# Is it necessary to consider the case $p=1$ separately?

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 $$1 (since $$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}

• 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. Apr 6, 2023 at 18:05
• @tomasz Ah I terribly misunderstood $L^\infty$. Could you elaborate on my second question, i.e., how $\lim \sup _{n \rightarrow \infty} \left|\int v_n \varphi-\bar{g} \int \varphi\right| =0$? Apr 6, 2023 at 18:08
• @Analyst: this is a result by Fejer Apr 6, 2023 at 18:18
• 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). Apr 8, 2023 at 19:01
• @Analyst: In your problem, it should be that either $T=1$, or the $L_p$ spaces involved are $L_p(0,T)$. Apr 8, 2023 at 20:09

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 $$\sigma(L_p,L_{p'})$$.