Showing that a sequence weakly converges but no subsequence converges strongly

Question

I have a sequence $(f_k)$ in $L_p(R)$ and I want to show that it weakly converges to $0$ but no subsequence converges strongly to $0$. The sequence is $$f_k = k^{1/p}1_{[0,1/k]}$$ and while I'm able to show weak convergence, I don't see how I can show that no subsequence $f_{k_j} \to 0$. I'm looking for a hint as to what theorem might be applicable.

Answer 1

Because $$\Vert f_k \Vert_p^p = \int_0^{1/k} |k^{1/p}|^p dx = \int_0^{1/k} k dx = 1$$ so $$\Vert f_k - 0 \Vert_p = \Vert f_k \Vert_p = 1 \not \to 0 ;$$ in particular, for any subsequence $\{ f_{k_i} \}_{i\in \Bbb{N}}$, the elements $f_{k_i}$ are still such that $$\Vert f_{k_i} \Vert_p = 1,$$ so we have the same conclusion.