The problem is the next:
$Let\ 1<p<\infty\ and \ let \ f_n=n^{1/p}f(nx)\ where\ f \ is\ a \ L^p\ function, \ examine\ the\ weak \ and\ strong\ convergence\ in \ L^p $
i could prove that $\parallel f_n\parallel_p=\parallel f\parallel_p$ , and $f_n$ is weakly convergent to $0$ but with the strong convergence i have doubts,
I don't know if it's okay, my idea is the following
we have that :
$\int_{\mathbb{R}}\liminf| f_n - f|^p dx \leq \liminf\parallel f-f_n\parallel_p \leq 2 \parallel f\parallel_p <\infty $
and writing the expression on this form :
$\int_{\mathbb{R}}| f_n - f|^p dx= \int_{\mathbb{R}}n| f(nx)|^p|1- \frac{f(x)}{f_n (x)}|^p dx$
and considering that exist $M$ such that $\lim_{n\rightarrow\infty} |f(nx)|^p < M$
we have that $\frac{f(x)}{f_n (x)}$ must necessarily tend to $1$ when $n\rightarrow\infty$
then $\lim |f_n (x) -f(x)|^p =0$
then, if the writing is correct, It would only be necessary to justify the passage of the limit under the integral and the exercise would be complete
Any help is welcome, best regards.
If $f=0$, no problems, so assume $f≠0$. Since you proved that $f_n \rightharpoonup 0$ and $\|f_n\|_{L^p} = \|f\|_{L^p}$, now it is easy to conclude.
If $f_n$ was strongly converging, then it would be strongly converging to $0$ (since it is the weak limit). But $\|f_n\|_{L^p} = \|f\|_{L^p} ≠ 0$ so $f_n$ does not converge strongly to $0$ ... contradiction.
Hence, $f_n$ does not converge strongly.