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 $x \in (0, 1)$.

Theorem $u_n \to \bar f$ in the weak topology $\sigma(L^p, L^{p'})$ where $p'$ is the Hölder conjugate of $p$.

I'm reading the proof of above theorem, i.e.,

Proof First, it is easy to check that $\int_a^b u_n(t) d t \rightarrow(b-a) \bar{f}$ (for every $\left.a, b \in(0,1)\right)$. This implies that $u_n \rightarrow \bar{f}$ weakly $\sigma(L^p, L^{p^{\prime}})$ whenever $1<p \leq \infty$ (since $p^{\prime}<\infty$, step functions are dense in $L^{p^{\prime}})$. When $p=1$, i.e., $f \in L_{\mathrm{loc}}^1(\mathbb{R})$, there is a $T$-periodic function $g \in L^{\infty}(\mathbb{R})$ such that $\frac{1}{T} \int_0^T|f-g|<\varepsilon$ (where $\varepsilon>0$ is fixed arbitrarily). Set $v_n(x)=g(n x), x \in(0,1)$ and let $\varphi \in L^{\infty}(0,1)$. We have $$ \left|\int u_n \varphi-\bar{f} \int \varphi\right| \leq 3 \varepsilon\|\varphi\|_{\infty}+\left|\int v_n \varphi-\bar{g} \int \varphi\right| $$ and thus $\lim \sup _{n \rightarrow \infty}\left|\int u_n \varphi-\bar{f} \int \varphi\right| \leq 3 \varepsilon\|\varphi\|_{\infty} \forall \varepsilon>0$. It follows that $u_n \rightarrow \bar{f}$ weakly $\sigma\left(L^1, L^{\infty}\right)$.

My question Clearly, $(0, 1)$ has finite Lebesgue measure, so $L^\infty (0, 1) \subset L^1(0, 1)$ and thus step functions are still dense in $L^\infty (0, 1)$. Could you explain why

  1. the author still considers the case $p=1$ separately?

  2. $\lim \sup _{n \rightarrow \infty} \left|\int v_n \varphi-\bar{g} \int \varphi\right| =0$ in the proof?

Update: I have added below steps for more clarity. We have $$ \begin{align} & \left|\int_0^1 u_n \varphi-\bar{f} \int_0^1 \varphi\right| \\ \le{} & \int_0^1 |u_n - v_n| \varphi + \bigg | \int_0^1 v_n \varphi - \bar g \int_0^1 \varphi \bigg | + \int |\bar g - \bar f| \varphi \\ \le{} & \|\varphi\|_\infty \int_0^1 |u_n - v_n| + \bigg | \int_0^1 v_n \varphi - \bar g \int_0^1 \varphi \bigg | + \|\varphi\|_\infty |\bar g - \bar f|. \end{align} $$

First, $|\bar g - \bar f| = \frac{1}{T} \int_0^T|f-g|< \varepsilon$. Let $m := \lfloor n/T \rfloor$. Then $$ \begin{align} \int_0^1 |u_n - v_n| &= \int_0^1 |f(nx)-g(nx)| \, dx = \frac{1}{n} \int_0^n |f-g| \\ &= \frac{1}{n} \int_0^{mT} |f-g| + \frac{1}{n} \int_{mT}^n |f-g| \\ &= \frac{m}{n} \int_0^{T} |f-g| + \frac{1}{n} \int_{mT}^n |f-g| \\ &\le \frac{(m+1)\varepsilon T}{n} = \big ( \big \lfloor \frac{n}{T} \big \rfloor +1 \big )\frac{\varepsilon T}{n} \\ &\le 2 \frac{n}{T} \frac{\varepsilon T}{n} = 2 \varepsilon. \end{align} $$

  • $\begingroup$ Step functions are not dense in $L^\infty(0,1)$. Consider a measurable function with values $0$ and $1$ which changes the value (essentially) infinitely many times. $\endgroup$
    – tomasz
    Apr 6, 2023 at 18:05
  • $\begingroup$ @tomasz Ah I terribly misunderstood $L^\infty$. Could you elaborate on my second question, i.e., how $\lim \sup _{n \rightarrow \infty} \left|\int v_n \varphi-\bar{g} \int \varphi\right| =0$? $\endgroup$
    – Analyst
    Apr 6, 2023 at 18:08
  • 1
    $\begingroup$ @Analyst: this is a result by Fejer $\endgroup$
    – Mittens
    Apr 6, 2023 at 18:18
  • $\begingroup$ Hi @Analyst! I added a posting which shows how to derive the theorem in your posting. directly from Fejer's lemma. That also indicates that the result is primarily an $L_1$ result (hence the separate considerations suggested by the writer of the proof you transcribed). $\endgroup$
    – Mittens
    Apr 8, 2023 at 19:01
  • $\begingroup$ @Analyst: In your problem, it should be that either $T=1$, or the $L_p$ spaces involved are $L_p(0,T)$. $\endgroup$
    – Mittens
    Apr 8, 2023 at 20:09

2 Answers 2

  1. The step functions are not dense in $L^\infty(0,1)$. For any non-trivial set $E$ you have $\|1-1_E\|_\infty=1$.

  2. Because $g\in L^\infty(0,1)\subset L^2(0,1)$, the previous case applies to tell you that $v_n\to\overline g$ weakly. This means that $$ \lim_n\int v_n\varphi=\overline g\,\int\varphi $$ for all $\varphi\in L^2(0,1)$; in particular, for $\varphi\in L^\infty(0,1)$. Hence from $$ \left|\int u_n \varphi-\bar{f} \int \varphi\right| \leq 3 \varepsilon\|\varphi\|_{\infty}+\left|\int v_n \varphi-\bar{g} \int \varphi\right| $$ you get $$ \limsup_n\left|\int u_n \varphi-\bar{f} \int \varphi\right| \leq 3 \varepsilon\|\varphi\|_{\infty}+\limsup_n\left|\int v_n \varphi-\bar{g} \int \varphi\right|=3\varepsilon\,\|\varphi\|_\infty. $$ As this can be done for every $\varepsilon>0$, you have shown that $$ \limsup_n\left|\int u_n \varphi-\bar{f} \int \varphi\right| =0, $$ which proves that $$ \lim_n\left|\int u_n \varphi-\bar{f} \int \varphi\right| =0. $$


The result in the OP is primarily a $L_1$ result, and the Theorem stated in the OP for $p>1$ follows from the $L_1$ case.

The $L_1$ result is due to Féjer, and can be formulated as follows:

(Féjer) Suppose $f$ is a bounded measurable $T$ periodic function ($T>0$). For any $\phi\in\mathcal{L}_1(\mathbb{R})$ and numeric sequence $\alpha_n\in\mathbb{R}$, $$ \lim_n\int_{\mathbb{R}} \phi(x)f(nx+\alpha_n)\,dx=\Big(\frac{1}{T}\int^T_0f\Big)\int_{\mathbb{R}} \phi \tag{1}\label{one} $$

The theorem in the OP can then be obtained as follows. For $L^{\text{loc}}_1$ functions $f,g$, let us denote $$I_n(g;f):=\int_{\mathbb{R}}g(x)f(nx+\alpha_n)\,dx,\qquad I(f)=\int_\mathbb{R}f\,dx, \qquad \overline{f}=\frac1T\int^T_0f$$ Notice that $I_n$ is linear in both the first and second argument. For $p\geq1$, let $p'$ be its convex conjugate: $\frac{1}{p}+\frac{1}{p'}=1$.

Suppose $f\in L^{\text{loc}}_p(\mathbb{R})$, and that $f$ is $T$-periodic. Let $h\in C([0,T])$ such that $\|f-h\|_{L^p[0,T]}<\varepsilon$, and extend $h$ as a $T$-periodic function on $\mathbb{R}$. For $g\in L_{p'}([0,T])$, extend $g$ to $\mathbb{R}$ as $g(x)=0$ for $|x|>T$. Clearly $g\mathbb{1}_{[0,T]}\in L_1(\mathbb{R})$. Since $f$ and $h$ are $T$-periodic, \begin{align} |I_n(\mathbb{1}_{[0,T]}g; f - h)|&\leq \|g\|_{L_{p'}([0,T]}\Big(\int^T_0|f(nx+\alpha_n)-h(nx+\alpha_n)|^p\,dx\Big)^{1/p}\\ &=\|g\|_{L_{p'}([0,T]}\Big(\frac1n\int^{nT+\alpha_n}_{\alpha_n}|f(u)-h(u)|^p\,du\Big)^{1/p}\\ &=\|g\|_{L_{p'}([0,T]}\|f-h\|_{L_p[0,T]}<\varepsilon\|g\|_{L_{p'}[0,T]} \end{align} Then \begin{align} \big|I_n(g\mathbb{1}_{[0,T]};f)-\overline{f} I(g\mathbb{1}_{[0,T]})\big|&\leq \big|I_n(g\mathbb{1}_{[0,T]}; f-h)\big|+|I_n(g\mathbb{1}_{[0,T]};h)-\overline{h}I(g\mathbb{1}_{[0,T]})| \\ &\qquad + |I(g\mathbb{1}_{[0,T]})(\overline{h}-\overline{f})|\\ &\leq\|g\|_{L_{p'}[0,T]}\varepsilon+\|g\|_{L_{p'}[0,T]}\frac{\varepsilon}{T^{1/p}} +|I_n(g\mathbb{1}_{[0,T]};h)-\overline{h}I(g\mathbb{1}_{[0,T]})| \end{align} Fejer's lemma implies that $\lim_n|I_n(g\mathbb{1}_{[0,T]};h)-\overline{h}I(g\mathbb{1}_{[0,T]})|=0$; this means that $$\limsup_n\big|I_n(g\mathbb{1}_{[0,T]};f)-\overline{f} I(g\mathbb{1}_{[0,T]})\big|\leq C\|g\|_{L^{p'}[0,T]}\varepsilon$$ for some constant $C$ independent of $f$ and $g$. The following result follows

Theorem: Suppose $f\in L_p[0,T]$ and $g\in L_{p'}[0,T]$, where $\frac1p+\frac1{p'}=1$, $p\geq1$. For any numeric sequence $(\alpha_n)\subset\mathbb{R}$, $$\lim_n\int^T_0 g(x) f(nx+\alpha_n)\,dx=\Big(\frac1T\int^T_0f\Big)\int^T_0 g$$ Equivalently, if $f_n(x)=f(nx+\alpha_n)$, then $f_n\xrightarrow{n\rightarrow\infty} \frac{1}{T}\int^T_0f$ in the weak topology $\sigma(L_p,L_{p'})$.


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