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A one-dimensional Sobolev representative is absolutely continuous. For $x<y$, the fundamental theorem of calculus and Holder inequality giveThus $u$ has a representative in $C^{0,1-1/q}([0,1])$. The qualification about representatives matters because a Sobolev space element is an almost-everywhere equivalence class.

In two dimensions, the Sobolev fundamental theorem of calculus on lines and Fubini's theorem imply that almost every horizontal and vertical slice belongs to $W^{1,q}(0,1)$ and has this one-dimensional Hölder continuity. The slice seminorm depends on the slice; this does not give one uniform pointwise estimate on the square. For $q>2$, Morrey's inequality additionally gives a globally Hölder continuous representative of exponent $1-2/q$. For $1<q\le 2$, global continuity need not hold. For example, with a smooth cutoff around an interior point, $u(x)=|x-x_0{|}^{-a}$ lies in $W^{1,q}$ when $0<a<2/q-1$, but is unbounded. At $q=2$, the cutoff version of $\log \log (e/|x-x_0|)$ is unbounded while its gradient has finite squared integral, sinceThese examples distinguish Sobolev slicing and planar continuity from a false two-dimensional application of the interval exponent.

Put $\Omega =(0,1{)}^2$. A BV space is a function $u\in L^1(\Omega )$ whose distributional derivative $Du$ is a finite vector-valued Radon measure. Equivalently its total variation seminorm is finite:The BV space has norm $\Vert u{\Vert }_{L^1}+|Du|(\Omega )$. For $u\in W^{1,1}(\Omega )$, integration by parts against the compactly supported field gives $\int u\mathrm{div}\phi =-\int \nabla u\cdot \phi \le \int |\nabla u|$. Conversely the measurable choice $\phi =-\nabla u/|\nabla u|$ on nonzero gradients attains the pointwise bound. Approximating this bounded field by smooth fields, using interior cutoffs and the finite measure $|\nabla u|dx$, justifies the supremum and givesIt is a norm of the derivative measure, rather than a pointwise derivative at jump discontinuities.

There is a genuine mismatch in the printed definition of the next functional. Its constraints on ${\phi }_0$ and $\phi$ are independent. Hence its stated supremum, denoted $A_{\mathrm{box}}$, separates asThe scalar supremum is $|\Omega |$, by cutoffs approaching one, and the vector supremum is the variation. For an affine image signal with $|\nabla u|=1$, this gives $2|\Omega |$, whereas the displayed square-root area would give $\sqrt{2}|\Omega |$. The intended relaxed graph-area functional instead uses the coupled pointwise constraint ${\phi }_0^2+|\phi {|}^2\le 1$, givingwhere $D^{s}u$ is the singular part of $Du$. Both readings have a minimizer, but their equations are different.

Here is the direct method in the calculus of variations for either reading. Let $A_{\bullet }$ be the literal $A_{\mathrm{box}}$ or the corrected $A$, and define the energy on $BV(\Omega )\cap L^2(\Omega )$, assigning infinity elsewhere. A minimizing sequence has bounded energy by comparison with $u=0$. Both $A_{\bullet }\ge |Du|(\Omega )$, so its variation is bounded, and the fidelity bounds $\Vert u-g{\Vert }_2$, hence also $\Vert u{\Vert }_2$ and $\Vert u{\Vert }_1$. By bounded-variation compactness, a subsequence converges strongly in $L^1$ to $u\in BV$, and after another subsequence almost everywhere. Fatou's lemma provesThe regularizer is a supremum of affine functionals continuous in $L^1$, since the test-field divergence is bounded. It is therefore lower semicontinuous. Combining the two lower bounds proves existence of a minimizer. In fact the convex regularizer and the strictly convex squared fidelity make the minimizer unique up to null sets. This does not assert that the minimizer must belong to $W^{1,1}$.

For the intended graph area, conditionally assume that the minimizer is in $W^{1,1}$. For $\eta \in C_c^{\infty }(\Omega )$, differentiate at $u+t\eta$. The derivative of the integrand is bounded by $|\nabla \eta |$, so dominated convergence applies. The weak equation isthat is,This is the graph-area Euler-Lagrange equation. Compactly supported variations impose no boundary condition in this statement.

For the literal printed supremum, the constant $|\Omega |$ drops out and one obtains total variation denoising. Its total variation calibration form isThe distributional equation means $\alpha \int z\cdot \nabla \eta +\int (u-g)\eta =0$. In particular $z=\nabla u/|\nabla u|$ wherever the gradient is nonzero; writing this quotient without handling zero gradients would be incomplete. Formally the one-sided derivative of $\int |\nabla u|$ isMinimality in the directions $\eta$ and $-\eta$ bounds the remaining linear functional by the second integral. The Hahn-Banach theorem extends it on that zero-gradient set to a bounded vector field of magnitude at most one, furnishing $z$ and the displayed weak equation. Thus the literal definition has a nonsmooth subgradient equation, not the square-root equation above.

We use the Bolzano-Weierstrass theorem: every bounded real sequence has a convergent subsequence. If $f$ were unbounded on $[a,b]$, there would be $x_n\in [a,b]$ with $|f(x_n)|\ge n$. The Bolzano-Weierstrass theorem gives a subsequence $x_{n_k}\to c$, and the closed interval contains $c$. Since $f$ is a continuous function, $f(x_{n_k})\to f(c)$, contradicting $|f(x_{n_k})|\ge n_k\to \infty$. Thus $f$ is bounded.

Let $s={\sup }_{x\in [a,b]}f(x)$. By the definition of the supremum, choose $y_n$ with $s-1/n<f(y_n)\le s$. Again the Bolzano-Weierstrass theorem gives $y_{n_k}\to d\in [a,b]$. Continuity yields $f(d)=s$. Applying the same argument to $-f$ gives a point where $f$ attains its infimum. Therefore both the supremum and the infimum are attained. This proves the extreme value theorem for a closed bounded interval.

The intermediate value theorem says that if $f:[a,b]\to \mathbb{R}$ is a continuous function, then every value between $f(a)$ and $f(b)$ is attained at some point of $[a,b]$.

Endpoint values are already attained. Suppose $f(a)<y<f(b)$ and define $E=\{x\in [a,b]:f(x)<y\}$. It is nonempty and bounded above, so $c=\sup E$ exists. Continuity at $a$ ensures that $E$ contains points to the right of $a$, and continuity at $b$ ensures that no point sufficiently close to $b$ belongs to $E$. Hence $a<c<b$.

By the definition of the supremum, there are points of $E$ approaching $c$. If $f(c)>y$, continuity would exclude all such nearby points from $E$, a contradiction. If $f(c)<y$, continuity would put some point to the right of $c$ in $E$, again a contradiction. Therefore $f(c)=y$. When $f(a)>f(b)$, apply this argument to $-f$. This proves the intermediate value theorem in all cases.

If two order automorphisms agree on $\mathbb{Q}$, their values at any real $x$ are the same supremum of the common values at rational points below $x$. This is the uniqueness argument for extension of an order automorphism from a dense subset; it does not require either automorphism to carry $\mathbb{Q}$ onto itself.

Their restrictions belong to ${\mathbb{R}}^{\mathbb{Q}}$, whose cardinality isThis gives the upper bound. The translations $x\mapsto x+r$, one for each real $r$, are distinct order automorphisms, giving the lower bound. Consequently

Use the order completeness of $\mathbb{R}$ to defineThe set in this supremum is nonempty and bounded above: choose $d_0,d_1\in D$ with $d_0<x<d_1$, using that $D$ is an order-dense subset. For $x\in D$, all its displayed values lie below $\phi (x)$. If $y<\phi (x)$, choose $e\in D$ with $y<e<\phi (x)$; then $d={\phi }^{-1}(e)<x$. Thus the supremum equals $\phi (x)$.

The extension is strictly increasing. If $x<y$, choose $d,e\in D$ with $x<d<e<y$; thenApply the same construction to ${\phi }^{-1}$, obtaining $\psi$. For $d<x$, choose $e\in D$ with $d<e<x$; this gives $\phi (d)<\tilde{\phi }(x)$. For $d>x$, a point of $D$ between $x$ and $d$ similarly gives $\phi (d)>\tilde{\phi }(x)$. HenceTaking inverse images and a supremum shows $\psi (\tilde{\phi }(x))=x$. The symmetric argument gives $\tilde{\phi }(\psi (y))=y$, so the extension is an order automorphism.

For uniqueness, any increasing extension $h$ has the same strict lower cut in $D$ at its value $h(x)$. Two distinct real numbers have different cuts in an order-dense subset, so $h(x)=\tilde{\phi }(x)$. The extension exists and is unique.

A Suslin line is a dense linear order without endpoints that is order-complete, has the countable chain condition for a linear order, and is not separable in its order topology. Thus every disjoint family of nonempty open intervals is countable, but there is no countable order-dense subset. The completeness condition says that every nonempty bounded-above subset has a supremum. These requirements distinguish a Suslin line from the real line, which has the countable order-dense subset $\mathbb{Q}$.

Put $\mathcal{V}=⟨V_{\kappa },\in ,R⟩$. Because $\kappa$ is inaccessible, $|V_{\xi }|<\kappa$ for every $\xi <\kappa$, and $\kappa$ is regular. Choose one witness in $V_{\kappa }$ for every existential formula of this expanded language and every finite parameter tuple for which a witness exists. These are Skolem functions; they can be chosen in the ambient universe even when $R$ is not definable inside $V_{\kappa }$.

For each $\xi <\kappa$, the ranks of all chosen witnesses with parameters from $V_{\xi }$ have supremum below $\kappa$: there are fewer than $\kappa$ such parameters and only countably many formulas. Choose $F(\xi )<\kappa$ strictly above those ranks. The limit closure pointsform a club set. At $\alpha \in C$, every existential assertion in $\mathcal{V}$ with parameters in $V_{\alpha }$ has a witness in $V_{\alpha }$. The Tarski-Vaught test yieldsThus the set $E$ of all such elementary levels is unbounded. It is also closed: the union at a limit of increasing elementary levels is an elementary substructure by the elementary chain theorem, and the union of their ranks is $V_{\alpha }$ with predicate $R\cap V_{\alpha }$. ThereforeThis is club reflection below an inaccessible cardinal.

The rank characterization holds for a set relation, or a set-like class relation. For such a well-founded relation, use well-founded recursion to defineThe predecessor set is a set, so the supremum is an ordinal. For a set-like class relation, the closure of the predecessors of any one point under finitely many predecessor steps is a set; perform the ordinary set recursion there. Uniqueness makes these local definitions agree, producing a class rank function. The formula immediately gives $xRy\Rightarrow \rho (x)<\rho (y)$.

Conversely, suppose such an ordinal-valued function exists. For any nonempty subset, or nonempty subclass in the class formulation, take an element whose rank is least among the ranks occurring. It has no predecessor in that subset or subclass, because a predecessor would have smaller rank. This proves well-foundedness.

There is a qualification in the printed class formulation. If well-founded class relations are defined to include set-likeness, it is already implicit and the assertion is correct. Under the minimal-element definition alone, it must be supplied. Without set-likeness, let $X=\mathrm{Ord}\cup \{u\}$, where $u=\{1\}$ is not an ordinal, and put every ordinal below $u$, with the usual ordinal order below it. Every nonempty subclass has a minimal element, so the relation is well founded. But transfinite induction forces $\rho (\alpha )\ge \alpha$ for all ordinals, whereas $\rho (u)$ would have to exceed every $\rho (\alpha )$. No such ordinal exists. Thus the unrestricted class assertion is false; set-likeness is necessary for this standard rank proof.

The Bolzano-Weierstrass theorem states that every bounded sequence of real numbers has a convergent subsequence. We first use it to show that a continuous function on $[a,b]$ is bounded above. Otherwise, choose $x_n\in [a,b]$ with $f(x_n)>n$. A convergent subsequence $x_{n_k}\to c$ exists, with $c\in [a,b]$ because the interval is closed. Continuity would give $f(x_{n_k})\to f(c)$, contradicting $f(x_{n_k})>n_k\to \infty$.

Let $L=\sup \{f(x):x\in [a,b]\}$, which is finite. By the definition of supremum, there are $y_n\in [a,b]$ with $L-1/n<f(y_n)\le L$. Another application of the Bolzano-Weierstrass theorem gives a subsequence $y_{n_k}\to c\in [a,b]$. By continuity,Thus the supremum is actually attained:This proves the maximum part of the extreme value theorem from sequential compactness rather than assuming it.

Now suppose $f^{\prime \prime }<0$ on $\mathbb{R}$. Applying the mean value theorem to $f^{\prime }$ shows that $f^{\prime }$ is strictly decreasing. At a local maximum of a differentiable function, the left difference quotients are nonnegative and the right difference quotients are nonpositive; existence of the derivative forces $f^{\prime }=0$. A strictly decreasing function has at most one zero. ThereforeIf that zero exists, $f^{\prime }$ is positive to its left and negative to its right, so the local maximum is also a global maximum.

For the uniform bound by $K<0$, defineIts derivative is $h^{\prime }(x)=f^{\prime \prime }(x)-K<0$, so $h$ is strictly decreasing by the mean value theorem. Comparing with $h(0)=f^{\prime }(0)$ givesBecause $K<0$, this makes $f^{\prime }$ negative at sufficiently large positive $x$ and positive at sufficiently large negative $x$. The function $f^{\prime }$ is continuous, since it is differentiable; the intermediate value theorem supplies a zero $c$ between such points. Its strict decrease makes this zero unique. Again using the mean value theorem for $f$, the derivative signs imply that $f$ increases up to $c$ and decreases after $c$:This illustrates uniform negative curvature and global maximization.

Merely requiring $f^{\prime \prime }<0$ does not force existence. TakeThen $f^{\prime }(x)=e^{-x}>0$ and $f^{\prime \prime }(x)=-e^{-x}<0$, but $f$ is strictly increasing, approaches $0$ as $x\to +\infty$, and never equals $0$. The answer to the final question is no, even for a function bounded above:The stronger uniform curvature assumption prevents this behaviour. Here $f^{\prime \prime }(x)\to 0$ at the end where the supremum is approached, so no uniform negative bound of that kind is available.

For a nonempty set $S\subseteq \mathbb{R}$, an upper bound in a partially ordered set is a real number $u$ such that $s\le u$ for every $s\in S$. Its supremum, or least upper bound, is an upper bound $L$ with $L\le u$ for every upper bound $u$. The least-upper-bound property asserts that every nonempty set of real numbers bounded above has a real supremum.

Let $(x_n)$ be a bounded, increasing sequence; here increasing allows $x_{n+1}=x_n$. Apply the least-upper-bound property to its set of terms and put $L=\sup \{x_n:n\ge 1\}$. Given $\epsilon >0$, the number $L-\epsilon$ cannot be an upper bound, so some $x_N$ satisfies $x_N>L-\epsilon$. The monotone sequence property then givesHence $|x_n-L|<\epsilon$ for all $n\ge N$. This proves the monotone bounded sequence result, with limit $\boxed{x_n⟶L}$.

For the two series, define their partial sums $A_N={\sum }_{n=1}^N{a}_n$ and $B_N={\sum }_{n=1}^N{b}_n$. Positivity makes both partial-sum sequences increasing, and termwise comparison gives $0<B_N\le A_N$. Since $(A_N)$ is a convergent sequence, it is bounded above, and therefore so is $(B_N)$. Applying the monotone bounded sequence result already proved shows that $(B_N)$ converges. By the definition of a convergent series, ${\sum }_{n=1}^{\infty }{b}_n$ converges. This is the comparison test for series for positive terms.

The sequence has uniform convergence on $X$ if there is a function $f:X\to \mathbb{R}$ such thatThe same $N$ must work for every $x$. Equivalently, the supremum of the error over $X$ tends to zero. In pointwise convergence, $N$ is allowed to depend on $x$.

For each fixed $0<x<1$, $x^n\to 0$, so the pointwise limit is $f(x)=1/(1-x)$. Howeverfor every fixed $n$. Hence the error has infinite supremum on $(0,1)$ and the convergence is not uniform. This does not contradict uniform convergence on any smaller closed interval $[0,r]$, $r<1$, where the error is at most $r^n/(1-r)$.

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.

If a nonempty subset $S\subset \mathbb{R}$ is not an interval, there are $a,b\in S$ and $c\notin S$ with $a<c<b$. The disjoint nonempty relatively open sets $S\cap (-\infty ,c)$ and $S\cap (c,\infty )$ form a separation, so $S$ is not connected.

Conversely, suppose an interval $I$ had a separation into relatively open nonempty sets $A,B$. Choose $a\in A,b\in B$, interchanging the names if necessary so $a<b$. Put $c=\sup (A\cap [a,b])$. Because $B$ contains a relative neighbourhood of $b$, $c<b$; because $A$ contains a relative neighbourhood of $a$, $c>a$. Also $c\in [a,b]\subset I$. If $c\in A$, relative openness supplies points of $A$ just above $c$, contradicting the supremum. If $c\in B$, a relative neighbourhood of $c$ excludes points of $A$ approaching $c$ from below, again contradicting the supremum. Thus every interval is connected, including a singleton and unbounded or open intervals.

TakeIt is closed, because its only accumulation point is $0$ and that point is included, and bounded. The Heine-Borel theorem makes it compact. The connected components of its complement are preciselyEach is an interval and therefore connected; none can be enlarged inside the complement across one of the intervening points of $K$.