# Convergence in $L^p(\lambda)$ of mean of backward shifted integrable functions

Let $$(\Omega, \mathcal{A}, \mu)$$ be a $$\sigma$$-finite measure space. I want to prove the following (Exercise 7.1.2 in Klenke's Probability book):

Let $$p \in (1, \infty), f \in \mathcal{L}^p(\lambda)$$, where $$\lambda$$ is the Lebesgue measure on $$\mathbb{R}$$. Let $$T: \mathbb{R} \to \mathbb{R}, x \to x + 1$$. Show that, \begin{align} \frac{1}{n}\sum_{k = 0}^{n - 1}f\circ T^k \to 0 \hspace{0.2cm} \text{as} \hspace{0.2cm} n \to \infty \hspace{0.2cm} \text{in} \hspace{0.2cm} L^p(\lambda). \end{align}

I am lost on how to prove this. I think I am at least able to prove that the sequence is a Cauchy sequence and thus converges, but even there I am not sure. Here is my attempt:

Let $$S_n = \frac{1}{n}\sum_{k = 0}^{n - 1}f\circ T^k$$. Then \begin{align} \|S_n - S_{n - 1}\|_p &= \|\frac{1}{n} f \circ T^{n-1} - \frac{1}{n(n-1)}\sum_{k = 0}^{n -2}f \circ T^k \|_p \\ & \leq \frac{1}{n} \|f \circ T^{n - 1}\|_p + \frac{1}{n(n - 1)}\sum_{k = 0}^{n -2} \|f \circ T^k \|_p\\ &= \frac{2M}{n} \to 0, \end{align} where $$M = \|f\circ T^k\|_p$$. Here I am obviously assuming $$\|f\circ T^i\|_p = \|f\circ T^j\|_p$$ for all $$i, j \in \{0, \ldots, n- 1\}$$, which I think is true, but can only provide intuitive (as opposed to formal) justification (Since $$f \in \mathcal{L}^p (\lambda)$$, then the integral of the function should not change by shifting it around horizontally. I guess the idea would be to invoke the translation invariance of the Lebesgue measure?).

I have two questions then:

(1) Is my proof that the sequence converges in $$L^p(\lambda)$$ correct? Is my above assumption correct? If yes, what would be the formal justification?

(2) How would one prove that the limit is $$0$$?

Here is an alternative solution.

The operator norm of $$S_n$$ is bounded above by $$1$$. Let $$g$$ be function in $$L^p$$. Then

$$\|S_n f\| \le \|S_n g\|+ \|S_n (f-g)\|\le \|S_n g \| + \|f-g\|.$$

Thus we will be done if we can show that for every $$\epsilon>0$$ we can find $$g$$ such that (i) $$\|f-g\|<\epsilon/2$$ and (ii) $$\|S_n g\|< \epsilon/2$$.

(i) For an integer $$M>0$$, let $$g_M= f {\bf 1}_{[-M,M)}$$. By dominated convergence, for $$M$$ large enough $$\|f-g_M\|<\epsilon/2$$.

Moreover, $$g_M$$ is the sum of $$2M$$ functions $$h_j$$, $$j=-M,\dots,M-1$$ where $$h_{j} = f {\bf 1}_{[j,j+1)}$$. Each of these functions is supported on an interval of length $$\le 1$$.

(ii) Because of the last fact, for each $$j$$, the functions $$h_j, h_j \circ T,\dots$$ have non-overlapping supports, and so (that's the key):

$$\|S_n h_j\|^p = n*n^{-p} \|h_ j\|^p \quad \Rightarrow \quad \|S_n h_j\|= n^{1/p-1} \|h_j\|.$$

Thus,

$$\|S_n g_M \|=\|S_n \sum h_j\| \le\sum \|S_n h_j\| \le n^{1/p-1} \sum_{j=-M}^{M-1} \|h_j\|.$$

Since $$p>1$$, $$n^{1/p-1}\to 0$$, and therefore we can choose $$n$$ large enough so that $$\|S_n g_M\|\le \epsilon/2$$.

$$\Box$$

• Thanks! Maybe I'm misunderstanding here, but why are the $h_j=g \mathbf{1}_{[j, j+1)}$ rather than $h_j=f \mathbf{1}_{[j, j+1)}$? Oct 17, 2022 at 16:10
• right. corrected! Oct 17, 2022 at 20:07

With regards to the questions of the OP
(1) your approach is not entirely correct since having $$\|S_nf-S_{n-1}f\|_p\xrightarrow{n\rightarrow\infty}0$$ does not imply that the sequence $$S_nf$$ is Cauchy in $$L_p$$

(2) Looking at the material presented in your textbook, the idea it seems is to show that $$\{|S_nf|^p:n\in\mathbb{N}\}$$ converges to $$0$$ in measure and also that is uniform integrable, at least for a collection of functions $$f$$ that is dense in $$L_p$$.

Here I outlined the ideas in more detail. The tricky part is proving uniform integrability of $$\{|S_nf|^p:n\in\mathbb{N}\}$$. I suggest an approach at the end. Finally, I should mentioned that applying tools from Ergodic Theorem (not covered in the textbook until much later and then only in the context of probability measures) give a short solution to the problem, in particular the Dunford-Hopf-Schwartz) ergodic theorem which states:

Theorem (Dunford--Hopf--Schwartz). Let $$(S,\mathscr{B},\mu)$$ a $$\sigma$$-finite measure space. Suppose $$T$$ is a positive $$\mathcal{L}_1(\mu)$$-$$\mathcal{L}_\infty(\mu)$$ contraction. Define $$A_n=\frac1n\sum^{n-1}_{k=1}T^k$$. If $$1\leq p<\infty$$, then

1. For any $$f\in\mathcal{L}_p(\mu)$$, $$Af:=\lim_nA_nf$$ exists $$\mu$$-a.s., and $$Af\in \mathcal{L}_p(\mu)$$.
2. The map $$A:f\mapsto Af$$ is a positive $$\mathcal{L}_1(\mu)$$--$$\mathcal{L}_{\infty-}(\mu)$$ contraction and $$AT=TA=A$$.
3. If $$1, $$\|A_nf-Af\|_p\xrightarrow{n\rightarrow\infty}0$$.

If $$\mu(S)<\infty$$, then:

1. $$A$$ is an $$\mathcal{L}_\infty(\mu)$$ contraction.
2. $$\|A_nf-Af\|_1\xrightarrow{n\rightarrow\infty}0$$ for all $$f\in\mathcal{L}_1(\mu)$$

In the case of the OP, the operator is $$Tf(x)=f(x+1)$$. Clearly $$Tf\geq0$$ when $$f\geq0$$, and $$\|Tf\|_p\leq\|f\|_p$$ (in fact equal) for any $$1\leq p\leq \infty$$ (This means that $$T$$ is an positive $$L_1-L_\infty$$ contraction).

Fix $$1, and as before, $$S_nf(x):=\frac1n\sum^{n-1}_{k=0}f(x+k)$$ for any function $$f\in L_p$$.

Recall that the space $$\mathcal{C}_{00}(\mathbb{R})$$ of continuous functions of compact support is dense in $$L_p(\lambda)$$. For $$g\in C_{00}(\mathbb{R})$$ $$S_ng(x)=\frac1n\sum^{n-1}_{k=1}g(x+k)\xrightarrow{n\rightarrow\infty}0$$ and so $$S_ng$$ also converges in measure to $$0$$.

Showing that $$\|S_ng\|_p\xrightarrow{n\rightarrow\infty}0$$ for all $$g\in C_{00}(\mathbb{R})$$ would imply the result for all $$f\in L_p$$. Indeed, given $$f\in L_p$$ and $$\varepsilon>0$$ choose $$g\in C_{00}(\mathbb{R})$$ such that $$\|f-g\|_p<\varepsilon$$. Then $$\|S_nf\|_p\leq \|S_n(f-g)\|_p+\|S_ng\|_p<\varepsilon+\|S_ng\|_p$$ whence we obtain $$\limsup_n\|S_nf\|_p\leq\varepsilon, \qquad\varepsilon>0$$ Thus, $$\|S_nf\|_p\xrightarrow{n\rightarrow\infty}0$$.

Edit:

After more thinking I realized that we can do without having to check for uniform integrability and instead directly prove that $$\|S_ng\|_p\xrightarrow{n\rightarrow\infty}0$$ for all $$g\in\mathcal{C}_{00}(\mathbb{R})$$. The support of $$g$$ is contained in some intercal of the form $$I:=(k,m]$$ for $$k,m\in\mathbb{Z}$$. Splitting $$I$$ in $$m-k$$ of length $$1$$ we write $$g=\sum^{m-k-1}_{\ell=0}g\mathbb{1}_{(k+\ell,k+\ell+1]}$$. Let $$\phi_{\ell}:=g\mathbb{1}_{(k+\ell,k+\ell+1]}$$. Since the operator $$S_n$$ is linear, we have that $$S_ng=\sum^{m-k-1}_{\ell=0}S_n(\phi_\ell)$$

The advantage of breaking $$g$$ in pieces of $$\phi_\ell$$ of support of length one is that successive integer translations yield functions whose supports whose interiors are pairwise disjoint, for $$\operatorname{supp}(\phi_\ell\circ T^j)=[k+\ell-j,k+\ell+1-j].$$ Then $$\|S_ng\|_p\leq\sum^{m-k-1}_{\ell}\|S_n\phi_\ell\|_p$$ and by translation invariance, for each $$0\leq \ell < m-k$$ \begin{align} \|S_n\phi_\ell\|^p_p&=\frac{1}{n^p}\int\Big|\sum^{n-1}_{j=0}\phi_\ell\circ T^k\Big|^p=\frac{1}{n^p}\int\sum^{n-1}_{j=0}|\phi_\ell\circ T^k|^p=\frac{1}{n^{p-1}}\|\phi_\ell\|^p_p \end{align} Putting things together, we obtain that $$\|S_ng\|_p\leq n^{\tfrac1p -1}(m-k)\|g\|_p\xrightarrow{n\rightarrow\infty}0$$