Fix $1 < p < \infty$. Given $f \in L^p(\mathbb{R})$ define $f_n(x) = n^{1/p}f(nx)$ for $n = 1, 2, \dots$. Prove that $f_n$ converges weakly to $0$ in $L^p$.
I'm really confised about this question. I know that I have to show that for every $g \in L^{p'}(\mathbb R)$, $\int_{\mathbb{R}} f_ng \to 0$, but I can't even visualize why $f_n$ would go to $0$ using any convergence. How do I "see" that things converge weakly? Also, how would I prove something like this?
Warning: very loose answer, just for intuition purposes.
The function is integrable over all of $\Re$. That means the as $|x| \rightarrow \infty$,$f(x) \rightarrow 0$. The function sequence looks like f(nx). That means for every $x \neq 0$, $nx \rightarrow \infty$, so $f(nx) \rightarrow 0$ as $nx \rightarrow \infty$. Now for the $n^{1/p}$. Let's look at p=1. If n*f(nx) did not go to zero, then f(x) would look like 1/x going to infinity. But the integral of that does not converge.