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Let $\nu$ a measure on $\mathbb{P}\mathbb{R}^2$. Let $A= \left( \begin{array}{ccc} \frac{1}{2} & 0 \\ 0 & 2 \end{array} \right)$ and $B= \left( \begin{array}{ccc} 0 & 1 \\ -1 & 0 \end{array} \right)$.

I want to find all the measure $\nu$ such that:

$$\nu=p(A)_*\nu+q(B)_*\nu$$

where p,q are two positive real numbers and $(A)_*\nu(C)=\nu(A^{-1}(C))$ for all $C\subset \mathbb{P}\mathbb{R}^2$.

The only think that I noticed is that the space $\mathbb{R}(0,1)$ $\mathbb{R}(1,0)$ are $A$ and $B$ stable.

I haven't any idea. Can you give me a hint please?

  • 0
    Let (1,0) and (0,1) two element of $\mathbb{P}\mathbb{R}^2$. I think that the measure $\delta_{(0,1)}$, $\delta_{(1,0)}$ and $1/2(\delta_{(0,1)}+\delta_{(1,0)})$ satisfies my relation. Is there any other measure? – 2017-01-02
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    Are you sure you are working with $\Bbb{P}\Bbb{R}^2$, not $\Bbb{P}\Bbb{R}^1$? – 2017-01-02
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    I'm working in the projective space of $\mathbb{R}^2$ – 2017-01-02
  • 2
    Aha, I was confused since $\Bbb{P}^2\Bbb{R} = \Bbb{R}^{3}-\{0\}/(\mathrm{x} \sim \lambda \mathrm{x})$. I guess you mean $\Bbb{P}(\Bbb{R}^2)$, which is in fact $\Bbb{P}^1\Bbb{R}$. – 2017-01-02
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    Do you have any idea to help me? – 2017-01-02
  • 0
    I have no prior knowledge related to measures on $\Bbb{P}(\Bbb{R}^2)$, so I am thinking about it. One idea I am currently trying is that $\Bbb{P}(\Bbb{R}^2)$ can be thought as $\Bbb{R} \cup \{\infty\}$ so that fractional linear transforms on $\Bbb{P}(\Bbb{R}^2)$ corresponds to Mobious transforms on $\Bbb{R} \cup \{\infty\}$. Then $A$ corresponds to $x \mapsto \frac{x}{4}$ and $B$ corresponds to $x \mapsto -\frac{1}{x}$. This probably makes things easier to work with. – 2017-01-02

1 Answers 1

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Here is a solution when $\nu$ is finite.

Let us identify $\Bbb{P}^1 \Bbb{R} \simeq \Bbb{R} \cup \{\infty\} =: \bar{\Bbb{R}}$ so that $(x,1)$ corresponds to $x$ and $(1, 0)$ corresponds to $\infty$. Suppose that $\nu$ is a finite Borel measure on $\bar{\Bbb{R}}$ that satisfies

$$ \nu = p (A_*\nu) + q (B_* \nu), \tag{*} $$

where $A(x) = x/4$ and $B(x) = -1/x$. Then it follows that $p +q = 1$.

Case 1. If $p = 0$, then $\nu = B_* \nu$ and thus $\nu$ is completely determined by its restriction on $[0, \infty)$.

To be precise, let $\mu$ be any finite Borel measure on $[0, \infty)$. Then $\nu = \mu + B_* \mu$ satisfies the desired property. On the other hand, if $\nu = B_* \nu$ then the restriction $\mu = \nu|_{[0,\infty)}$ is a finite Borel measure on $[0, \infty)$ such that $\nu = \mu + B_* \mu$.

Case 2. If $p = 1$, then $\nu(S) = (A_* \nu)(S) = \nu(4S)$ and it easily follows that $\nu$ is supported on $\{0,\infty\}$. Then $\nu = a \delta_0 + b \delta_{\infty}$ for some $a, b \geq 0$.

Case 3. When $0 < p < 1$, we claim that $\nu = a (\delta_0 + \delta_{\infty})$. Once we prove that $\nu$ is supported on the set $\{0, \infty\}$ so that $\nu = a \delta_0 + b \delta_{\infty}$ for some $a, b \geq 0$, the conclusion immediate follows from

$$ a = \nu(\{0\}) = p\nu(\{0\}) + q\nu(\{\infty\}) = pa + qb. $$

Let $X = (X_n : n \geq 1)$ be a sequence of i.i.d. random variables such that $\Bbb{P}(X_n = 0) = p$ and $\Bbb{P}(X_n = 1) = q$. If we write $A = A_0$ and $B = A_1$, then the stationarity condition $\text{(*)}$ is interpreted as

$$\nu(C) = \Bbb{E}[(A_{X_n})_* \nu(C)], \qquad \forall C \in \mathcal{B}(\bar{\Bbb{R}}).$$

Now fix a Borel set $C \in \mathcal{B}(\bar{\Bbb{R}})$ and write $U_n = (A_{X_n}\cdots A_{X_1})_* \nu(C)$. Then $(U_n)$ is a martingale w.r.t. the filtration $\mathcal{F}_n = \sigma(X_1, \cdots, X_n)$, since

\begin{align*} \Bbb{E}[U_{n+1} \mid \mathcal{F}_n] &= \Bbb{E}[ (A_{X_{n+1}})_* \nu( (A_{X_n} \cdots A_{X_1})^{-1} C) \mid \mathcal{F}_n] \\ &= \nu( (A_{X_n} \cdots A_{X_1})^{-1} C) \\ &= U_n. \end{align*}

Let $(T_k : k \geq 0)$ be stopping times w.r.t. $\mathcal{F}_n$ defined by

$$T_0 = 0, \qquad T_k = \inf\{n > T_{k-1} : X_{T_n} = 1 \}, \qquad k \geq 1. $$

That is, $T_k$ is the $k$-th arrival time of $1$. If we write $Y_k = T_k - T_{k-1}$, then $(Y_k : k \geq 1)$ is a sequence of i.i.d. geometric random variables. Now from the observation that $BA^kB(x) = 4^{-k}x$, it follows that

\begin{align*} U_{T_{2k}} &= (B A^{Y_{2k}-1} B A^{Y_{2k-1}-1} \cdots B A^{Y_2 - 1} B A^{Y_1 - 1})_* \nu(C) \\ &= \nu(4^{(Y_2 - Y_1) + \cdots + (Y_{2k} - Y_{2k-1})}C). \end{align*}

In particular, since $U_n$ is bounded and each $T_{2k}$ is integrable, optional stopping theorem tells that

$$ \nu(C) = \Bbb{E}[U_{T_{2k}}] = \Bbb{E}[\nu(4^{(Y_2 - Y_1) + \cdots + (Y_{2k} - Y_{2k-1})}C)]. $$

Now here is a crucial observation: if we write $S_k = (Y_2 - Y_1) + \cdots + (Y_{2k} - Y_{2k-1})$, then by the classical CLT, we have $\Bbb{P}(|S_k| \leq R) \to 0$ as $k \to \infty$ for each fixed $R > 0$.

If we write $C_k := (-4^k, -\frac{1}{4^k}) \cup (\frac{1}{4^k}, 4^k)$ and $\Bbb{R}^{\times} = \Bbb{R} \setminus\{0\}$, then for each $k, l \geq 1$ we have

\begin{align*} \nu(C_k) &= \Bbb{E}[\nu(4^{S_n}C_k)] \\ &= \Bbb{E}[\nu(4^{S_n}C_k) \ ; \ (4^{S_n}C_k) \cap C_l = \varnothing] + \Bbb{E}[\nu(4^{S_n}C_k) \ ; \ (4^{S_n}C_k) \cap C_l \neq \varnothing] \\ &\leq \nu(\Bbb{R}^{\times}\setminus C_l) + \nu(\bar{\Bbb{R}})\Bbb{P}(|S_n| \leq k+l) \end{align*}

Then by taking $n\to\infty$ followed by $k, l\to\infty$ it follows that $\nu(\Bbb{R}^{\times}) = 0$. Therefore $\nu$ is supported on $\{0,\infty\}$ and the claim follows.