Diagonal argument

$L^{p^{\prime }}$ has a countable dense subset, say $\mathcal{G}$. [Recall an approximation result from Chapter 8.]

Append to $\mathcal{G}$ all finite rational linear combinations of its elements. We will refer to this augmented collection as $\mathcal{G}$; $\mathcal{G}$ [the augmented collection] is still countable. Next, let $\left\{ f_{n}\right\}_{n=1}^{\infty }$ denote a sequence in $B_{1}^{p}$.

For every $g \in L^{p^{\prime }}$, $\left\{ \int f_{n} g \, \mathrm{d} \mu \right\}_{n=1}^{\infty }$ is a bounded sequence of real numbers with bound $\left\| g\right\|_{p^{\prime }}$. From this we may deduce that $\left\{ \int f_{n} g \, \mathrm{d} \mu \right\}_{n=1}^{\infty }$ is a bounded sequence of real numbers with bound $\left\| g\right\|_{p^{\prime }}$ for every $g \in \mathcal{G}$.

Since $\mathcal{G}$ is countable, there is a subsequence $\left\{ n_{k}\right\}_{k=1}^{\infty }$ of the sequence $1, 2, \cdots$ such that $\lim_{k} \int f_{n_{k}} g \, \mathrm{d} \mu$ exists for every $g \in \mathcal{G}$. [This is a direct application of the diagonal method from advanced calculus: The array whose $r$th row is $\left\{ \int f_{r} g_{n} \, \mathrm{d} \mu \right\}_{n=1}^{\infty }$, with $\left\{ g_{n}\right\}_{n=1}^{\infty }$ an enumeration of $\mathcal{G}$, is bounded row by row, so Exercise 1 of this section applies.] Write $L_{\mathcal{G}}(g)=\lim_{k} \int f_{n_{k}} g \, \mathrm{d} \mu$ for every $g \in \mathcal{G}$; Hölder's Inequality gives $\left|L_{\mathcal{G}}(g)\right| \leqslant \left\| g\right\|_{p^{\prime }}$ for every $g \in \mathcal{G}$.

$L_{\mathcal{G}}$ is uniformly continuous on its domain $\mathcal{G}$. [We have

for every $g, g^{\prime } \in \mathcal{G}$.]

There exists a unique continuous function $L$ on $L^{p^{\prime }}$ such that $L=L_{\mathcal{G}}$ on $\mathcal{G}$. [From advanced calculus, we know that a uniformly continuous function $\psi$ defined on a dense subset $D$ of a metric space $(X, \rho )$ has a unique continuous extension to all of $X$.]

For every $g, g^{\prime } \in \mathcal{G}$ we have $L_{\mathcal{G}}(g+g^{\prime })=L_{\mathcal{G}}(g)+L_{\mathcal{G}}(g^{\prime })$. [Direct verification.]

For every $h, h^{\prime } \in L^{p^{\prime }}$ we have $L(h+h^{\prime })=L(h)+L(h^{\prime })$. [Let $\left\{ g_{n}\right\}_{n=1}^{\infty }$ denote a sequence in the dense collection $\mathcal{G}$ with $g_{n} \xrightarrow {p^{\prime }} h$; let $\left\{ g_{n}^{\prime }\right\}_{n=1}^{\infty }$ denote a sequence in $\mathcal{G}$ with $g_{n}^{\prime } \xrightarrow {p^{\prime }} h^{\prime }$. Thus $\left\{ g_{n}+g_{n}^{\prime }\right\}_{n=1}^{\infty }$ is a sequence in $\mathcal{G}$ with $g_{n}+g_{n}^{\prime } \xrightarrow {p^{\prime }} h+h^{\prime }$. Use the continuity of $L$ and (e) to verify the equation

]

Directly verify that $L_{\mathcal{G}}(cg)=cL_{\mathcal{G}}(g)$ for every $g \in \mathcal{G}$ and $c \in \mathbb {R}$. This in turn shows that $L(cg)=cL(g)$ for every $c \in \mathbb {R}$ and $g \in L^{p^{\prime }}$. [The first part is easy. The second part is obtained in the same spirit as was (f).]

By (g), $L$ is a linear functional on $L^{p^{\prime }}$. Show in addition that $L$ is bounded with $\left\| L\right\| \leqslant 1$. [We already know that $L$ is continuous from (d); use the relevant claim from Section 8.8 to achieve boundedness. Next, pick $h \in L^{p^{\prime }}$; let $\left\{ g_{n}\right\}_{n=1}^{\infty }$ denote a sequence in $\mathcal{G}$ with $g_{n} \xrightarrow {p^{\prime }} h$. Verify that

]

We now invoke the Riesz Representation Theorem for the bounded linear functional $L$ on $L^{p^{\prime }}$, obtaining a function $f \in L^{p}$ such that $L(g)=\int fg \, \mathrm{d} \mu$ for every $g \in L^{p^{\prime }}$. Since $\left\| L\right\| \leqslant 1$, we have $\left\| f\right\|_{p}=\left\| L\right\| \leqslant 1$ by Claim 1 of Section 8.9, hence $f \in B_{1}^{p}$. Recall the subsequence $\left\{ n_{k}\right\}_{k=1}^{\infty }$ given after part (b). Observe that for every $g \in \mathcal{G}$ we have

We have that $\left\{ f_{n_{k}}\right\}_{k=1}^{\infty }$ converges weakly to $f$, completing the proof. [Pick $g^{\prime }$ in $L^{p^{\prime }}$ and $\varepsilon >0$; find $g \in \mathcal{G}$ with $\left\| g-g^{\prime }\right\|_{p^{\prime }}<\frac{\varepsilon }{3}$. Define $\alpha_{k}=\left|\int f_{n_{k}} g \, \mathrm{d} \mu -\int fg \, \mathrm{d} \mu \right|$ for each $k \in \mathbb {N}$, and observe that the above shows $\lim_{k} \alpha_{k}=0$. Verify that

For large enough $k$ we thus have $\left|\int fg^{\prime } \, \mathrm{d} \mu -\int f_{n_{k}} g^{\prime } \, \mathrm{d} \mu \right|<\varepsilon$. Since $g^{\prime } \in L^{p^{\prime }}$ is arbitrary, the entire proof is complete.]

Show that there exists an increasing sequence $n_{1}, n_{2}, \ldots$ of positive integers such that $\lim_{r \rightarrow \infty } x_{i}\left(n_{r}\right)$ exists for all $i$.

Use this result to prove that, for an irreducible Markov chain, if it is not the case that $p_{i j}(n) \rightarrow 0$ as $n \rightarrow \infty$ for all $i$ and $j$, then there exists a sequence $(n_{r}: r \geq 1)$ and a vector $\mathbf{\alpha }( \neq \mathbf{0})$ such that $p_{i j}\left(n_{r}\right) \rightarrow \alpha_{j}$ as $r \rightarrow \infty$ for all $i$ and $j$.

Let $\left\{ z_{n}: n \in \mathbb {N}\right\}$ be a sequence with $z_{n} \in H_{n}$. Use compactness of each $K_{i}$ to find subsequences $\mathbb {N}_{1} \supseteq \mathbb {N}_{2} \supseteq \mathbb {N}_{3} \supseteq \ldots$ of $\mathbb {N}$ for which $y(i): = \lim_{n \in \mathbb {N}_{i}} \pi_{i} z_{n}$ exists, as a point of $K_{i}$. Define $\mathbb {N}_{\infty }$ to be the subsequence whose $i$ th member equals the $i$ th member of $\mathbb {N}_{i}$. Show that $\lim_{n \in \mathbb {N}_{\infty }} \pi_{i} z_{n} = y(i)$ for every $i$.

Show that the first $i$ components of $y(i+1)$ coincide with $y(i)$ for each $i$. Deduce that there exists a $y$ in $\Omega$ for which $\pi_{i} y = y(i) \in K_{i}$ for every $i$. Deduce that $y \in H$, and hence $H \neq \emptyset$.

Specialize to the case where $z_{n} \in H$ for every $n$. Show that there is a subsequence that converges to a point of $H$.