prove a series weakly converges to its mean value Let $u ∈ L^∞ (\Bbb R)$ such that $u(x + 1) = u(x)$ almost for every $x ∈ \Bbb R$.
Let $ū = \int_0^1 u(y) dy$ and let's define $(u_n )_{n∈\Bbb N}$ by $u_n (x) = u(nx)$ $\forall n ∈ \Bbb N$ and for almost every $x ∈ \Bbb R$.
Prove that $u_n \rightharpoonup ū$ weakly in $L^p (0, 1)$.
My attempt:
Let $f\in L^{p'}$ where $1/p+1/p'=1$:
$|\langle f,u_n\rangle  - \langle rf,ū\rangle | \le ||u_n-ū||_{L^P(0,1)}||f||_{L^{p'}(0,1)}$ by Holder.
Now I should prove this quantity vanishes to 0 when $n\to \infty$ but I have no clue on how to do it?
I am not sure what limitations do we have due to the "almost everywhere" 1-periodicity of $u$ and $u_n$.
Thank you for your help.
 A: Using Holder's inequality at the beginning is too crude here because in order to conclude, you would need to prove that $u_n \to \overline u$ strongly in $L^p(0,1)$ which is not necessarily the case. 
Instead notice that \begin{align*}\int^1_0 u(nx)f(x)dx &= \frac 1 n\int^n_0 u(y) f(y/n)dy \\&= \frac 1 n\sum_{k=1}^{n} \int_{k-1}^k u(y)f(y/n)dy\\
&=\frac 1 n\sum^n_{k=1} \int^1_0 u(k-1 + z) f((k-1+z)/n) dz\\
&= \frac 1 n\sum^n_{k=1} \int^1_0 u(z) f((k-1+z)/n) dz\\
&= \int^1_0 u(z) \left[ \frac 1 n\sum^n_{k=1} f\left(\frac{z+k-1}n \right)\right] dz.
\end{align*} Now write $$\int^1_0 \overline u f(x)dx = \int^1_0 \int^1_0 u(z) f(x)dx\,dz.$$ Thus \begin{align*} \lvert \langle u_n,f \rangle -\langle \overline u, f\rangle \rvert &= \left \lvert \int^1_0 u(nx) f(x)dx - \int^1_0 \overline u f(x)dx \right \rvert \\
&= \left \lvert \int^1_0 u(z) \left[ \frac 1 n\sum^n_{k=1} f\left(\frac{z+k-1}n \right)\right] dz - \int^1_0 \int^1_0 u(z) f(x) dx\,dz \right\rvert \\ 
&= \left\lvert \int^1_0 u(z) \left[\frac 1 n\sum^n_{k=1} f\left(\frac{z+k-1}n\right)  - \int^1_0 f(x) dx \right]  dz\right\rvert\\
&\le \int^1_0 \lvert u(z) \rvert \left \lvert \frac 1 n\sum^n_{k=1} f\left(\frac{z+k-1}n\right)  - \int^1_0 f(x) dx \right \rvert dz.
\end{align*} Now you can use Holder's inequality, and setting $$f_n(z) = \frac 1 n\sum^n_{k=1} f\left(\frac{z+k-1}n\right), \,\,\,\,\, z \in [0,1],$$ you need to prove that for $f \in L^{p'}(0,1)$, we have $f_n \to \overline f$ in $L^{p'}(0,1)$. Since this $f_n$ is simply a Riemann approximation to $\overline f$, it is fairly straightforward to conclude by first considering $f \in C[0,1]$, and then using a density argument.
