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The printed assertion is false as a supremum bound uniform over shrinking bandwidths. The supplied hint proves a pointwise stochastic order with constants uniform in $x$. Taking the supremum inside the probability changes the problem: a growing number of spatial windows produces a logarithmic cost. We first establish the valid consequence of the hint, then give a counterexample satisfying the printed assumptions.

For $I=I_{x,h}=[x-h,x+h]$, let $p=\mu (I)$, $A_T={\int }_0^T{1}_I(X_t)dt$ and $N_T={\int }_0^T{1}_I(X_t)\sigma (X_t)d{W}_t$. The Invariant distribution of an Itô diffusion has a probability density function bounded above and bounded away from zero on the required compact interval. Choosing $h_0<M/4$ therefore gives constants $c_0,C_0>0$ such that $c_{0}h\le p\le C_{0}h$, uniformly for $|x|\le M/2$ and $0<h<h_0$.

For the centered function $f=1_I-p$, $\int |f|d\mu =2p(1-p)\le 2p$. Outside $[-M,M]$, $f=-p$. The given second-moment estimate and the Chebyshev inequality yieldOn the event $A_T\ge Tp/2$, the drift coefficient contributes an average whose error is at most $R{h}^{\alpha }$, by Hölder continuity. The Itô isometry and stationarity give $\mathbb{E}{N}_T^2\le \Vert \sigma {\Vert }_{\infty }^{2}Tp$. Another use of the Chebyshev inequality provesThis also accounts for the zero-denominator convention through the first exceptional event. The bound proves pointwise $R{h}^{\alpha }+O_{\mathbb{P}}((Th{)}^{-1/2})$, uniformly in the location's tail probabilities. It supplies no bound for the probability of a supremum over all locations.

For an explicit counterexample, take $\sigma \equiv 1$ and the bounded globally Lipschitz functionIt has Lipschitz constant $R=1$, exponent $\alpha =1$, and satisfies the inward-drift conditions with $M=2$, $\gamma =2$. The stochastic differential equation has a unique strong solution of a stochastic differential equation. Its Invariant distribution of an Itô diffusion has densityThe exponent is zero on $[-1,1]$, equals $-(|x|-1{)}^2$ when $1<|x|<2$, and equals $3-2|x|$ outside $[-2,2]$. This verifies integrability, positivity and a constant density ${\rho }_0=Z^{-1}$ on the central interval. Start the Itô diffusion in this Invariant distribution of an Itô diffusion to obtain the required stationary process.

Set $h=T^{-1/3}$ and choose $N=\lfloor (4h{)}^{-1}\rfloor$ windows with centers $x_j=-1/2+(4j-2)h$, $1\le j\le N$. Their intervals $I_j=[x_j-h,x_j+h]$ are disjoint and lie in $[-1,1]$, where the drift is zero. WriteThen ${\hat{b}}_T(x_j,h)=M_j(T)/A_j(T)$ whenever $A_j(T)>0$. The same occupation bound as above gives $\mathbb{E}(A_j(T)-q{)}^2\le C_{1}T{h}^2$ for $T\ge 1$. With $\delta =T^{-1/6}$, a union bound and the Chebyshev inequality implyThe quadratic variations are $⟨M_j⟩=A_j$, and the cross quadratic covariations vanish since the windows are disjoint. Each clock tends to infinity almost surely: for any fixed window, the occupation estimate at times $2^k$, followed by the Borel-Cantelli lemmas, gives $A_j(2^k)/(2^k\mu (I_j))\to 1$. The Knight theorem for orthogonal martingales therefore represents $M_j(t)=B_j(A_j(t))$ using independent standard Brownian motions $B_1,\dots ,B_N$. This theorem is applied to the finite collection for each $T$; no growing-dimensional central limit theorem is being assumed.

On the preceding clock event, compare each time-changed Brownian motion with its value at deterministic time $q$. The Brownian reflection principle and a union bound show that, for every fixed $\epsilon >0$,This bound does not require the Brownian motions to be independent of their clocks. In particular, ${\max }_j|M_j(T)-B_j(q)|/\sqrt{q}\to 0$ in probability. The variables $B_j(q)/\sqrt{q}$ are independent $N(0,1)$ random variables. Their Gaussian maximum exceeds $\sqrt{\log N}$ with probability tending to one: for $r\ge 1$, integration of the normal density over $[r,r+1/r]$ gives $\mathbb{P}(|Z|>r)\ge c{r}^{-1}{e}^{-r^2/2}$, and henceSince $A_j(T)\le (1+\delta )q$ on the clock event, it follows that, for some $c_3>0$,But $\sqrt{Th}Rh=1$ for this bandwidth. Thus the error after subtraction of $Rh$ and multiplication by $\sqrt{Th}$ is not bounded in probability, contradicting the printed stochastic order. The valid pointwise bound above is provable; the uniform shrinking-bandwidth assertion is not. This counterexample shows why a uniform rate needs at least a $\sqrt{\log (1/h)}$ enlargement of the noise scale. If $h$ is instead held fixed and constants may depend on $h$, this counterexample does not contradict that different interpretation.

Let $m=\mathbb{E}\Vert X_1{\Vert }_2$ and $q=\sqrt{n}K\ge 2$. The Lipschitz continuity of cosine and the Cauchy-Schwarz inequality implyThe centered expectation term contributes $nm$; omitting it would leave the Lipschitz bound incomplete. In particular $S_n$ has continuous sample paths on the compact cube, so its maximum exists and is measurable.

PutChoose a Cartesian finite net $G\subset [-K,K{]}^d$ whose coordinate spacings are at most $\eta$, including the endpoints. Every point of the cube is within Euclidean norm distance $\sqrt{d}\eta$ of $G$, andOn the event $L_n\le n{q}^a$, replacing any $u$ by a nearby grid point changes $S_n$ by at most $\sqrt{dn}$. Since $R>8\sqrt{d}$ and $\log (n{K}^2)\ge \log 4$, this is less than $\frac{R}{4}\sqrt{n\log (n{K}^2)}$. ThereforeThe Markov inequality bounds the first event by $2m{q}^{-a}$. At a fixed $v$, the summands are centered, independent and identically distributed, and have absolute value at most $2$. The supplied Hoeffding inequality, with $M=2$, bounds each grid event byA union bound now givesThe choice of $a$ balances the cost of a large random Lipschitz constant against the covering number of the cube: $d(1+a)-R^2/64=-a$. Consequently $C=2m+2\cdot 4^d$ works, withThis uses only the first moment $m$ and has a constant independent of $n,K,R$. The finite net is deterministic even though the Lipschitz constant is random.

Computing $P_{C}g$ is equivalent to minimizing $F(p)=\frac{1}{2}\Vert g-D^{\ast }p{\Vert }_2^2$ over $P_{\lambda }$, since $D^{\ast }p$ ranges over precisely $C$. Differentiating with the adjoint operator gives $\nabla F(p)=D(D^{\ast }p-g)$, whose Lipschitz constant is $L=\Vert D{D}^{\ast }\Vert =\Vert D{\Vert }^2$. Starting with any $p^0\in P_{\lambda }$, for example zero, the projected-gradient dual total variation algorithm isThe block formula is the Euclidean projection onto a convex set $P_{\lambda }$, since it clips each two-dimensional block to a Euclidean ball of radius $\lambda$. Both components must be clipped together for the isotropic block mixed norm.

For the unscaled forward differences, $\Vert D_x^{+}u{\Vert }_2^2\le 4\Vert u{\Vert }_2^2$ and likewise in the other direction, so $L\le 8$. More exactly, for $N>1$ the one-dimensional path difference has eigenvalues $4{\sin }^2(k\pi /(2N))$, $0\le k<N$, and the square-grid operator sums two such spectra. Thus $L=8{\cos }^2(\pi /(2N))$. Finite-dimensional projected gradient descent for a convex objective with an $L$-Lipschitz gradient converges from every feasible starting point to a dual minimizer for fixed $0<\tau <2/L$. The theorem assumes a nonempty closed convex set, a convex differentiable objective with globally Lipschitz gradient, and a nonempty minimizer set. All hold here: $P_{\lambda }$ is compact and convex, and $F$ is a continuous convex quadratic. Finite-dimensional convergence is of the full sequence, rather than only its objective values.

Consequently any fixed step with $0<\tau <1/4$ is safe on this grid; the more conservative $0<\tau \le 1/8$ is also safe. The dual minimizer need not be unique because $D^{\ast }$ has a null space, but $D^{\ast }{p}^k\to P_{C}g$ and $u^k\to u$ uniquely. For $N=1$, the gradient penalty vanishes and $u=g$, so no positive-$L$ step restriction is needed.

A map $f:A\to \mathbb{R}$ is Lipschitz continuous with Lipschitz constant $L$ whenfor all $y,z\in A$.

Fix $y_0\in A$. The Lipschitz inequality and the triangle inequality implywhile choosing $y=y_0$ gives a finite upper bound. Hence the infimum defining $F(x)$ is a real number.

If $x\in A$, then $f(x)\le f(y)+Ld(x,y)$ for every $y\in A$, while $y=x$ gives the reverse inequality after taking the infimum. Thus $F(x)=f(x)$. For arbitrary $x,z\in X$,Taking infima gives $F(x)\le F(z)+Ld(x,z)$; interchanging $x,z$ givesThis is the McShane extension theorem: $F$ extends $f$ without increasing its Lipschitz constant.

Let $X,Y$ be two continuous solutions on the same filtered probability space, driven by the same Brownian motion and with $X_0=Y_0$ almost surely. Write $D=X-Y$ and choose a common Lipschitz constant $L$ for $b,\sigma$. Introduce the stopping time ${\tau }_n=\inf \{t\ge 0:|X_t|+|Y_t|\ge n\}$. On $\{{\tau }_n=0\}$ the stopped difference is zero; before ${\tau }_n$, both coefficients and the difference are bounded.

The Itô formula and the zero expectation of the stopped square-integrable Itô integral yieldThe Gronwall inequality implies $\mathbb{E}{D}_{t\wedge {\tau }_n}^2=0$. Continuous solution paths are bounded on each compact time interval, so ${\tau }_n\uparrow \infty$ almost surely. Letting $n\to \infty$ proves $X_t=Y_t$ almost surely at each fixed time, and a countable dense set of times together with continuity gives indistinguishability of stochastic processes. Therefore the stochastic differential equation has pathwise uniqueness.

For $1\le k\le n$, the nearest neighbour distance ${\rho }_{(k)}(x)$ is the $k$th order statistic of $D_i=|X_i-x|$. Each $D_i$ has continuous distribution function $p_x(r)$ on $[0,\infty )$. The probability integral transform makes $U_i=p_x(D_i)$ independent and identically distributed random variables with the uniform distribution on $(0,1)$. Monotonicity of $p_x$ gives $P=U_{(k)}$, a uniform order statistic. Strict monotonicity is not needed at this stage.

For $0<s<1$, counting observations below $s$ givesDifferentiating and cancelling consecutive terms gives the Beta distribution:The Gamma function identity $\Gamma (j)=(j-1)!$ identifies this with the stated normalization. Ratios of Beta function integrals yield the momentsThis is the probability content of a nearest neighbour ball identity, specialized to intervals on the real line.

Now write $f_0=f(x)>0$. The Lipschitz bound givesThusStrict positivity of the probability density function makes $p_x$ continuous and strictly increasing from zero to one, so its inverse function is defined for every $s\in (0,1)$.

For the requested conditional calculation, suppose $f_0^2\ge L$. At $r_0=s/f_0$ the preceding bound givesTherefore $r=p_x^{-1}(s)\le s/f_0$, and substituting in the approximation proves the inverse probability content bound for a Lipschitz density:The same argument works, without $f_0^2\ge L$, on the meaningful local range $0<s<1$ with $s\le f_0^2/L$.

Since the nearest neighbour density estimator obeys $f(x)/{\hat{f}}_{(k)}(x)=2(n+1)f_0{p}_x^{-1}(P)/k$, its conditional bound follows from the Beta distribution moments:Hence, under the printed location condition,The conditional inverse function bound also bounds $p_x^{-1}(P)$, so the expected value in this calculation is finite under those formal hypotheses.

There is, however, a genuine flaw in the original PDF: the printed condition $f(x)\ge \sqrt{L}$ has no admissible points when $f$ is strictly positive on all of $\mathbb{R}$. A probability density function on all of $\mathbb{R}$ cannot have Lipschitz constant zero, so $L>0$. Its Lipschitz bound forcesIntegrating this triangular lower envelope gives the Lipschitz density height boundThe inequality is strict because the probability density function has positive mass outside the finite interval. Thus $f(x)<\sqrt{L}$ everywhere. The last two conditional conclusions above are valid implications but vacuous as printed. Dropping strict positivity gives only $f(x)\le \sqrt{L}$, with equality forcing the entire probability density function to be the normalized triangular envelope. The local inverse bound does not justify replacing the printed condition by its reverse in the final expected value bound, since $P$ ranges over all of $(0,1)$.

The Kantorovich–Rubinstein theorem concerns a Polish space $(X,\rho )$ and probability measures $\mu ,\nu$ with finite first moments, meaning $\int \rho (x,x_0)d\mu (x)+\int \rho (x,x_0)d\nu (x)<\infty$ for one, hence every, $x_0\in X$. It statesHere $\mathrm{Lip}(f)\le 1$ means $|f(x)-f(y)|\le \rho (x,y)$ for every $x,y$. The Wasserstein distance with ground cost $\rho$ therefore equals a supremum over functions with Lipschitz constant at most one. One may normalize $f(x_0)=0$, since adding a constant does not change the difference of integrals. This normalization bounds $|f(x)|$ by $\rho (x,x_0)$, so finite first moments ensure integrability. The supremum is unchanged if an absolute value is placed around the difference, because $-f$ is admissible whenever $f$ is.

The map $F$ itself has a horseshoe only for $\mu =2$. Indeed, its Lipschitz constant is $\mu$. If two intervals $J_0,J_1$ form a horseshoe and their convex hull has length $L$, thenwhich forces $\mu \ge 2$. At $\mu =2$, take $J=(-1,1)$, $K_0=(-1,0)$ and $K_1=(0,1)$; both $K_i$ map onto $J$.

The Lipschitz constant of $F^2$ is ${\mu }^2$, so the same length argument makes $\mu \ge \sqrt{2}$ necessary for an $F^2$ horseshoe. It is sufficient. PutFor $\mu \ge \sqrt{2}$, one has $a<1/\mu \le b\le 1$. Withthe two relevant monotone branches satisfyThus $K_0,K_1\subset J$ are disjoint and $F^2(K_0)=F^2(K_1)=J$. Therefore

Let $U\subset \mathbb{R}\times {\mathbb{R}}^d$ be an open set and $f:U\to {\mathbb{R}}^d$ a continuous function. Suppose $f$ is locally Lipschitz in the state $y$, with a Lipschitz constant uniform on a sufficiently small time-state neighborhood of each point. Then each $(t_0,y_0)\in U$ determines a unique local solution of a differential equation to the displayed initial value problem, on an interval around $t_0$. This is the local theorem in Ordinary Differential Equations and Dynamical Systems by Gerald Teschl.

For example, $y^{\prime }=y$, $y(0)=1$, has the solution of a differential equation $y(t)=e^t$: its derivative is itself and its starting value is $1$. The Lipschitz constant is $L=1$, since $|u-v|=|f(t,u)-f(t,v)|$. Thus uniqueness follows even before computing $y(1)=e\approx 2.71828183$. Picard iteration below constructs this exponential function one polynomial at a time.

The Lipschitz condition has real content. For the continuous function involving a square root,each waiting time $c\ge 0$ gives a different forward solution of a differential equation:The constant solution of a differential equation $y(t)=0$ is another choice. At $t=2$, delays $c=0,1,2$ give $y=4,1,0$ respectively. Before the join, both sides of the ordinary differential equation are zero; afterwards, $y_c^{\prime }=2(t-c)=2\sqrt{y_c}$. Both one-sided derivatives at $c$ are zero, so the join is valid. All these trajectories can also be extended by zero to $t<0$. The failed Lipschitz condition is visible in the quotientFor $\epsilon =10^{-2},10^{-4},10^{-6}$ this quotient is $20,200,2000$. Continuity still allows local existence by the Peano existence theorem; it cannot choose a waiting time. MIT's 18.034 Lecture 2, Example 2.3 develops the same waiting-time mechanism for a different ordinary differential equation, $y^{\prime }=3y^{2/3}$.

The useful promise is that an initial condition and a sufficiently regular local rule select one trajectory. It justifies comparing models and following trajectories without an arbitrary choice at each instant. A locally Lipschitz continuous rule need not keep that trajectory bounded: $y^{\prime }=1+y^2$ meets the theorem's hypotheses everywhere but has finite-time blowup. Longer existence is governed by the finite-dimensional continuation criterion.

Minimize $F(p)=\frac{1}{2}\Vert g-D^{\ast }p{\Vert }_2^2$ over $P=\{p:|p_i{|}_2\le \lambda \}$ usingThe pointwise Euclidean projection onto a convex set sends a block $r$ to $r/\max (1,|r{|}_2/\lambda )$. The gradient of $F$ has Lipschitz constant $\Vert D{\Vert }^2$, so finite-dimensional projected gradient descent converges for $0<\tau <2/\Vert D{\Vert }^2$. On an unscaled square grid with $N>1$, $0<\tau <1/4$ is safe. Reconstruct $u^k=g-D^{\ast }{p}^k$; unused dual directions may make the minimizing $p$ nonunique while the primal minimizer is unique.

The continuous map projecting a real number onto the closed interval $[a,b]$, with $a\le b$. It equals $a$ below the interval, equals the input inside it, and equals $b$ above it. It is nondecreasing and has Lipschitz constant one.

For an injective function that does not increase distances, one stretch-factor convention measures the largest compression that must be undone:If the sharp Lipschitz constant of $f$ is one, this equals bi-Lipschitz distortion. Fixing the scale matters: uniform compression $f(x)=x/2$ has stretch factor two in this convention, but bi-Lipschitz distortion one. In a geometric spanner, the corresponding stretch factor is the largest graph-route distance divided by direct Euclidean distance; the graph-to-Euclidean identification does not increase distances. ^[ref].

A uniformly continuous function has one input tolerance valid throughout its domain of a function. For $f:D\to \mathbb{R}$, with $D\subseteq \mathbb{R}$, uniform continuity meansChoose the output tolerance $\epsilon$ first, then a positive $\delta$ valid for every pair of function inputs. It may depend on $\epsilon$, the function and its domain of a function, but not on the pair’s location. The absolute values measure input and output distance.^[ref]

Compare the quantifier order with the epsilon-delta definition of continuity:Both lines end with $|x-a|<\delta \Rightarrow |f(x)-f(a)|<\epsilon$. In the first, $\delta$ can depend on the point $a$; in the second, it is chosen before $a$. Thus uniform continuity implies continuity at every point. Each requested $\epsilon$ gets its own shared tolerance.

For an exact example, take the polynomial function $f(x)=x^2$ on the closed interval $D=[0,2]$. For every pair of function inputs in that interval, polynomial factorization givesConsequently $\delta =\epsilon /4$ works everywhere on this domain of a function. With $\epsilon =0.4$, the common tolerance is $\delta =0.1$. Near the right endpoint, for example, $x=1.92$ and $y=2$ have input distance $0.08$ and output distance $4-1.92^2=0.3136<0.4$. The bound is sufficient; it does not claim the largest admissible input window at each point.

On the whole real line, the same polynomial function is a continuous function but fails uniform continuity. Take the sequences $x_n=n$ and $y_n=n+1/n$, for positive integers $n$. ThenFor any proposed $\delta >0$, choose an integer $n>1/\delta$. The inputs are closer than $\delta$, yet their outputs violate the fixed tolerance $\epsilon =1$. The sequences need not themselves have a finite limit: their mutual distance is what tends to zero. This is the obstruction captured by the sequential criterion for uniform continuity. Restricting the domain of a function to $[0,2]$ removes these escaping pairs.

A uniformly continuous function need not be a bounded function. The identity function $f(x)=x$ on the real line has $|f(x)-f(y)|=|x-y|$, so $\delta =\epsilon$ always works, although its image of a function is the unbounded real line. Uniform continuity controls changes between nearby inputs, not the overall size of outputs. Likewise, an unbounded domain of a function does not by itself prevent uniform continuity.

The square root $g(x)=\sqrt{x}$ supplies a different common tolerance on the interval $[0,\infty )$. Assuming $x\ge y\ge 0$, put $u=\sqrt{x}$ and $v=\sqrt{y}$. Since $0\le u-v\le u+v$,Interchanging the inputs when necessary provesThus $\delta ={\epsilon }^2$ guarantees uniform continuity on the entire domain of a function, including zero. For $\epsilon =0.4$, any input distance below $0.16$ suffices. The pairs $(0,0.09)$ and $(4,4.09)$ both have input distance $0.09$; their output distances are respectively $0.3$ and $0.09/(\sqrt{4.09}+2)<0.0225$. The common guarantee accommodates quite different local sensitivities.

This Hölder continuity bound allows a curved tolerance rule. It is weaker than a Lipschitz bound: at zero the output change divided by the input change isNo finite Lipschitz constant works near zero, even though uniform continuity does. A Lipschitz function does have uniform continuity, choosing $\delta =\epsilon /L$ when its Lipschitz bound $L$ is positive; when $L=0$ it is a constant function.^[ref]

For general metric spaces, replace the absolute values by the respective distance functions. The Heine-Cantor theorem gives an existence guarantee: every continuous function on a compact metric space is uniformly continuous. In particular, continuity on a finite closed interval provides a common tolerance, even when no simple formula for it is known.^[ref]

For a short proof outline, fix $\epsilon >0$. Continuity gives each point $a$ an open ball of radius $r_a>0$ within which outputs differ from $f(a)$ by less than $\epsilon /2$. The open balls of half these radii still cover the domain of a function. Compactness supplies a finite subcover. Take $\delta$ to be the smallest half-radius in this finite set. Two inputs closer than $\delta$ lie together in one of the original open balls, so the triangle inequality bounds their output distance by $\epsilon$. The beautiful step is that the compact set turns local choices into finitely many choices, whose minimum stays positive.

A shared tolerance also makes uniformly continuous functions preserve Cauchy sequences: after choosing $\delta$ for $\epsilon$, all sufficiently late input pairs are closer than $\delta$, so all corresponding output pairs are closer than $\epsilon$. Unlike a convergent sequence argument, this does not require an input limit inside the domain of a function. It is why uniform continuity is useful for extending a function toward missing endpoints and for the decay argument below.