Compute $\lim_n \|u_n-\bar f\|_p$

Question

Let $p \in [1, \infty]$. Let $f \in L^p_{\text{loc}} (\mathbb R)$ be $T$-periodic, i.e., $f(x+T) = f(x)$ a.e. $x \in \mathbb R$. Let $$ \bar f := \frac{1}{T} \int_0^T f (t) \, dt. $$

We define a sequence $(u_n) \subset L^p(0, 1)$ by $u_n (x) := f(nx)$ for all $x \in (0, 1)$. I'm trying to compute $\lim_n \|u_n-\bar f\|_p$. Could you have a check on my below attempt?

Let $\varphi(n) := \lfloor n/T \rfloor$. Then $$ \begin{align} \|u_n-\bar f\|_p^p &= \int_0^1 |f(nx)-\bar f|^p \, dx \\ &= \frac{1}{n} \int_0^n |f-\bar f|^p \\ &= \frac{\varphi(n)}{n} \int_0^{T} |f-\bar f|^p + \frac{1}{n} \int_{\varphi(n)T}^n |f-\bar f|^p. \end{align} $$

Clearly, $\frac{1}{n} \displaystyle\int_{\varphi(n)T}^n |f-\bar f|^p \xrightarrow{n \to \infty}0$. On the other hand, $$ \frac{\varphi(n)}{n} = \big \lfloor \frac{n}{T} \big \rfloor \frac{1}{n} \xrightarrow{n \to \infty} \frac{1}{T}. $$

It follows that $$ \lim_n \|u_n-\bar f\|_p = \bigg ( \frac{1}{T} \int_0^{T} |f-\bar f|^p \bigg)^{1/p}. $$

Mittens · Accepted Answer · 2023-04-08T19:35:21.893

1

Te calculation is actually much simpler that what you have. Notice that if $\phi$ is $T$ periodic and say, locally integrable, then for any $a\in\mathbb{R}$, $$\int^{T+a}_a\phi=\int^T_0\phi$$

From that observation, along with the change of variable $u=nx+b$, one gets \begin{align} \int^T_0\phi(nx+b)\,dx&=\frac{1}{n}\int^{nT+b}_b\phi(u)\,du\\ &=\frac{1}{n}\sum^{n-1}_{k=0}\int^{(k+1)T+b}_{kT+b}\phi(u)\,du\\ &=\frac{1}{n}\sum^{n-1}_{k=0}\int^{(k+1)T}_{kT}\phi(u)\,du\\ &=\int^T_0\phi(u)\,dx\end{align} In your case, take $\phi(x)=|f(x)-\overline{f}|$, where $f$ is a $1$-periodic function ($T=1$).

edited Apr 08 '23 at 19:35

answered Apr 08 '23 at 19:12

Mittens

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I could not see how your $\phi$ and your calculation are related to my question... – Analyst Apr 08 '23 at 19:17
Let $\phi(x)=|f(x)-\overline{f}|$. If $f$ is $T$ periodic (in your case $T=1$), then so is $\phi$ . Notice $\phi(nx)=|f(nx)-\overline{f}|$ – Mittens Apr 08 '23 at 19:22
But in my case, I integrate on $[0, 1]$, whereas you integrate on $[0, T]$. – Analyst Apr 08 '23 at 19:28
@Analyst: Take $T=1$, see my edit – Mittens Apr 08 '23 at 19:29
That's why my calculation is longer than yours, i.e., $\varphi(n) := \lfloor n/T \rfloor$ may be not an integer, so I must take care of the "decimal" part. – Analyst Apr 08 '23 at 19:32
You don't need $\varphi(n)=\lfloor n/T \rfloor$ at all! It is a direct computation. – Mittens Apr 08 '23 at 19:34
What do you mean by "a direct computation"? – Analyst Apr 08 '23 at 19:36
@Analyst: either the $T$ in your OP should be $1$, to the interval of interest should be $(0,T)$. It seems that you are using this posting to fill in the details of your previous posting. There they use $L_p(0,1)$ and the periodic functions are all $1$-periodic. The results in your posting hold $1$ is replaced by $T$, i.e., if one considers $L_p(0,T)$ and $T$-periodic functions. – Mittens Apr 08 '23 at 19:38
Could you give me the link to my previous question you are mentioning? – Analyst Apr 08 '23 at 19:40
Let us continue this discussion in chat. – Mittens Apr 08 '23 at 19:42

Compute $\lim_n \|u_n-\bar f\|_p$

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