Proving that a sequence converges weak-star in $L^\infty(\mathbb R)$

Question

Let $f \in L^\infty(\mathbb R)$ be a periodic function with period $1$. How can one prove that the sequence $\{f_n\}_{n \in \mathbb N}$ of functions given by $$f_n(x)=f(nx)$$ converges weak-star in $L^\infty(\mathbb R)$ where the limit is the constant function $\overline f$ with $$\overline f \equiv\int_0^1f(x) \, dx \, ?$$

Answer 1

Recall that $L^\infty(\mathbb{R})=\Big( L^1(\mathbb{R}) \Big)^*$, and that for a Banach space $X$ a sequence $y_n\in X^*$ converges to $y$ weak-* if $y_n(x)\rightarrow y(x)$ for all $x\in X$. Hence what you need to show is that $$ \int_\mathbb{R}g\cdot f_n dx \rightarrow \int_\mathbb{R}g\cdot \overline{f}dx \qquad \forall g\in L^1(\mathbb{R}). $$

We can notice by periodicity of $f$ that $\int_{k}^{k+1} \big( f(x)-\overline{f} \big)dx=0$.

Let $g\in L^1(\mathbb{R})\cap C_c(\mathbb{R})$. Hence: $$ \int_\mathbb{R}g(x) f(nx) dx = \int_{-K}^Kg(x)f(nx)dx$$ and $$\int_\mathbb{R}g(x)\overline{f}dx = \overline{f}\int_{-K}^Kg(x)dx $$ for some $K\in \mathbb{N}$.

Rewriting this we get that: \begin{align} \int_\mathbb{R}g(x) \big(f(nx)-\overline{f} \big) dx & = \frac{1}{n}\int_{-nK}^{nK} g\left(\frac{x}{n}\right)\cdot \big(f(x)-\overline{f}\big)dx \\ & = \frac{1}{n} \overset{nK-1}{\underset{j=-nK}{\sum}}\int_{j}^{j+1}g\left(\frac{x}{n}\right)\cdot \big(f(x)-\overline{f}\big)dx. \end{align}

Since $g$ is continuous and compactly supported, we know that $-M \le g(x)\le M$ for some $M>0$.

Then it holds that \begin{align} \frac{1}{n} \overset{nK-1}{\underset{j=-nK}{\sum}}\int_{j}^{j+1}g\left(\frac{x}{n}\right)\cdot\big(f(x)-\overline{f}\big)dx & \le \frac{1}{n} \overset{nK-1}{\underset{j=-nK}{\sum}}\int_{j}^{j+1}M \cdot\big(f(x)-\overline{f}\big)dx \\ & = \frac{1}{n} \overset{nK-1}{\underset{j=-nK}{\sum}}M\int_{j}^{j+1}\big(f(x)-\overline{f}\big)dx = 0. \end{align}

Similarly $\frac{1}{n} \overset{nK-1}{\underset{j=-nK}{\sum}}\int_{j}^{j+1}g\left(\frac{x}{n}\right)\cdot\big(f(x)-\overline{f}\big)dx \ge 0$, hence for $\frac{1}{n}< \delta$ we obtain that $$\int_\mathbb{R}g(x) \big(f(nx)-\overline{f} \big) dx=0.$$

Since $C_c(\mathbb{R})$ is dense in $L^1$, this implies that for all $g\in L^1(\mathbb{R})$ there exists a sequence $g_m\in L^1(\mathbb{R})\cap C(\mathbb{R})$, $g_m \overset{L^1}{\rightarrow}g$ and $$ 0 =\int_\mathbb{R}g_m(x) \big(f(nx)-\overline{f} \big) dx \to \int_\mathbb{R}g(x) \big(f(nx)-\overline{f} \big) dx. $$

This argument is true for all $g\in L^1$ hence, $f_n$ converges weak-* to $f$.

Edit:(Answer to viktor question in the comments)

I think this could work. Given your rational edgepoints $a,b$, you can write for all $n$, $$ m_1(n)= \lceil{na} \rceil \quad m_2(n)=\lfloor nb \rfloor $$ where $na=m_1(n)-r_1(n)$ and $nb=m_2(n)+r_2(n)$. When $r_1(n)$ and $r_2(n)$ are zero you have your identity $$ \frac{1}{n}\int_{na}^{nb}f(u) du=(b-a)\int_0^1f(u)du. $$ You also have

$$ \begin{align} \frac{1}{n}\int_{na}^{nb}f(u) du&= \frac{1}{n}\int_{na}^{m_1(n)}f(u)du +\frac{1}{n}\int_{m_1(n)}^{m_2(n)}f(u)du+ \frac{1}{n}\int_{m_2(n)}^{nb}f(u)du \end{align}$$

Because $f\in L^\infty$, we can conclude

$$ \Bigg\vert\frac{1}{n}\int_{na}^{m_1(n)}f(u)du\Bigg\vert \leq \frac{1}{n} \Vert f \Vert_\infty \cdot \left( m_1(n)-na \right) \leq \frac{1}{n} \Vert f \Vert_\infty \cdot 1 \rightarrow 0,$$ and similarly

$$ \Bigg\vert\frac{1}{n}\int_{m_2(n)}^{nb}f(u)du\Bigg\vert \leq \frac{1}{n} \Vert f \Vert_\infty \cdot \left( nb-m_2(n) \right) \leq \frac{1}{n} \Vert f \Vert_\infty \cdot 1 \rightarrow 0. $$

Notice that $na-nb=m_2(n)-m_1(n) +\Big(r_2(n)+r_1(n) \Big)$, hence $$ \lim \frac{1}{n}\int_{na}^{nb}f(u) du = \lim \frac{m_2(n)-m_1(n)}{n}\int_0^1 f(u)du=(b-a)\int_0^1f(u)du $$