Double algebraic genericity of universal harmonic functions on trees

1. (i)

   This was proved as a Lemma in [3], without using any of the other claims stated in this proposition.

2. (ii)

   It suffices to prove that if $f:\partial T\to E$ and $g:\partial T\to E$ are two $\mathcal{M}$-measurable functions and $a\in\mathbb{C}$ is a complex number, then the functions $f+g$ and $af$ are $\mathcal{M}$-measurable. Since the set $\{h_{n}:n\in\mathbb{N}\}$ is dense in $L^{0}(\partial T,E)$ we can find a sequence in that set that convergences to $f$ in probability. We can then find a subsequence of that sequence that convergences to $f$ pointwise almost everywhere. By reindexing we can find a sequence $\{f_{n}\}_{n=1}^{\infty}$ in the set $\{h_{n}:n\in\mathbb{N}\}$ such that $f_{n}\to f$ pointwise almost everywhere and each $f_{n}$ is $\mathcal{M}_{\tau(n)}$-measurable for some $\tau(n)\in\mathbb{N}$. By the same argument we can find a sequence $\{g_{n}\}_{n=1}^{\infty}$ in the set $\{h_{n}:n\in\mathbb{N}\}$ such that $g_{n}\to g$ pointwise almost everywhere and each $g_{n}$ is $\mathcal{M}_{\ell(n)}$-measurable for some $\ell(n)\in\mathbb{N}$. Thus we can find a set $A\in\mathcal{M}$ with $\mathbb{P}(A)=1$ such that $f_{n}(e)\to f(e)$ and $g_{n}(e)\to g(e)$ for all $e\in A$. Since addition on $E$ is continuous we have that $f_{n}(e)+g_{n}(e)\to f(e)+g(e)$ for all $e\in A$. By setting $r(n)=\max\{\tau(n),\ell(n)\}$ for all $n\in\mathbb{N}$ we have that $\mathcal{M}_{\tau(n)}\subseteq\mathcal{M}_{r(n)}$ and $\mathcal{M}_{\ell(n)}\subseteq\mathcal{M}_{r(n)}$ for all $n\in\mathbb{N}$. Therefore, both $f_{n}$ and $g_{n}$ are $\mathcal{M}_{r(n)}$-measurable for all $n\in\mathbb{N}$ and so they are constant on all the sets $B_{x}$ for $x\in T_{r(n)}$. This implies that the sum $f_{n}+g_{n}$ is constant on all the sets $B_{x}$ for $x\in T_{r(n)}$ and thus, the sum $f_{n}+g_{n}$ is $\mathcal{M}_{r(n)}$-measurable and therefore $\mathcal{M}$-measurable. So, we have that $f+g$ is an almost everywhere pointwise limit of the sequence of $\mathcal{M}$-measurable functions $\{f_{n}+g_{n}\}_{n=1}^{\infty}$, the values of which are elements of the metric space $(E,d)$, and the measure space $(\partial T,\mathcal{M},\mathbb{P})$ is complete. This implies that the function $f+g$ is $\mathcal{M}$-measurable. The proof that $af$ is $\mathcal{M}$-measurable is similar, it requires the fact that multiplication on $E$ is continuous, and is omitted.

3. (iii)

   The proof of this claim is similar to the proof of the respective claim about real-valued functions, which is well known. It suffices to replace the absolute value in $\mathbb{R}$ by a translation invariant metric $d$. Thus, the proof is omitted.

4. (iv)

   The proof of this claim is similar to the proof of the respective claim about real-valued functions, which is well known. It suffices to replace the absolute value in $\mathbb{R}$ by a norm induced metric $d$. Thus, the proof is omitted.

∎

Let $a_{1}f_{1}+\dots+a_{s}f_{s}$ be an arbitrary element of the set $\operatorname{span}\{f_{n}:n\in\mathbb{N}\}$ with $a_{s}\neq 0$. Let $V$ be a nonempty open subset of $L^{0}(\partial T,E)$. There exist some $\varepsilon>0$ and $h\in V$ such that $B(h,\varepsilon)\subseteq V$. For $i\in\{1,\dots,s\}$ we define

$$b_{i}=\begin{cases}a_{i},&\text{if }a_{i}\neq 0\\ 1,&\text{if }a_{i}=0\end{cases}.$$

We set $\delta=\displaystyle{\frac{1}{s\cdot\varepsilon}}$ and consider the set

$$\widehat{V}=\frac{1}{b_{1}}B(0,\delta)\times\frac{1}{b_{2}}B(0,\delta)\times\cdots\times\frac{1}{b_{s}}B(h,\delta)\times L^{0}(\partial T,E)\times\cdots.$$

We now fix some $n\in\mathbb{N}$ and suppose that $\omega_{n}(f)\in\widehat{V}$. Then for every $i\in\{1,\dots,s-1\}$ we have

$$\omega_{n}(f_{i})\in\frac{1}{b_{i}}B(0,\delta)\text{~{}~{}~{}and~{}~{}~{}}\omega_{n}(f_{s})\in\frac{1}{b_{s}}B(h,\delta).$$

By the definition of $b_{1},\dots,b_{n}$ and the linearity of $\omega_{n}(\cdot)$ we conclude that for all $i\in\{1,\dots,s-1\}$ we have

$$P(\omega_{n}(a_{i}f_{i}),0)<\delta\text{~{}~{}~{}and~{}~{}~{}}P(\omega_{n}(a_{s}f_{s}),h)<\delta.$$

The metric $P$ is translation invariant, since the metric $d$ is translation invariant, thus

	$\displaystyle P(\omega_{n}(a_{1}f_{1}+\dots+a_{s}f_{s}),h)$	$\displaystyle\leqslant P(\omega_{n}(a_{1}f_{1}),0)+\dots+P(\omega_{n}(a_{s-1}f_{s-1}),0)+P(\omega_{n}(a_{s}f_{s}),h)$
		$\displaystyle<s\cdot\delta.$

Since $s\cdot\delta=\varepsilon$ we conclude that $\omega_{n}(a_{1}f_{1}+\dots+a_{s}f_{s})\in B(h,\varepsilon)$, where $B(h,\varepsilon)\subseteq V$. Therefore, we have proved that

$$\{n\in\mathbb{N}:\omega_{n}(f)\in\widehat{V}\}\subseteq\{n\in\mathbb{N}:\omega_{n}(a_{1}f_{1}+\dots+a_{s}f_{s})\in V\}.$$

Notice that convergence in probability in $L^{0}(\partial T,E^{\mathbb{N}})$, which is the same set as $L^{0}(\partial T,E)^{\mathbb{N}}$, is equivalent to convergence in the product topology when considering $L^{0}(\partial T,E)$ with the topology of convergence in probability. Therefore $\widehat{V}$ is an open subset of $L^{0}(\partial T,E^{\mathbb{N}})$ and it is also nonempty. Since $f\in X(T,E)$, the set $\{n\in\mathbb{N}:\omega_{n}(f)\in\widehat{V}\}$ has upper density equal to one, which implies that the set $\{n\in\mathbb{N}:\omega_{n}(a_{1}f_{1}+\dots+a_{s}f_{s})\in V\}$ has upper density equal to one. Thus $a_{1}f_{1}+\dots a_{s}f_{s}\in X(T,E)$. ∎

The metric space $H(T,E)$ is separable, so let $\{\varphi_{n}:n\in\mathbb{N}\}$ be a dense sequence in $H(T,E)$. Since $E^{\mathbb{N}}$ when considered with the product metric $\hat{d}$ is a separable topological vector space, Theorem 2.7 applies. Thus, the class $X(T,E^{\mathbb{N}})$ is dense in $H(T,E^{\mathbb{N}})$ and in particular nonempty, so let $f=(f_{1},f_{2},\dots)$ be an element of $X(T,E^{\mathbb{N}})$. We now fix some $n\in\mathbb{N}$. There exists some $j_{0}(n)\in\mathbb{N}$ such that

$$\sum_{j=j_{0}(n)}^{\infty}{\frac{1}{2^{j}}}<\frac{1}{n}.$$

There also exists some $N(n)\in\mathbb{N}$ such that

$$\{z_{0},\dots,z_{j_{0}(n)}\}\subseteq\bigcup_{k=0}^{N(n)}{T_{k}},$$

where $\{z_{n}:n\in\mathbb{N}\}$ is the denumeration of $T$ through which the metric $\rho$ was defined. We set $g_{n}(x)=\varphi_{n}(x)-f_{n}(x)$ for all $x\in\bigcup_{k=0}^{N(n)}{T_{k}}$ and we set $g_{n}(x)=g_{n}(x^{-})$ for all $x\in\bigcup_{k=N(n)+1}^{\infty}{T_{k}}$, where $x^{-}$ is the father of $x$. Notice that the function $g_{n}:T\to E$ defined this way is a generalized harmonic function such that $\omega_{k}(g_{n})=\omega_{N(n)}(g_{n})$ for all $k>N(n)$ and $\rho(\varphi_{n}-f_{n},g_{n})<1/n$. Since the metric $d$ is translation invariant it follows that the metric $\rho$ is translation invariant, thus $\rho(f_{n}+g_{n},g_{n})<1/n$ for all $n\in\mathbb{N}$. Since $H(T,E)$ has no isolated points, the sequence $\{f_{n}+g_{n}\}_{n=1}^{\infty}$ is dense in $H(T,E)$. We will now show that the function $f+g=(f_{1}+g_{1},f_{2}+g_{2},\dots)\in H(T,E^{\mathbb{N}})$ belongs to the class $X(T,E^{\mathbb{N}})$ which completes the proof. It suffices to show that for every nonempty open basic set $V\subseteq L^{0}(\partial T,E)$ the set $\{n\in\mathbb{N}:\omega_{n}(f+g)\in V\}$ has upper density equal to one. Since convergence in probability in $L^{0}(\partial T,E^{\mathbb{N}})$, which is the same set as $L^{0}(\partial T,E)^{\mathbb{N}}$, is equivalent to convergence in the product topology when considering $L^{0}(\partial T,E)$ with the topology of convergence in probability, a nonempty basic open set of $L^{0}(\partial T,E^{\mathbb{N}})$ can be written as

$$V=V_{1}\times\cdots\times V_{s}\times L^{0}(\partial T,E)\times\cdots,$$

where $V_{1},\dots,V_{s}$ are nonempty open subsets of $L^{0}(\partial T,E)$. We set $L=\max\{N(1),\dots,N(s)\}$. Then, for all $n\geqslant L$ and all $j\in\{1,\dots,s\}$ we have $\omega_{n}(g_{j})=\omega_{L}(g_{j})$. We fix some $n\in\mathbb{N}$ with $n\geqslant L$ and suppose that $\omega_{n}(f)\in V-\omega_{L}(g)$. Then $\omega_{n}(f_{j})\in V_{j}-\omega_{L}(f_{j})$ for all $j\in\{1,\dots,s\}$, which implies that $\omega_{n}(f_{j})+\omega_{n}(g_{j})\in V_{j}$. But $\omega_{n}(f_{j}+g_{j})=\omega_{n}(f_{j})+\omega_{n}(g_{j})$ for all $j\in\{1,\dots,n\}$, so we have that $\omega_{n}(f_{j}+g_{j})\in V_{j}$ for all $j\in\{1,\dots,n\}$, which in turn implies that $\omega_{n}(f+g)\in V$. Thus, we have proved that

$$\{n\geqslant L:\omega_{n}(f)\in V-\omega_{L}(g)\}\subseteq\{n\in\mathbb{N}:\omega_{n}(f+g)\in V\}.$$

But the set $V-\omega_{L}(g)$ is a nonempty open subset of $L^{0}(\partial T,E^{\mathbb{N}})$ and $f\in X(T,E^{\mathbb{N}})$, therefore the set $\{n\in\mathbb{N}:\omega_{n}(f)\in V-\omega_{L}(g)\}$ has upper density equal to one. This implies that the set $\{n\geqslant L:\omega_{n}(f)\in V-\omega_{L}(g)\}$ has upper density equal to one and therefore the set $\{n\in\mathbb{N}:\omega_{n}(f+g)\in V\}$ has upper density equal to one. ∎

By Lemma 3.3 there exists some $f=(f_{1},f_{2},\dots)\in X(T,E^{\mathbb{N}})$ such that the sequence $\{f_{n}\}_{n=1}^{\infty}$ is dense in $H(T,E)$. By Lemma 3.2 the vector space $\operatorname{span}\{f_{n}:n\in\mathbb{N}\}$ is contained in the set $X(T,E)\cup\{0\}$ and the sequence $\{f_{n}\}_{n=1}^{\infty}$ is contained in the vector space $\operatorname{span}\{f_{n}:n\in\mathbb{N}\}$, which implies that vector space $\operatorname{span}\{f_{n}:n\in\mathbb{N}\}$ is dense in $H(T,E)$. ∎

By Theorem 3.1 there exists a vector space $F_{1}$ contained in $U_{FM}(T,E)\cup\{0\}$, which is a subset of $U(T,E)\cup\{0\}$, that is dense in $H(T,E)$. By Theorem 3.4 there exists a vector space $F_{2}$ contained in $X(T,E)\cup\{0\}$, which is a subset of $U(T,E)\cup\{0\}$, that is dense in $H(T,E)$. By Proposition 2.8 we have that $F_{1}\cap F_{2}=\{0\}$. ∎