# Does $P^n f(x)=\frac{1}{2^n}\sum_{k=0}^n {n\choose k} f\left(\frac{x}{2^k}\right)$ converge uniformly on compact sets, for $f\in C_b (\mathbb{R})$?

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.

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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.