In steady flow, the supplied Euler equations for an inviscid fluid givesTaking its scalar product with $\mathbf{u}$ gives $\mathbf{u}\cdot \nabla H=0$. Along a streamline with tangent $\mathbf{u}$, this is precisely zero derivative of $H$. Hence $H$ is constant along each streamline, with possibly different constants on different streamlines.

If the flow is also irrotational, $\mathbf{\omega }=0$ and thus $\nabla H=0$. Therefore $H$ has one constant value throughout each connected fluid region. The stronger conclusion uses vanishing of the entire gradient, not just its component along the velocity.

Choose $Z=(A,{\mathcal{T}}_A)$ and let $f$ be the identity set map $A\to A$. The composition $i\circ f$ is the inclusion and is continuous by part (i). The hypothesized universal implication therefore says that the identity mapis continuous. Its preimage of a set $V\in {\mathcal{T}}^{\prime }$ is $V$ itself, so every such $V$ lies in ${\mathcal{T}}_A$. ThereforeCombining this with part (ii), when both properties hold, characterizes the subspace topology uniquely.

Every open set of the subspace is $A\cap U$ for some $U\in \mathcal{T}$. Since $f$ takes values in $A$,The right side is open in $Z$ by the assumed continuity of the composition. Hence preimages of every ${\mathcal{T}}_A$-open set are open, proving $f:Z\to (A,{\mathcal{T}}_A)$ is continuous.

If the inclusion is continuous from $(A,{\mathcal{T}}^{\prime })$, then $i^{-1}(U)=A\cap U$ belongs to ${\mathcal{T}}^{\prime }$ for every $U\in \mathcal{T}$. These intersections are precisely all members of ${\mathcal{T}}_A$, soThus the subspace topology is the coarsest topology on $A$ making the inclusion continuous.

The subspace topology is ${\mathcal{T}}_A=\{A\cap U:U\in \mathcal{T}\}$. For every open $U\subseteq X$,The inverse-image definition of continuity therefore gives the inclusion is continuous from $(A,{\mathcal{T}}_A)$ to $(X,\mathcal{T})$.

A metric space is sequentially compact if every sequence in it has a subsequence converging to a point of the same space. The sequence $x_n=1/(n+1)$ lies in $(0,1)$, but every subsequence converges in $\mathbb{R}$ to zero. Uniqueness of limits excludes an alternative limit inside $(0,1)$, so the open interval is not sequentially compact.

For the product, take any sequence $(x_n,y_n)$. sequential compactness of $X$ gives a subsequence $x_{n_k}\to x\in X$. Apply sequential compactness of $Y$ to the corresponding sequence $y_{n_k}$ to obtain a further subsequence $y_{n_{k_j}}\to y\in Y$. Passing to a subsequence preserves the convergence of the first coordinate, and thereforeThus $X\times Y$ is sequentially compact with the stated sum metric. Taking the second coordinate subsequence inside the first one is essential; two unrelated subsequences would not suffice.

Convergence to a point $x\in X$ meansThe limit must belong to the metric space itself. The definition is equivalent to $d(x_n,x)\to 0$ as a real sequence.

Similarly set $g_n(x)=x^{1/2}{1}_{[1/n,1]}(x)$. These nonnegative measurable functions increase to $g$ almost everywhere. The fundamental theorem of calculus givesBy monotone convergence,The integral is finite and $g\ge 0$, so $g$ is Lebesgue integrable. Identifying Lebesgue and ordinary integrals of continuous functions on compact intervals justifies each truncated calculation.

Use the increasing nonnegative measurable functions $f_n(x)=x^{-2}{1}_{[1/n,1]}(x)$. Their pointwise limit is the given function except at irrelevant endpoint choices, and the fundamental theorem of calculus on $[1/n,1]$ givesmonotone convergence therefore gives $\int f=\infty$. Since $f\ge 0$, this is also $\int |f|$, and $f$ is not Lebesgue integrable. Measurability alone does not make a finite integral.

The monotone convergence theorem states that if measurable functions $f_n$ are nonnegative and $f_n\uparrow f$ pointwise, then $f$ is measurable andallowing either side to be infinite. The same conclusion holds under almost-everywhere monotone convergence after changing values on a null set. No assumption that the integrals stay bounded is required.

The precise conclusion isTo prove it, observe that the maximum rises exactly when the new height difference is positive:A block landing in a shorter column produces $D_m=0$ and no rise; every other case raises the maximum and leaves $D_m>0$. Starting from $H_0=0$, telescoping givesThe difference chain is irreducible and positive recurrent by part (i). Apply the countable-state Markov chain ergodic theorem to the bounded indicator function $g(k)=1_{\{k>0\}}$. Its stationary mean is ${\sum }_{k\ge 1}{\pi }_k=1-{\pi }_0=2/3$, giving the asserted almost sure convergence. No aperiodicity assumption is needed for this time average. This establishes the growth rate of two-column sticky deposition.

The difference process $D_n$ is a Markov chain. If the columns are equal, the next block creates a height difference of one. If the difference is $k\ge 1$, landing in the taller column increases it to $k+1$, while landing in the shorter column makes it zero. Therefore its only nonzero transition probabilities areThese probabilities depend only on the present difference, not on the absolute heights or their history. This is the reset-or-increment Markov chain with $p=1/2$.

For its stationary distribution, the incoming-flow equations give ${\pi }_1={\pi }_0$ and ${\pi }_k={\pi }_{k-1}/2$ for $k\ge 2$. Normalization uses a geometric series:ThusThe remaining balance equation is also satisfied: the flow into zero is $(1-{\pi }_0)/2=1/3$. Every state can reach zero and every positive state can be reached from zero, so it is an irreducible Markov chain. Its invariant probability distribution makes it positive recurrent, and that distribution is unique.

The maximum-height process $H_n$ is not Markov. Starting empty, condition first on the positive-probability history $(H_1,H_2,H_3)=(1,1,2)$. After two blocks the columns were equal, so after three their difference is one. Hence $\mathbb{P}(H_4=3\mid H_1=1,H_2=1,H_3=2)=1/2$.

For the history $(1,2,2)$, the first two blocks built one column and the third caught the other column up, leaving equal heights. Hence $\mathbb{P}(H_4=3\mid H_1=1,H_2=2,H_3=2)=1$. Both histories have the same time and present height $H_3=2$, but different predictions. This violates the Markov property and illustrates why a function of a Markov chain need not be Markov.

For an irreducible Markov chain on a countable state space, the set has either zero or one element. By uniqueness of stationary distributions for an irreducible countable chain, a stationary probability distribution exists exactly in the positive recurrent case, and thenfor the first return time $T_i^{+}$. The right-hand side fixes every entry, proving uniqueness.

For a finite state space, existence is guaranteed, so there is exactly one. A countably infinite state space can have none: the simple symmetric random walk on the integers is irreducible and null recurrent. Periodicity does not affect this existence-and-uniqueness statement.

A stationary distribution is a nonnegative row vector $\pi =({\pi }_i{)}_{i\in S}$ with total mass one that is preserved by one transition:In matrix notation, $\pi P=\pi$. If the initial state has distribution $\pi$, so does the state at each later time.

A transition matrix $P$ defines an irreducible Markov chain if every state is accessible from every other state:Equivalently, all states belong to one communicating class. When $i\ne j$, a positive number of steps is necessarily used.

Use the deterministic weights $a_{n,k}=\sqrt{k}/n$. Their squared sum and maximum areTo apply the quadratic expansion of a cumulant-generating function rigorously, write $\log M_X(u)=u^2/2+r(u)$, with $r(u)=o(u^2)$. Fix $t\ne 0$ and letAll arguments $t\sqrt{k}/n$ lie in this shrinking interval. ConsequentlyTaking logs of the moment-generating function and using independence of random variables now givesThe case $t=0$ is immediate. Since $e^{t^2/4}$ is the moment-generating function of a normal distribution of variance $1/2$, the convergence theorem for moment-generating functions givesThis is a weighted central limit theorem with negligible maximum weight. Its limit variance also agrees with the exact value $\mathrm{Var}(T_n)=(n+1)/(2n)$ from variance additivity for independent random variables.

Write $S_n=n^{-1/2}{\sum }_{k=1}^n{X}_k$. By independence of random variables, its moment-generating function isFor each fixed real $t$, the expansion in part (i) yieldsFor $t=0$ the equality is immediate. Exponentiation gives $M_{S_n}(t)\to e^{t^2/2}$, the moment-generating function of a standard normal variable. The functions are finite on a common neighborhood of zero, so the convergence theorem for moment-generating functions provesThis proves the central limit theorem under the exponential-moment hypothesis of this part.

The given expected value and variance imply $M_X(0)=1$, $M_X^{\prime }(0)=0$ and $M_X^{\prime \prime }(0)=1$. The Taylor expansion therefore givesSince $\log (1+u)=u+o(u)$ as $u\to 0$, it follows thatThis is the quadratic expansion of a cumulant-generating function for a centered variable of variance one.

Let $Y_1,Y_2,\dots$ be real independent and identically distributed random variables with finite expected value $\mu$ and variance $0<{\sigma }^2<\infty$. The central limit theorem statesEquivalently, the cumulative distribution functions of these normalized sums converge at every real point to the standard normal distribution function. The theorem itself does not require a finite moment-generating function.

Integrate against the standard normal density by completing the square:Thus the moment-generating function of a standard normal variable is