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For a compact H-hull and a Borel subset of the intrinsic boundary of a simply connected domain, hydrodynamic normalization at infinity and the Poisson kernel for the upper half-plane give this limit as $y\to \infty$ with $x/y\to 0$. After multiplying the kernel by $\pi y$, it tends pointwise to one and is eventually uniformly bounded. Dominated convergence proves the finite-measure case and Fatou's lemma proves the infinite-measure case.

For a simple predictable process, define its stochastic integral byCommon refinement of the deterministic partitions makes this definition independent of the representation. Each summand is a martingale increment multiplied by information already available at its left endpoint, so the resulting process is a continuous square-integrable martingale starting at zero. Its terminal value is the same sum with $t$ replaced by $t_n$.

Write ${\Delta }_{k}M=M_{t_{k+1}}-M_{t_k}$. If $j<k$, then $Z_{t_j}{\Delta }_{j}M{Z}_{t_k}$ is ${\mathcal{F}}_{t_k}$-measurable. By the conditional expectation identity for increments of a martingale,The conditional bracket isometry for stopped martingale increments givesFor completeness, this identity follows by applying the optional stopping theorem to $M$ and $M^2-[M]$ and expanding the square. The latter process is a true uniformly integrable martingale here: the Doob L2 maximal inequality bounds $\mathbb{E}{\sup }_t|M_t{|}^2$, stopped square identities and Fatou's lemma give $\mathbb{E}[M{]}_{\infty }<\infty$, and its absolute supremum is bounded by ${\sup }_t|M_t{|}^2+[M{]}_{\infty }$. Consequently all the stopped identities pass to the limit.

Multiplying the conditional identity by $Z_{t_k}^2$, taking expectations, and using the vanished cross terms proves the Itô isometry:

To extend the stochastic integral, use the finite quadratic-variation measure ${\nu }_M(C)=\mathbb{E}\int 1_{C}d[M]$ on the predictable sigma-algebra. The density of simple predictable processes for finite measures gives approximations to every $H\in L^2({\nu }_M)$. The Itô isometry makes the corresponding terminal integrals Cauchy in $L^2$, and the Doob L2 maximal inequality makes the whole processes Cauchy in expected squared supremum. Their limit defines a continuous martingale, and the isometry persists. For a locally bounded predictable process, choose increasing stopping times tending to infinity on which it is bounded, also restrict the time horizon, and perform this construction after stopping. The isometry shows that the stopped definitions agree on overlaps, so they patch into a uniquely defined continuous local martingale $H\cdot M$.

First stop the solution on leaving $[-n,n]$, at ${\tau }_n$, and set $F_n(t)=\mathbb{E}{\sup }_{u\le t}|X_{u\wedge {\tau }_n}{|}^2$. The stochastic integral representation, the Doob L2 maximal inequality and the Cauchy-Schwarz inequality give, for $0\le t\le 1$,Here stopping ensures square integrability before the calculation, and $8{\sigma }^2+2t{b}^2\le 8({\sigma }^2+b^2)$ explains the constant. The Gronwall inequality gives $F_n(t)\le e^{8At}-1$. For globally Lipschitz functions as coefficients, the global existence theorem for stochastic differential equations with Lipschitz coefficients supplies a nonexplosive solution. Letting $n$ increase and using Fatou's lemma proves the slightly stronger maximal second-moment bound under linear growth:

For coefficients that are only locally Lipschitz functions, let ${\pi }_n(x)=\max (-n,\min (x,n))$ and replace the coefficients by $\sigma ({\pi }_n(x))$ and $b({\pi }_n(x))$. These are globally Lipschitz functions, agree with the originals on $[-n,n]$, and obey the same linear growth condition for an SDE, since $|{\pi }_n(x)|\le |x|$. The globally defined solutions agree until their common exits from $[-n,n]$ by pathwise uniqueness. The uniform preceding estimate and Markov's inequality implyPatch the solutions before their increasing exit times. Their limiting lifetime exceeds $1$ almost surely by this bound. Thus a pathwise unique strong solution exists throughout $[0,1]$ even for locally Lipschitz coefficients.

Write $Z_t=⟨{\kappa }_t⟩$, where the angle brackets average only over walk paths, and use $\mathbb{E}$ only for the environment. Every nearest-neighbor path of length $t$ has probability $(2d{)}^{-t}$, so the finite path sum is exactly the walk expectation. It depends only on environment layers through time $t$, and hence is ${\mathcal{F}}_t$-measurable. Its factors lie in $[1-\epsilon ,1+\epsilon ]$, so $Z_t$ is positive and integrable at each fixed time.

For any fixed path through time $t+1$, the environment variable $h(t+1,{\xi }_{t+1})$ is independent of ${\mathcal{F}}_t$ and has mean zero. Conditional expectation therefore removes its last factor. Summing over the $2d$ possible last steps of each length-$t$ path cancels the extra factor $2d$ in the path probability. ThusThis is the positive random-environment path-weight martingale. It is nonnegative and bounded in $L^1$, so the martingale convergence theorem gives a finite limit $\zeta \ge 0$ almost surely. At this stage one can conclude $\mathbb{E}\zeta \le 1$ by Fatou's lemma, but not yet equality; the second-moment bound in part (b) will prevent loss of mean.

Use the natural filtration ${\mathcal{G}}_n=\sigma ({\xi }_1,\dots ,{\xi }_n)$. Define the deterministic tail $r_n={\sum }_{j=n}^{\infty }{\delta }_j$ and put $Z_n={\xi }_n+r_n$. This is nonnegative and integrable. Since $r_n={\delta }_n+r_{n+1}$,Thus the deterministic summable drift correction produces a nonnegative supermartingale. The almost sure supermartingale convergence theorem gives a finite limit $Z_{\infty }$ almost surely. Its finiteness is also reflected in the bound $\mathbb{E}{Z}_n\le \mathbb{E}{Z}_1$ and Fatou's lemma, which gives $\mathbb{E}{Z}_{\infty }\le \mathbb{E}{Z}_1<\infty$. The convergence theorem follows from the upcrossing argument: bounded expected negative parts of $-Z$ rule out infinitely many crossings of rational intervals, and the expectation bound excludes an infinite limit.

Since $r_n\to 0$, the required conclusion is $\boxed{{\xi }_n⟶Z_{\infty }\text{ almost surely, with }{Z}_{\infty }<\infty }$. The original sequence need not itself be a supermartingale; subtracting the vanishing deterministic tail recovers it from one that is.

The almost sure submartingale convergence theorem states that a discrete-time submartingale $Y_n$ satisfying ${\sup }_n\mathbb{E}{Y}_n^{+}<\infty$ converges almost surely to a finite integrable limit. In particular, a martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$ converges almost surely to an integrable $M_{\infty }$, and $\mathbb{E}|M_{\infty }|\le {\sup }_n\mathbb{E}|M_n|$ by Fatou's lemma. This is the martingale convergence theorem. Uniform integrability gives the stronger conclusion of L1 convergence; it is not implied by bounded first moments alone.

Completeness means that every norm-Cauchy sequence has a limit belonging to the same Lp space, with $\Vert f_n-f{\Vert }_p\to 0$. To prove it, choose a subsequence $f_{n_j}$ with $\Vert f_{n_{j+1}}-f_{n_j}{\Vert }_p\le 2^{-j}$. The functionshave $p$-norm bounded by $\Vert f_{n_1}{\Vert }_p+1$, by Minkowski inequality. They increase to $H$, and Fatou's lemma gives $\int H^p\le (\Vert f_{n_1}{\Vert }_p+1{)}^p$. Thus $H$ is finite almost everywhere, so the subsequence telescopes absolutely there to a measurable limit $f$ with $|f|\le H$ and $f\in L^p$.

The same argument on the tails gives $\Vert f-f_{n_j}{\Vert }_p\le {\sum }_{k\ge j}{2}^{-k}\to 0$. Given $ϵ$, choose such a subsequence term beyond the Cauchy threshold and use $\Vert f_n-f{\Vert }_p\le \Vert f_n-f_{n_j}{\Vert }_p+\Vert f_{n_j}-f{\Vert }_p$. This proves convergence of the full sequence and completeness for every $1\le p<\infty$, on an arbitrary measure space.

For independent observations in a natural exponential family, the log likelihood isIn a minimal exponential family the Observed Fisher information matrix is positive definite, so the log likelihood is strictly concave. Any finite solution of the likelihood equation $\nabla \kappa (\hat{\theta })=\bar{Y}$ is therefore the unique global maximum likelihood estimator. Existence is a separate issue: the observed sufficient-statistic mean must belong to the image of the mean parameter map $\nabla \kappa$.

For a full regular minimal natural exponential family this image is the interior of the convex support of an exponential family, $C=\overline{\mathrm{conv}(\mathrm{supp}(h\nu ))}$. The necessity follows from a supporting hyperplane argument. If a finite parameter had its mean on a boundary face, some nonzero $v$ would satisfy $v^{T}Y\le b$ almost surely and $\mathbb{E}(v^{T}Y)=b$. Equality of the expectation forces $v^{T}Y=b$ almost surely. Exponential tilting preserves the base-measure null sets, so this contradicts minimality.

For sufficiency, let $\bar{Y}$ be an interior point of $C$. Choose finitely many points $y_1,\dots ,y_m$ of the support whose convex hull contains a ball about $\bar{Y}$, say of radius $2\epsilon$. Such points exist because $\bar{Y}$ is interior to the closed convex hull; one may approximate the vertices of a small simplex surrounding it by finite convex combinations of support points. Take support neighbourhoods of radius at most $\epsilon$ with positive finite base-measure masses $a_j$. These masses can be taken finite by restricting to bounded neighbourhoods and using finiteness of the normalizing integral at any fixed interior parameter. For every vector $\theta$, at least one $j$ has ${\theta }^T(y_j-\bar{Y})\ge 2\epsilon \Vert \theta \Vert$. Integrating over that neighbourhood yieldsThus the negative log likelihood grows to infinity as $\Vert \theta \Vert \to \infty$. At a finite boundary point of the full open natural parameter space, the normalizing integral is infinite, since otherwise that point would itself belong to the full domain. Fatou's lemma makes $\kappa (\theta )$ diverge on approach to such a boundary. A minimizing sequence for $\kappa (\theta )-{\theta }^T\bar{Y}$ therefore stays in a bounded subset away from the boundary; it has an interior limit where the minimum is attained. Its gradient vanishes, giving the required likelihood equation. Strict convexity gives uniqueness.

Hence, under these full, regular and minimal conventions,This is maximum likelihood existence in a full regular natural exponential family. Regularity alone does not make every sample admissible: in a Bernoulli distribution an all-zero or all-one sample has its sample mean on the boundary of $C=[0,1]$, and the canonical maximum is approached only as $\theta \to -\infty$ or $+\infty$. The usual probability parameter then has a boundary maximum in an enlarged model, but no finite canonical maximizer. A regular restriction of the natural parameter space requires membership in its actual mean-map image, rather than merely in $\mathrm{int}C$. Without minimality, affine redundancies may prevent uniqueness of the canonical parameter even when the fitted probability distribution is unique.

Fatou's lemma states that nonnegative measurable functions satisfyTo prove it, put $h_n={\inf }_{k\ge n}{f}_k$. These measurable nonnegative functions increase to $\mathrm{liminf}{f}_n$. By the monotone convergence theorem, $\int \mathrm{liminf}{f}_n={\lim }_n\int h_n$. Since $h_n\le f_k$ for every $k\ge n$, one has $\int h_n\le {\inf }_{k\ge n}\int f_k$. Taking the limit proves the inequality, including extended infinite integrals.

Relative to a filtration $({\mathcal{F}}_t)$, a martingale is an adapted integrable process satisfying $\mathbb{E}[M_t\mid {\mathcal{F}}_s]=M_s$ for $s\le t$. The martingale convergence theorem gives an almost-sure finite limit for a nonnegative martingale, with $\mathbb{E}[M_{\infty }]\le \mathbb{E}[M_0]$ by Fatou's lemma. In continuous time take the usual right-continuous version and filtration hypotheses. Convergence in $L^1$, and preservation of expectation at infinity, additionally require uniform integrability.

For $a<b$, let $U_N[a,b]$ be the number of completed upcrossings by the observations $X_1,\dots ,X_N$: buy when the value is at most $a$, sell at the first subsequent value at least $b$, and repeat. With $z^{-}=\max \{-z,0\}$, one form of the Doob upcrossing inequality for a martingale isFor completeness, hold either zero or one unit according to this rule. The holding at each step is measurable before the next martingale increment, so the total gain $G_N$ has expected value zero. Completed trades earn at least $b-a$ each; an unfinished trade loses at most $(X_N-a{)}^{-}$. Thus $G_N\ge (b-a)U_N[a,b]-(X_N-a{)}^{-}$, proving the inequality.

Suppose $C={\sup }_n\mathbb{E}|X_n|<\infty$. For each rational pair $a<b$ the inequality and monotone convergence giveEvery rational interval therefore has only finitely many upcrossings, simultaneously outside a countable union of null sets. If $\mathrm{liminf}{X}_n<\mathrm{limsup}{X}_n$, there is a rational interval strictly between them and infinitely many upcrossings, a contradiction. Thus the limit $X_{\infty }$ exists in the extended real line almost surely. Fatou's lemma now givesThe limit is finite almost surely and integrable. This proves the martingale convergence theorem rather than using it as the deduction.

The exact extra condition for convergence in L1 is uniform integrability:To see sufficiency, first discard the tails of $X_n$ uniformly and the tail of the integrable $X_{\infty }$. On the set where both absolute values are at most $K$, their difference is bounded by $2K$ and tends to zero almost surely; dominated convergence finishes the proof. Conversely, if $X_n\to X_{\infty }$ in $L^1$, thenMake the first term uniformly small for all sufficiently large $n$, then control the finitely many remaining integrable variables by increasing $K$. Hence uniform integrability is necessary too. In this case the conditional expectation identity $X_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ follows by taking the $L^1$ limit in $X_n=\mathbb{E}[X_m\mid {\mathcal{F}}_n]$ for $m\ge n$.

The coin-doubling martingale shows why an $L^1$ bound is not enough. Let ${\xi }_1,{\xi }_2,\dots$ be independent fair coin indicators and setConditional on the first $n$ coins, the next value is either $2X_n$ or zero with equal probabilities, so this is a nonnegative martingale. Its expected value is always $1$, while a first failure occurs almost surely, giving $X_n\to 0$ almost surely. Yet $\mathbb{E}|X_n-0|=1$ for every $n$. Indeed, whenever $2^n>K$, its tail expected value above $K$ is still $1$. Thus convergence in L1 fails because uniform integrability fails.

Write $A_n=\int f_n^{+}$, $B_n=\int f_n^{-}$, $A=\int f^{+}$ and $B=\int f^{-}$. Almost-everywhere convergence gives $f_n^{\pm }\to f^{\pm }$, so Fatou's lemma gives $A\le \mathrm{liminf}{A}_n$ and $B\le \mathrm{liminf}{B}_n$. The given norm condition is $A_n+B_n\to A+B$. Consequentlyand the symmetric argument bounds $\mathrm{limsup}{B}_n\le B$. Combining upper and lower limits provesNo common integrable dominating function for $f_n$ has been assumed.

Fatou's lemma states that for nonnegative measurable $h_n$,Indeed $g_n={\inf }_{k\ge n}{h}_k$ is measurable, increases to $\mathrm{liminf}{h}_n$, and satisfies $g_n\le h_k$ for every $k\ge n$. The monotone convergence theorem gives $\int \mathrm{liminf}{h}_n={\lim }_n\int g_n\le {\lim }_n{\inf }_{k\ge n}\int h_k$, proving the claim, including infinite values.

The dominated convergence theorem states that if measurable $h_n\to h$ almost everywhere and $|h_n|\le g$ almost everywhere for one integrable nonnegative $g$, then $h$ is integrable and $\int h_n\to \int h$; in fact $\int |h_n-h|\to 0$. Integrability follows from $|h|\le g$. For real functions, Fatou applied to $g+h_n$ gives $\int h\le \mathrm{liminf}\int h_n$, and applied to $g-h_n$ gives the reverse inequality against the upper limit. Thus the integrals converge. To obtain the stronger conclusion, Fatou applied to $2g-|h_n-h|\ge 0$ gives $2\int g\le 2\int g-\mathrm{limsup}\int |h_n-h|$, forcing the latter limit to be zero. Complex functions may be treated through their real and imaginary parts or the same absolute-difference argument.

Use the intrinsic boundary of a simply connected domain, so different approaches to the two sides of a slit remain distinct. The mapping-out function $g_K$ extends as a homeomorphism from this intrinsic compactification to that of the complex upper half-plane. Set $T=T(H)$. The imaginary coordinate of planar Brownian motion hits zero in finite time almost surely, and $T$ is no larger than that time. Thus $T<\infty$ and continuity gives a finite Euclidean exit point.

By conformal invariance of planar Brownian motion, $g_K(B_t)$ is a planar Brownian motion in the complex upper half-plane, run with clockThe terminal clock cannot be infinite: that would make the transformed Brownian motion stay in the upper half-plane forever. It cannot stop while the transformed path is in the interior either, since continuity of $g_K^{-1}$ would then put the original exit point inside $H$. Hence the terminal clock is precisely the transformed Brownian exit time. The transformed path converges to a real boundary point, and applying the extended inverse proves almost sure convergence to a point of the intrinsic boundary.

Write $g_K(x+iy)=u+iv$, and let $E=g_K(S)\cap \mathbb{R}$. The point at infinity has zero harmonic measure. Conformal invariance of planar Brownian motion and the Poisson kernel for the upper half-plane giveThe hydrodynamic normalization at infinity gives $g_K(z)=z+O(1/z)$, so along the specified approach $v/y\to 1$ and $u/y\to 0$. For each fixed real $t$,If $\mathrm{Leb}(E)<\infty$, dominated convergence applies because $y/v$ is eventually bounded. If $\mathrm{Leb}(E)=\infty$, Fatou's lemma makes the limit infinite. This proves the harmonic-measure asymptotic at infinitywith the equality understood in the extended nonnegative reals.

For $t\in \mathbb{R}$, define the set functionalThe ROF level-set formulation states that the unique minimizer of $J(u)=\Vert u-f{\Vert }_2^2/2+\alpha \mathrm{TV}(u)$ has superlevel sets minimizing $F_t$ for almost every $t$. Conversely, an admissible $u$ with minimizing superlevel sets is the ROF minimizer. Thus the concise characterization is

The signed layer-cake identity for quadratic fidelity and the coarea formula for BV functions giveIndeed $u^2/2-fu={\int }_{\mathbb{R}}(t-f)({\chi }_{\{u>t\}}-{\chi }_{\{0>t\}})dt$ pointwise. Its absolute integral is bounded by $u^2/2+|fu|$, so Fubini's theorem applies for $u,f\in L^2$. The negative-level baseline is essential; integrating the unadjusted $F_t$ would generally diverge.

Each $F_t$ attains its minimum. A minimizing sequence of indicator functions has bounded $L^1$ norm, and comparison with the empty set bounds its perimeter by an $L^1$ forcing bound. Bounded-variation compactness supplies a limiting indicator function. The forcing integral converges by dominated convergence, while relative perimeter is sequentially lower semicontinuous.

For $s<t$, the supplied comparison lemma applied to $(f-t)/\alpha <(f-s)/\alpha$ makes every selected minimizer $A_t$ contained in $A_s$ up to a null set. The comparison of perimeter minimizers with ordered forcing also follows directly: compare $A_t$ with $A_t\cap A_s$, compare $A_s$ with $A_t\cup A_s$, add and use submodularity of relative perimeter to obtain $(t-s)|A_t\setminus A_s|\le 0$.

Select minimizers at rational levels, remove their countably many exceptional null sets, and reconstruct $v(x)=\sup \{q\in \mathbb{Q}:x\in A_q\}$. The strict superlevel set is $\{v>t\}={\bigcup }_{q>t}{A}_q$. This union also minimizes $F_t$: take $q\downarrow t$, use monotone $L^1$ convergence of its indicator functions and sequential lower semicontinuity, and note that $m(t)={\min }_E{F}_t(E)$ satisfies $|m(t)-m(s)|\le |\Omega ||t-s|$.

To justify finite energy before assuming it, clip this reconstruction to $v_M\in [-M,M]$. Comparison with the empty set gives $\alpha \mathrm{Per}(E_t;\Omega )\le {\int }_{\Omega }|t-f|$ uniformly on a bounded interval of levels. Since $v_M=-M+{\int }_{-M}^M{\chi }_{E_t}dt$, pairing with compactly supported test-field divergences bounds its total variation seminorm by ${\int }_{-M}^M\mathrm{Per}(E_t;\Omega )dt<\infty$. Thus $v_M$ belongs to the BV space before applying the coarea formula for BV functions. The layer-cake argument on the finite interval $[-M,M]$ shows $J(v_M)\le J(T_{M}w)$ for every finite-energy competitor $w$, where $T_M$ is clipping. In particular $w=0$ bounds the $L^2$ norms and variations uniformly. Fatou's lemma excludes infinite values of $v$ on a positive-measure set. Bounded-variation compactness and weak convergence in $L^2$ identify an admissible limit $v$. For any fixed competitor, $T_{M}w\to w$ in $L^2$ and its variation tends to that of $w$, by bounded-variation contraction under clipping and sequential lower semicontinuity. Thus $J(v)\le J(w)$.

The quadratic fidelity is strictly convex, so every ROF minimizer equals $v$ almost everywhere and has the selected minimizing superlevel sets. Conversely, for any $u$ whose levels minimize $F_t$, integrate the levelwise inequality against any competitor's levels in the displayed identity to obtain $J(u)\le J(w)$. This proves both directions for signed as well as nonnegative data.

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.

The weak topology of probability measures is the smallest topology making every map $\mu \mapsto \int hd\mu$ continuous for bounded continuous real $h$ on $X$. Define the bounded-Lipschitz metric byUsing the maximum of the two bounds instead of their sum changes the metric by at most a factor of two and gives the same topology. Symmetry and the triangle inequality follow from the supremum formula. Bounded Lipschitz functions distinguish probability measures by the open-set approximations below, so $\beta$ is a metric. On a separable metric space, $\beta$ induces the weak topology.

Here are the essential details. If $\beta ({\mu }_n,\mu )\to 0$, integrals of every bounded Lipschitz function converge after rescaling its Lipschitz bound. For an open set $A\ne X$, the functionsincrease to $1_A$; for $A=X$ take $h_m=1$. Thus $\mu (A)={\sup }_m\int h_{m}d\mu$. It follows both that $\mu \mapsto \mu (A)$ is lower semicontinuous for the weak topology, as a supremum of continuous maps, and that beta convergence impliesThis is the open-set Portmanteau criterion. Its closed-set equivalent, by complements, isEither family of inequalities is necessary and sufficient for weak convergence of probability measures. For sufficiency of the open inequalities, apply the layer-cake formula and Fatou's lemma to the open superlevel sets of a continuous function $0\le h\le 1$, obtaining $\mathrm{liminf}\int hd{\mu }_n\ge \int hd\mu$. Applying the same argument to $1-h$ gives the reverse bound. Scaling handles every bounded continuous function.

Conversely, weak convergence implies beta convergence. Choose a compact set $K$ of arbitrarily large $\mu$-mass, using tightness of a probability measure on a Polish space. The unit bounded-Lipschitz class is uniformly bounded and equicontinuous on $K$, so finitely many of its members approximate all others uniformly there. On the open $\delta$-neighborhood $K^{\delta }$, approximation errors increase by at most $2\delta$. The open-set inequality makes ${\mu }_n(K^{\delta })$ large for all sufficiently large $n$. Weak convergence for the finitely many selected functions, combined with these approximation errors and the small mass outside $K^{\delta }$, bounds the supremum defining $\beta$ by an arbitrarily small number. The same finite-test and open-mass bounds define weak neighborhoods, so the argument gives equality of the topologies, not only their convergent sequences. It does not first assume uniform tightness of the entire sequence.

The map $j=i_{\ast }$ is injective and continuous: for $h\in C(\tilde{X})$, $h\circ i$ is bounded continuous on $X$, and $\int hdj(\mu )=\int h\circ id\mu$. To prove inverse continuity on its image, suppose $j({\mu }_n)\to j(\mu )$. For every open $A\subset X$, there is an open $U\subset \tilde{X}$ with $i(A)=U\cap i(X)$. Hence ${\mu }_n(A)=j({\mu }_n)(U)$ and $\mu (A)=j(\mu )(U)$. The open-set criterion on $\tilde{X}$ now gives the criterion on $X$, so ${\mu }_n\to \mu$. Both spaces are metrizable, and this sequential argument provesThis is the probability pushforward embedding theorem. The printed claim needs the words “onto its image”: it is generally not onto all $P(\tilde{X})$, since a Dirac mass at a point outside $i(X)$ is not in its image. By Question 1, the completely metrizable subspace $i(X)$ is a Borel G-delta set, and

Now let $S$ be weakly closed and uniformly tight. Take any sequence $({\mu }_n)$ in $S$. The space $P(\tilde{X})$ is compact for the weak topology. For completeness, this compactness follows by choosing a countable uniformly dense family in $C(\tilde{X})$, extracting a diagonal subsequence of its bounded integrals, and extending the limits to a positive normalized functional on $C(\tilde{X})$; the Riesz-Markov-Kakutani representation theorem gives the limiting probability measure. Thus, along a subsequence, $j({\mu }_n)\to \lambda$.

For every $m$, choose compact $K_m\subset X$ with $\mu (K_m)\ge 1-2^{-m}$ for every $\mu \in S$. The set $i(K_m)$ is compact and closed in $\tilde{X}$, so the closed-set criterion givesTherefore $\lambda (i(X))=1$, and $\lambda =j(\mu )$ for some $\mu \in P(X)$. Inverse continuity gives ${\mu }_n\to \mu$, and closedness puts $\mu \in S$. Hence $S$ is sequentially compact, and metrizability makes it compact. This proves the required closed uniformly tight compactness criterion, rather than assuming it from Prokhorov's theorem.

Such a compact metrizable extension is always available: for a countable dense set $(x_k)$, the coordinates $d(x,x_k)/(1+d(x,x_k))$ embed $X$ homeomorphically into $[0,1{]}^{\mathbb{N}}$. Injectivity and inverse continuity follow by choosing $x_k$ close to a specified point and using the triangle inequality. The closure of the image gives $\tilde{X}$.

To show separability of $P(X)$, use finite sums of Dirac measures on a countable dense subset of $X$ with nonnegative rational weights summing to one. This is a countable family. Given $\mu$, cover a large-mass compact set by finitely many small balls centered in that subset, move the mass in each piece to its center, and move the remaining small mass to one fixed center. The bounded-Lipschitz error is at most the ball radius plus twice the exceptional mass. Approximate the resulting weights by rational weights. Thus these atomic measures are beta-dense.

Finally use a complete compatible metric $d$ on the Polish space $X$, as required for the granted total-boundedness-to-tightness assertion. A beta-Cauchy sequence is beta-totally bounded, and so is its closure in $P(X)$. By the allowed assertion this closure is uniformly tight; it is weakly closed because the weak and beta topologies agree. The compactness result just proved gives a convergent subsequence, and the Cauchy property forces the whole sequence to converge in beta. Consequently beta is complete for this choice of $d$, andIf the originally displayed compatible metric is incomplete, replace it by a complete compatible metric for this last argument; the weak topology itself is unchanged.