Let $1 < p < \infty$. Let $u \in L^{\infty}(\mathbb{R})$ be a 1-periodic function, that is $\forall x \in \mathbb{R}, u(x+1) = u(x)$. For all $n \in \mathbb{N}$, consider $f_n(x) = u(nx)$ for all $x \in (0,1)$. Let $f: (0,1) \rightarrow \mathbb{R}$ be the constant function defined by $f(x) = \int_{0}^{1} u(t) dt$.
- Prove that $f_n \rightharpoonup f$ in $L^p(0,1)$.
I have already demonstrated that $f_n \rightharpoonup f$ in $L^p(0,1) \Leftrightarrow$ the sequence $(f_n)$ is bounded in $L^p(0,1)$ and $\forall x \in [0,1], \int_{0}^{x} f_n(y) dy \rightarrow \int_{0}^{x} f(y) dy$.
First of all observe that, $u \in L^{\infty}(\mathbb{R}) \Rightarrow f_n \in L^{\infty}(0,1), \forall n$ (this follows from the fact that $\sup_{x \in \mathbb{R}} |u(x)| = \sup_{x \in (0,1)} |u(nx)|, \forall n \in \mathbb{N}$). Since $\mathcal{L}^1 (0,1) < \infty$, we have that $f_n \in L^p(0,1), \forall 1 < p < \infty$. The next step is to show that $(f_n)_n$ is bounded in $L^p(0,1)$. However, note that $u \in L^\infty(\mathbb{R}) \Rightarrow \exists C > 0 : ||u||_{L^{\infty}(\mathbb{R})} < C$ and hence $ ||f_n||_{L^p(0,1)} \le ||f_n||_{L^{\infty}(0,1)} < C, \forall 1<p<\infty$. Finally, one has to prove the convergence of the integral. Fix $x \in [0,1]$, then $\int_{0}^{x} f_n(y) dy = \int_{0}^{x} u(ny) dy = \frac{1}{n} \int_{0}^{nx} u(t) dt$. For $n \rightarrow \infty$, I want to use the periodicity of $u$ to conclude that $\frac{1}{n} \int_{0}^{nx} u(t) dt \rightarrow x (\int_{0}^{1} u(t) dt) = \int_{0}^{x} dx \, (\int_{0}^{1} u(t) dt)$, but I am struggling to justify formally the reasoning.
- Prove that $f_n \rightarrow f$ in $L^p(0,1)$ $\Leftrightarrow u = const.$ a.e.
One implication is obvious: if $u$ is a.e. equal to a constant $c$, then $(f_n)_n$ is a constant sequence, $f_n(x) = c$ a.e. in $(0,1)$, and thus it strongly converges in $L^p(0,1)$ to this constant $f(x)= c$.
I'm having more troubles in demonstrating the direct implication. Being the weak limit $f$ a constant and assuming strong convergence to $f$, eventually the sequence will be constant. In order to prove that $u$ is constant a.e. I think I have to use once again the periodicity of $u$, to conclude that also the first terms of the sequence must be constant. I don't know if there exists a more straight-forward way to prove it. However, I cannot formalize my reasoning.
If anyone could give me a hint on how to proceed as well as checking if I am correct in my reasoning, it would be extremely appreciated.
(1) Let $U:= \int_0^1 u(x)dx$. Since $u$ is one-periodic, $$ \frac1n \int_0^{nx} u(t)dt = \frac1n\left( \lfloor nx \rfloor \cdot U + r_n\right), $$ where $r_n$ is an integral of $u$ on a interval of length $\le1$: $$ r_n = \int_{\lfloor nx \rfloor}^{nx} u(s)ds \le \|u\|_{L^\infty}. $$ Since $$ nx-1\le \lfloor nx \rfloor \le nx, $$ it follows $\frac1n \lfloor nx \rfloor \to x$.
(2) Using the coordinate transform $(0,1)\mapsto (0,n)$ one can prove $\|f_n\|_{L^p}=\|u\|_{L^p}$. Similarly one can show $$ meas\{x: \ f_n(x)\ge c\} = meas\{x:\ u(x) \ge c\} $$ for all $c$. As strong convergence in $L^p$ implies convergence in measure, this should help to prove that $u=const$ in case the convergence is strong.