Showing that the mean of translations of a function approaches 0 in $L_p$

Question

Given $p \in (1,\infty)$, $f \in L^p(\Bbb R)$ and $T: \Bbb R \to \Bbb R,x \mapsto x+1$. How do I show that for $n \to \infty$ $$\frac{1}{n}\sum_{k=0}^n f \circ T^k \to 0$$ in $L^p$? I see that for very large $N$ $f$ and $f\circ T^N$ have their masses concentrated in different areas and hence one has $||f+T^N\circ f||_p^p \approx ||f||_p^p+||T^N\circ f||_p^p$, but I don't see how to make this explicit.

Answer 1

In other words, you want to prove that the sequence of ergodic mean operators $$ U_n:L^p(\mathbb{R})\longrightarrow L^p(\mathbb{R}) \qquad U_nf:=\frac{1}{n}\sum_{k=0}^{n-1}f\circ T^k $$ converges pointwise to $0$.

First note that $U_T:f\longmapsto f\circ T$ is a surjective isometry from $L^p$ to itself. As a consequence, $\|U_n\|\leq 1$ for every $n$. So it suffices to prove the result for simple functions, by density. And if the property is true for $f_1,\ldots,f_k$, it is true for every linear combination of them. So it suffices to prove the result for characteristic functions $1_A$ for bounded measurable sets $A$. Finally, up to splitting $A$, we can assume that $A\subseteq (j,j+1)$. It is easy in this case. See below if needed.

The property is false for $p=1$, as shown by the example $f=1_{(0,1)}$. And false as well for $p=\infty$, as shown by $f=1$.

Now for every $k$, $1_A\circ T^k=1_{T^{-k}A}\subseteq (j-k,j-k+1)$. Since the latter are disjoint, it follows that $\|U_n1_A\|_p\leq \frac{1}{n}n^\frac{1}{p}\longrightarrow 0$ for $p>1$.

Answer 2

Here are some hints.

If $f$ is supported inside an interval of length 1 then all the $f \circ T^n$ have disjoint support and this is easy, as you note.

If $f$ is supported in a bounded interval, you can write it as a finite sum of functions supported inside intervals of length 1.

For general $f \in L^p$, most of the mass of $f$ is inside a bounded interval. That is, we can write $f = f_1 + f_2$ where $f_1$ is supported inside a bounded interval, and $\|f_2\|_p < \epsilon$.