i have following problem:
Be $1<p<\infty$, $T \in \mathbb{R}$, $f \in L^{\infty}( \mathbb{R})$ with $f(x+T) = f(x)$ for $x \in \mathbb{R}$
$f'= \frac{1}{T} \int_0^T f(t) dt$
with a sequence $u_n \subset L^p (0,1)$ and $u_n =f(xn)$
I want to show
$i)$ for $u_n$ we have weak convergence in $L^p (0,1)$ against $f'$
$ii)$ find $lim ||u_n - f'||_{L^p}$
$iii)$ T=1 and $\alpha, \beta \in \mathbb{R}$. Find weak limit of $u_n$ for $$f = \left\{\begin{array}{rcl} \alpha &\text{on}& (0,1/2) \\ \beta &\text{on}& (1/,2) \\ \end{array} \right. $$
I think i solved i). However i dont know how to go on with ii) and iii). I guess its $\frac {\alpha +\beta }{2}$ for ii)
Anyone can help?
The answer to ii) is $||f-f'||_p$. In fact $||u_n-f'||_p$ does not depend on n!. To see this follow these simple steps: split $\int |u_n-f'|^{p}$ into integrals over the intervals $(kT,(k+1)T)$, k varying over all integers. In each term make the change of variable $y=nx$. Now observe that if a function g has period T then its integral over $(NnT,(N+1)nT)$ equals n times the integral over $(NT,(N+1)T)$. [The reason for this is we can split the integral into a sum of terms where each term is an integral over an interval of length $NT$; by periodicity g has the same integral over any of these intervals. Now just sum the terms you get as an integral over $\mathbb R$.