Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

One can show that the Iterated Function System consisting of transformations $$S_1(x)=x, \;\;\ S_2(x)=\frac{1}{2}\;\;\; (x\in\mathbb{R})$$ with constant probabilities $$p_1=p_2=\frac{1}{2}$$ is asymptotically stable with the Dirac measure $\delta_0$ as a unique invariant measure, that is for its Markov (dual) operator $P$ given by $$Pf(x)=\frac{1}{2} \left(f(x) + f\left(\frac12 x\right)\right)\;\;\; (x\in \mathbb{R},\; f\in C_b(\mathbb{R})),$$ where $C_b(\mathbb{R})$ stands for the set of all bounded continuous real-valued functions, we have $$P^n f(x)\to \int_{\mathbb{R}} f(y)\,\delta_0(dy)=f(0)\;\;\; (f\in C_b(\mathbb{R})).$$ My question is: Does the sequence $(P^n f)_{n\geq 1}$ converge uniformly to $f(0)$ on compact sets, for any $f\in C(\mathbb{R})$? I calculated that $$P^n f(x)=\frac{1}{2^n}\sum_{k=0}^n {n\choose k} f\left(\frac{x}{2^k}\right)$$

I suppose it does not, but I can't find an appropriate function. I have tried $x\mapsto (\sin x) /x$ (and 1 at $x=0$) but I do not know how to show it.

share|cite|improve this question
up vote 3 down vote accepted

Let $K\subset \Bbb R$ compact, $R$ such that $|x|\leq R$ for all $x\in K$ and $\varepsilon>0$. Let $\delta$ in the definition of continuity at $0$, and $k_0$ such that $R\cdot 2^{—k_0}\leq\delta$. Then we have \begin{align} |P^nf(x)-f(0)|&\leq \frac 1{2^n}\sum_{k=0}^{k_0}\binom nk\left|f\left(\frac x{2^k}\right)-f(0)\right|+\frac 1{2^n}\sum_{k=k_0+1}^n\binom nk\left|f\left(\frac x{2^k}\right)-f(0)\right|\\ &\leq\frac 1{2^n}\sum_{k=0}^{k_0}\binom nk\left|f\left(\frac x{2^k}\right)-f(0)\right|+\varepsilon\\ &\leq 2\sup_{t\in K}|f(t)|\frac 1{2^n}\sum_{k=0}^{k_0}\binom nk+\varepsilon. \end{align} Since $\sum_{k=0}^{k_0}\binom nk$ is polynomial in $n$, we get the wanted result.

share|cite|improve this answer
Thank you for the quick response – dawid Sep 5 '12 at 16:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.