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The almost sure submartingale convergence theorem says that if a submartingale $(Y_n)$ satisfiesthen there is an integrable random variable $Y_{\infty }$ such thatIndeed the positive-part bound also bounds the negative parts, since $\mathbb{E}{Y}_n\ge \mathbb{E}{Y}_0$; the theorem gives a finite limit. In particular, the Martingale convergence theorem applies to any martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$.

For a nonnegative martingale, $\mathbb{E}{M}_n=\mathbb{E}{M}_0$ supplies this bound automatically. Its limit obeys $\mathbb{E}{M}_{\infty }\le \mathbb{E}{M}_0$ by the Fatou lemma. Almost sure convergence alone need not preserve the expectation or give convergence in L1. Uniform integrability is the additional condition that supplies convergence in L1 and the identity $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$.

For the divergence-form elliptic operator use the weak subsolution convention $\int a^{ij}{D}_{j}u{D}_i\phi \le 0$ for nonnegative test functions. Set $u_{\epsilon }=u+\epsilon$ and test with $\phi ={\eta }^2{u}_{\epsilon }^{\beta -1}$, where $\beta >1$. The Sobolev chain rule makes this test admissible; boundedness and the positive regularization control the derivative of its power. Ellipticity and the Cauchy-Schwarz inequality give, with $X=\int |Du{|}^2{u}_{\epsilon }^{\beta -2}{\eta }^2$ and $Y=\int u_{\epsilon }^{\beta }|D\eta {|}^2$,Division when $X>0$, and the trivial case $X=0$, establish the power Caccioppoli inequality. Letting $\epsilon \downarrow 0$ givesFor $1<\beta <2$ this limit uses the Fatou lemma; the gradient of a Sobolev function vanishes on a level set, so the weighted integrand is assigned zero on $\{u=0\}$. The same regularization justifies differentiating $u^{\beta /2}\eta$ in $W^{1,2}$.

The PDF bounds individual entries by $\Lambda$. Then ${\Lambda }_{\ast }\le n\Lambda$: dimensional factors are included in the ellipticity constants when the first estimate is written as $C(\lambda ,\Lambda )$. With literal entrywise bounds, the explicit valid constant displayed here is $4n^2{\Lambda }^2/{\lambda }^2$. If $\Lambda$ bounds the operator norm instead, it is $4{\Lambda }^2/{\lambda }^2$. This distinction does not alter any later conclusion, whose constants explicitly depend on $n$.

For $n\ge 3$, put $\sigma =n/(n-2)$. Apply the Sobolev inequality to $g=u^{\beta /2}\eta$ and useFor $\beta \ge p>1$, $\beta /(\beta -1)\le p/(p-1)$, so the resulting constant is uniform in $\beta$. Taking a cutoff equal to one on $B_r$, supported in $B_R$, with $|D\eta |\le C/(R-r)$, yieldsAt $R=1$ a compactly supported cutoff can still be chosen with this bound by leaving an outer margin; alternatively pass to the limit from smaller radii.

For Moser iteration, set ${\beta }_j=p{\sigma }^j$ and $r_j=1/2+2^{-j-1}$. Repeated use of the preceding estimate bounds the successive norms byThis product is finite becauseAs the exponents tend to infinity, their norms on the Euclidean ball of radius $1/2$ tend to its essential supremum. ThereforeThe power $\beta$ is independent of the exponents of Hölder continuity in earlier questions.

The Martingale convergence theorem in its $L^1$-bounded form says that a martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$ has an integrable limit $M_{\infty }$ andThe norm bound on the limit follows from Fatou lemma. Boundedness in $L^1$ by itself does not imply convergence in L1. The uniformly integrable martingale convergence theorem gives the stronger conclusion: if $(M_n)$ is uniformly integrable, then $M_n\to M_{\infty }$ both almost surely and in L1 norm, and $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$. Conversely, convergence in L1 implies uniform integrability.

For the distinction, the fair-coin doubling martingale $M_n=2^n{1}_{\{\text{first }n\text{ tosses are heads}\}}$ has expectation one for every $n$ but converges almost surely to zero. Its L1 norm remains one, so its convergence is not in L1 norm. These two formulations specify exactly which hypothesis is needed in part (d).

Put $S=H\cdot M$ and $A_t={\int }_0^t{H}_s^{2}d[M{]}_s$. The candidate $A$ is continuous, adapted, nondecreasing, starts at zero and has integrable terminal value. We prove $S^2-A$ is a martingale, rather than assuming the desired quadratic variation of a stochastic integral.

The assumed boundedness of $M$ also makes $M^2-[M]$ a true martingale on every finite horizon. Indeed, apply its local martingale identity at a localizing sequence; bounded stopped squares and the Fatou lemma show that $[M{]}_t$ is integrable. The stopped expressions are then dominated by ${\sup }_s|M_s{|}^2+[M{]}_t$, so the dominated convergence theorem gives the unrestricted conditional identity on that horizon.

First let $H$ be a bounded simple predictable process. On each interval with coefficient $h$, an increment of $S$ is $h(M_t-M_a)$. The defining quadratic variation identity for $M$, together with the martingale property of $M$, makesa martingale for $t\ge a$. Expanding $S_t^2-S_a^2$ gives a term $2S_{a}h(M_t-M_a)$ and the square term above; their conditional expectations yield the desired identity on that interval. Integrability follows because the variables are square-integrable; concatenating the finitely many intervals gives $S^2-A$ a martingale.

For general bounded predictable $H$, choose bounded simple predictable processes $H^n\to H$ in $L^2(M)$, which is possible since they generate the predictable sigma-algebra. Set $S^n=H^n\cdot M$ and $A_t^n={\int }_0^t(H_s^n{)}^{2}d[M{]}_s$. The Itô isometry and Doob L2 maximal inequality imply $S_t^n\to S_t$ in $L^2$; the Cauchy-Schwarz inequality gives $(S_t^n{)}^2\to S_t^2$ in $L^1$ andPass the conditional expectation identity for $(S^n{)}^2-A^n$ to the $L^1$ limit. Thus $S^2-A$ is a martingale. Uniqueness in the defining quadratic variation property now gives

For $1<p<\infty$, apply Fatou lemma to the nonnegative sequence $|f_n{|}^p$, which converges pointwise to $|f{|}^p$:For $p=\infty$, each inequality $\Vert f_n{\Vert }_{\infty }\le M$ fails pointwise only on a null set. Outside the union of these countably many null sets, $|f_n(x)|\le M$ for every $n$, so the pointwise limit also satisfies $|f(x)|\le M$. Therefore in both cases

The Martingale convergence theorem says that if a discrete-time martingale $(M_n)$ satisfies $C={\sup }_n\mathbb{E}|M_n|<\infty$, then there is an integrable random variable $M_{\infty }$ such thatThe hypothesis does not by itself ensure convergence in L1.

For each pair of rational numbers $a<b$, the Doob upcrossing inequality gives $\mathbb{E}{U}_N[a,b]\le (C+|a|)/(b-a)$. The monotone convergence theorem therefore gives a finite expected value for $U_{\infty }[a,b]={\lim }_N{U}_N[a,b]$, so this upcrossing count is finite almost surely. There are only countably many such pairs, hence all their upcrossing counts are simultaneously finite outside one event of probability zero.

If ${\mathrm{liminf}}_n{M}_n<{\mathrm{limsup}}_n{M}_n$, the density of the rational numbers supplies $a<b$ strictly between them, forcing infinitely many upcrossings. Consequently $M_n$ has a limit in the extended real numbers almost surely. By the Fatou lemma,A limit of either $+\infty$ or $-\infty$ would make this lower limit infinite, so the limit is finite almost surely, and the same inequality proves its integrability.

Use the standard definition of a local martingale, including an integrable initial value, and let $T_n\uparrow \infty$ be a localizing sequence. If $X\ge 0$, the Fatou lemma gives $\mathbb{E}{X}_t\le {\mathrm{liminf}}_n\mathbb{E}{X}_{t\wedge T_n}=\mathbb{E}{X}_0<\infty$. For $s\le t$, the Conditional Fatou lemma givesFor clarity, the Conditional Fatou lemma itself follows by setting $U_m={\inf }_{n\ge m}{U}_n$ for nonnegative $U_n$, applying the conditional version of the monotone convergence theorem to $U_m\uparrow \mathrm{liminf}{U}_n$, and observing $\mathbb{E}(U_m\mid {\mathcal{F}}_s)\le {\inf }_{n\ge m}\mathbb{E}(U_n\mid {\mathcal{F}}_s)$. Hence a nonnegative local martingale is a supermartingale.

If $|X_t|\le C$ simultaneously for all $t$, with a deterministic $C<\infty$, the dominated convergence theorem applied to the stopped martingale identity gives, for each $A\in {\mathcal{F}}_s$,This is exactly the conditional expectation identity, so a bounded local martingale is a true martingale. The same proof works for deterministic bounds on every finite time interval; merely having a finite random pathwise bound supplies no dominating integrable random variable and is a different convention.

Under the standard local martingale definition, $X_0$ is integrable. For a nonnegative $X$, the Fatou lemma and a localizing sequence giveHence every $X_t$ is integrable, and part (a) applies. Therefore a nonnegative discrete-time local martingale is a true martingale. This is stronger than the continuous-time supermartingale conclusion.

From convergence in probability, choose a subsequence $X_{n_k}\to X$ almost surely. The Fatou lemma then givesso $X\in L^2$.

The bounded second moments make $(X_n)$ uniformly integrable, and $X$ is integrable. Convergence in probability together with uniform integrability therefore givesConvergence in $L^2$ need not follow. On $[0,1]$ with Lebesgue measure, letThen $X_n\to 0$ in probability and $\mathbb{E}|X_n{|}^2=1$, but $\Vert X_n{\Vert }_2=1$ for every $n$. This is the bounded second moments do not upgrade convergence in probability to convergence in L2 example.

At time zero, ${{\rm Y}}_0(z)=\mathrm{Im}z$. Compactness of $K\subset \mathbb{H}$ givesAlso $M_0(z)$ is uniformly bounded above on $K$. Onone has $M_{{\tau }_{ϵ}}\ge {ϵ}^{-p}{2}^{-q}$. Optional stopping, Fatou lemma, and the assumed conditional angular estimate giveThusThe exponent $(8-\kappa )/\kappa$ requested in the question does not follow and is false as written. The SLE Green-function estimate gives probability comparable to ${ϵ}^{1-\kappa /8}$, confirming that the denominator in the requested exponent should be $8$.

We prove the contrapositive. LetThe function $F$ is constant along $N$, so minimize it on $N^{\perp }$. Suppose there is no $H$ with $H\cdot X\ge 0$ almost surely and strict inequality with positive probability. If a sequence $h_k\in N^{\perp }$ satisfies $\Vert h_k\Vert \to \infty$, pass to a subsequence withBecause $H\notin N$ and there is no arbitrage direction, $\mathbb{P}(H\cdot X<0)>0$. On that event, $e^{-h_k\cdot X}\zeta \to \infty$, and Fatou lemma gives ${\mathrm{liminf}}_{k}F(h_k)=\infty$. Hence every finite sublevel set of $F$ in $N^{\perp }$ is bounded. It is also closed, so $F$ attains its infimum there and has a bounded minimizing sequence. This contradicts the assumption. Therefore there is a unit vector $H$ satisfying

For every rational $0\le a<b$, part a implies $U[a,b]<\infty$ almost surely. The intersection of these probability-one events over the countable collection of rational pairs still has probability one. On this event, ifsome rational interval $[a,b]$ lies strictly between them, forcing infinitely many upcrossings, a contradiction. Thus $X_n$ has an extended limit almost surely.

The limit cannot be $+\infty$ on a set of positive probability: Fatou lemma and the supermartingale property giveNonnegativity excludes $-\infty$. Therefore $X_n$ converges almost surely to a finite random variable, proving the almost sure supermartingale convergence theorem in this case.

Fatou lemma states that for any sequence $(f_n)$ of nonnegative measurable functions on a measure space,

For the proof, defineThen $(g_n)$ is a nonnegative increasing sequence andThe monotone convergence theorem therefore givesFor each $k\ge n$, we have $g_n\le f_k$, so monotonicity of the Lebesgue integral yieldsTaking the limit in $n$ proves the asserted inequality and the proof of Fatou lemma.

Yes. Almost-sure convergence givesalmost surely. Since the random variables are nonnegative, Fatou lemma applies and givesThe inequalities remain valid in the extended nonnegative reals, so no integrability assumption beyond nonnegativity is required.

First replace $u$ by $u+\epsilon$ and later let $\epsilon \downarrow 0$. In the supersolution inequality use the nonnegative test functionwhere $\eta$ is compactly supported. Ellipticity and Young inequality give the Logarithmic Caccioppoli inequalityChoose $\eta =1$ on $B_{\rho }$, supported in $B$, with $|D\eta |\le C(1-\rho {)}^{-1}$. Letting $\epsilon \downarrow 0$ and using Fatou lemma gives

Define $Z_n={\sup }_{m\ge n}|X_m-X|$. Then $0\le Z_n\le 2$ and $Z_n\downarrow 0$ almost surely. The dominated convergence theorem gives $\mathbb{E}{Z}_n\to 0$.

Set $W_n=\mathbb{E}[Z_n\mid {\mathcal{F}}_n]$. Since $Z_{n+1}\le Z_n$, the tower property of conditional expectation givesso $(W_n)$ is a nonnegative supermartingale. The almost sure supermartingale convergence theorem gives $W_n\to W_{\infty }$ almost surely, and Fatou lemma yields $\mathbb{E}{W}_{\infty }\le {\mathrm{liminf}}_n\mathbb{E}{W}_n=0$. Hence $W_n\to 0$ almost surely; because $\mathbb{E}{W}_n=\mathbb{E}{Z}_n\to 0$, convergence also holds in $L^1$.

Finally,The second term tends to zero almost surely and in $L^1$ by part b. The first does so by the preceding argument, proving the moving-variable conditional-expectation convergence.

Let $A_n=\{X_m=0\text{ for some }m\ge n\}$. By the Markov property,The events $A_n$ decrease to the event that the walk visits zero infinitely often, which has probability zero by the stated transience assumption. HencePart a and the almost sure supermartingale convergence theorem give an almost-sure limit $V\ge 0$. By Fatou lemma, $\mathbb{E}V\le {\mathrm{liminf}}_n\mathbb{E}[v(X_n)]=0$, so $V=0$ almost surely.

Let $T$ be a measure-preserving transformation of $(X,\mathcal{F},\mu )$ and let $f\in L^1(\mu )$. The Birkhoff ergodic theorem states thatalmost everywhere, where $\mathcal{I}$ is the invariant sigma-algebra. The limit is $T$-invariant and

Assume now that $\mu (X)=1$. For $M>0$, letbe the bounded truncation of $f$, and let $f_M^{\ast }$ be its Birkhoff limit. Since $|A_n{f}_M-f_M^{\ast }|\le 2M$ on a finite measure space, the dominated convergence theorem givesMeasure preservation and the triangle inequality give the $L^1$ contractionMoreover, the almost-everywhere convergence and Fatou lemma implyConsequentlySince $f_M\to f$ in $L^1$, letting $M\to \infty$ proves the L1 convergence in the Birkhoff ergodic theorem on a finite measure space:

If $u=0$ on $\Sigma$ and $|\Sigma |>0$, thenPart (b.i) consequently givesLetting $\epsilon \downarrow 0$ and using Fatou lemma yields ${\int }_{B_1}{u}^{p_0}=0$. Since $u\ge 0$ is continuous,

Assume ${\sup }_n\mathbb{E}|M_n|<\infty$. For every pair of rationals $a<b$, Doob upcrossing inequality givesby monotone convergence theorem. Thus every rational interval is upcrossed only finitely often almost surely. If a real sequence has distinct limit inferior and limit superior, it completes infinitely many upcrossings of some rational interval between them. Hence $M_n$ converges in the extended real line almost surely.

Fatou lemma givesso the limit is finite almost surely and integrable. This is the $L^1$-bounded form of the Martingale convergence theorem.