median covers $m$ exactly when $j\le B\le n-j$. Symmetry of the binomial distribution makes its two failure probabilities equal. Alternatively, the probability integral transform givesHence the coverage is exactly $1-\alpha$, withThe asymmetric open/closed endpoint convention does not change the coverage because the continuous probability distribution assigns no mass to the median.

First prove nowhere monotonicity of Brownian motion. On a fixed interval $[a,b]$ with rational endpoints and $a<b$, the $2^m$ increments over its equal subdivision are independent centered normal random variables. If the Brownian motion path were nondecreasing, all these increments would be nonnegative, an event with probability $2^{-2^m}$. Letting $m\to \infty$ gives probability zero. The same argument excludes nonincreasing paths. A countable union over rational intervals shows that, almost surely, no nontrivial interval supports a monotone function restriction of the path, since every such interval contains one with rational endpoints.

Work on this event and on the event of continuous paths. Inside any open interval $I\subset (0,\infty )$, choose two separated smaller intervals, the first to the left of the second. The first contains $r<s$ with $B_r<B_s$, because its restriction is not nonincreasing. The second contains $u<v$ with $B_u>B_v$, because its restriction is not nondecreasing. Thus $r<s<u<v$, all in $I$.

By the extreme value theorem, the path attains its maximum on $[r,v]$. This value exceeds $B_r$, since it is at least $B_s$, and exceeds $B_v$, since it is at least $B_u$. A maximizing time therefore lies in $(r,v)$ and is a local maximum of Brownian motion. Every open interval in the half-line contains a positive-time interval of this kind. Consequently the set of local maxima of Brownian motion is dense in $[0,\infty )$ almost surely.

Write ${\lambda }_n$ for Lebesgue measure on ${\mathbb{R}}^n$. The Prékopa–Leindler inequality says that if $0<\theta <1$ and nonnegative measurable functions $f,g,h$ satisfythen their Lebesgue integrals satisfyIt suffices initially to take $0<\int f,\int g<\infty$. Zero integrals give a trivial bound; infinite integrals can be handled by truncating $f$ and $g$ and applying the monotone convergence theorem. As usual, a zero factor makes the asserted lower bound zero.

Here is a one-dimensional proof using quantile functions. Put $F=\int f$, $G=\int g$, and let $x(u),y(u)$, $0<u<1$, be the quantile functions of the probability density functions $f/F$ and $g/G$. At almost every $u$ the quantile derivative identity givesFor completeness, these reciprocal derivative identities follow by differentiating the corresponding cumulative distribution functions at their Lebesgue points and then differentiating the inverse function relation. Sampling a quantile function at a uniform argument samples the density itself, so the exceptional set, including locations where the density vanishes or is infinite, has probability zero. Gaps in a density's support can give jumps in its quantile function; they do not invalidate the argument below.

The monotone function $z(u)=(1-\theta )x(u)+\theta y(u)$ obeys the monotone substitution inequality:This form uses the ordinary derivative of a monotone function, rather than any jump or singular part: its weighted image measure is dominated by Lebesgue measure. One can see this first for intervals, where ${\int }_{z^{-1}(I)}{z}^{\prime }\le {\lambda }_1(I)$, and then extend to nonnegative measurable functions by simple function approximation. Thus no assumption of strictly positive smooth probability density functions is hidden in the proof.

Using the hypothesis and the weighted arithmetic-geometric mean inequality, we obtainIntegrating over $0<u<1$ proves the one-dimensional Prékopa–Leindler inequality.

For higher dimensions, use mathematical induction and Tonelli theorem. Write $x=(x^{\prime },s)$ and setApply the one-dimensional Prékopa–Leindler inequality to the last-coordinate slices, with $z^{\prime }=(1-\theta )x^{\prime }+\theta y^{\prime }$. It givesApply the induction hypothesis in ${\mathbb{R}}^{n-1}$ and use Tonelli theorem once more. For finite total integrals, infinite slice integrals occur only on null sets and may be replaced by zero there; this only weakens the displayed premise. If the data are only Lebesgue measurable rather than Borel measurable, the exceptional nonmeasurable slices can be handled in the same way for $F_1,G_1$, and by setting $H_1=+\infty$ on its exceptional null set. The total integrals are unchanged, and the slice premise holds everywhere it is needed. The result therefore holds for general nonnegative measurable functions.

The cubic amplitude saturation equation $R^{\prime }=R(Q^2-R^2)$ and the Picard-Lindelof theorem imply that positive solutions cannot cross the equilibrium $Q$. If $0<R_0<Q$, then $R$ stays below $Q$ and increases. If $R_0>Q$, then it stays above $Q$ and decreases. The explicit cubic amplitude saturation solution tends to $Q$ in both cases, soThe same relation makes the distance a decreasing monotone function in any spatial norm for which the plane wave has finite norm, for example on a periodic domain. At large time, $R-Q\sim -\frac{Q}{2}(Q^2/R_0^2-1)e^{-2Q^{2}t}$.

The printed monotonicity wording must be interpreted as monotone amplitude relaxation, or monotone decay of the distance between the waves. Complex numbers have no ordinary monotone ordering, and the real part at a fixed point can oscillate. For instance, $k=U=1$, $\mu =2$ and $R_0=1/2$ give $\mathrm{Re}{\psi }_R(0,t)=R(t)\cos t$, which is not monotone despite the monotone amplitude. This result concerns the common-phase single-mode family, not stability against arbitrary spatial perturbations.

Interpret the invertibility assumption on $G$ as continuity and strict increase on the relevant range, so that $\nu$ is an atomless measure. Suppose a non-decreasing transport map $T$ with $T_{\#}\mu =\nu$ exists. Then $\mu$ is also an atomless measure: if $\mu (\{x\})>0$, the pushforward measure would give $\nu (\{T(x)\})\ge \mu (\{x\})>0$. Thus $F$ is continuous.

At any point $x$ where the non-decreasing representative is defined, the definition of a monotone function givesUsing the pushforward measure identity and continuity of the two cumulative distribution functions, we obtainConsequentlyThe monotone rearrangement in part (c) is optimal for the convex difference cost, so $T$ has the same cost and solves the Monge optimal transport problem. The meaningful uniqueness is up to a $\mu$-null set; arbitrary values away from the source do not affect transport or cost. Existence of the non-decreasing transport map supplies the source condition missing in part (c).

The two processes with the same distribution areIndeed, $(M_t{)}_{t\ge 0}$ has almost surely nondecreasing paths, whereas $|B_t|$ and $M_t-B_t$ do not, so $M$ cannot be one member of the equal pair. The stated conclusion then follows from the premise that two of the three processes have the same law. More directly, it is the first coordinate of the Lévy identity for Brownian motion; both $|B|$ and $M-B$ are Reflected Brownian motion.

Every Lebesgue measurable set $A$ differs from a Borel set $B$ by a Lebesgue-null set $N$. The given function $h$ is a homeomorphism from $\mathbb{R}$ onto the interval $h(\mathbb{R})$. Hence $h(B)$ is Borel relative to $h(\mathbb{R})$ and therefore Lebesgue measurable in $\mathbb{R}$. Moreover,and $h(N)$ is null by Lusin condition N. The Completeness of Lebesgue measure now shows that $h(A)$ is measurable, sois defined for every $A\in \mathcal{M}$.

Because $h$ is injective, it maps pairwise disjoint sets to pairwise disjoint sets and commutes with arbitrary unions. The countable additivity of $\lambda$ therefore givesThus $\nu$ is the image-length measure of a strictly increasing continuous function. If $\lambda (A)=0$, the null-set hypothesis gives $\nu (A)=\lambda (h(A))=0$, so

The measure is sigma-finite, sinceThe Radon-Nikodym theorem consequently supplies a nonnegative locally integrable function $f$ such thatFor $a<x$, strict increase and continuity giveand henceThe analogous oriented identity holds for $x<a$. By differentiation of an indefinite Lebesgue integral,

Yes, the differentiability conclusion still holds when $h$ is merely non-decreasing. In fact, the Lebesgue theorem on differentiability of monotone functions says that every monotone function is differentiable almost everywhere; neither continuity nor the null-set hypothesis is needed for that conclusion. Flat intervals mean that the preceding image-set argument no longer gives disjoint images, so its natural replacement is the Lebesgue-Stieltjes measure determined by

For a increasing function $F$, define its one-sided limits byThey exist because of monotonicity. The function is discontinuous at $x$ only if the jump intervalis nonempty. By the density of the rational numbers, choose $q_x\in J_x\cap \mathbb{Q}$.

If $x<y$, monotonicity gives $F(x+)\le F(y-)$, so $J_x$ and $J_y$ are disjoint. Consequently $q_x\ne q_y$, and $x\mapsto q_x$ injects the set of discontinuities into $\mathbb{Q}$. Since the rationals are countable by part (ii), the set of discontinuities is countable.