I have to prove the following:
Proposition: Let $$\mu := \mathcal{L}^1 \big|_{[0,1]} $$ and $1<p< \infty$. Consider the sequence of functions $ \{f_h \}_{h>0} \subset L^p(\mathbb{R}, \mu)$ where $f_h(x)=f(hx)$ with $$ f(x) := \begin{cases} 1 \quad & \text{ if } \quad 0 \le \{x\} < \frac{1}{2} \\ -1 \quad &\text{ if } \quad \frac{1}{2} \le \{x\} < 1 \end{cases}$$ where $\{x\}$ is the fractional part of $x$. Let $\nu:= \frac{1}{2} ( \delta_1 + \delta_{-1})$. Then
- For each $h>0$ it holds $$ ((f_h)_{\#}) (\mu) = \nu $$
- $f_h \overset{\ast}{\rightharpoonup} 0$ as $h \to 0$
where $((f_h)_{\#}) (\mu)(B) = \mu ( f_h^{-1}(B))$ for any borel set $B$.
I'm not sure they are true; I suspect it should be $h>1$ and $h \to + \infty$. Indeed:
$ ((f_h)_{\#}) (\mu) = \nu \Leftrightarrow \mu ( f_h^{-1}((a,b))= \nu((a,b))$ for every $-\infty< a <b <+ \infty$. If both (or none of) $1$ and $-1$ belong to $(a,b)$ the equality is easy. Suppose $1 \in (a,b)$ and $-1 \notin (a,b)$, and take for example $h=1/2$ then $$\nu((a,b))= \frac{1}{2} $$ and $$ \mu (f_{1/2}^{-1}((a,b))) = \mu (\{ x \in [0,1] \mid f_{1/2}(x) = 1 \}) = \mu ( \{x \in [0,1] \mid 0 \le \{ \frac{1}{2} x \} < 1/2 \} ) = 1$$ On the other hand, if $h \ge 1$ I think the equality holds. Am I wrong?
If $h < 1/2$ and $\varphi \in L^q(\mathbb{R}, \mu)$ with $1/p + 1/q =1$ then
$$ \int_{\mathbb{R}} f_h \varphi d \mu = \int_{ \{ x \in [0,1] \mid 0 \le \{ hx \} < 1/2 \} } \varphi d \mathcal{L}^1 = \int_{[0,1]} \varphi d \mathcal{L}^1 = \int_{[0,1]} \varphi d \mu $$ and then I have $f_h \overset{\ast}{\rightharpoonup} 1$ as $h \to 0$. I think I can prove that, if $h \to + \infty$, the set on which $f_h = 1$ becomes union of smaller and smaller disjoint intervals which total measure is always 1/2. The same for the set on which $f_h = -1$. But, even if it is true, I don't know how to prove statement 2 (with $h \to + \infty$).