Let $1<p_0<\infty$. Find an example of a sequence $\{f_k\}$ such that $f_k\in L^p$ for $1\le p <\infty$, $f_k\to0$ in $L^p$ for $1\le p < p_0$, but $f_k$ does not converge in $L^{p_0}$.
I thought of $f_k = k^{1/p_0}\chi_{(0,1/k)}$. Then $||f_k||_p=k^{1/p_0- 1/p}$. So $f_k\in L^p$ for $1\le p <\infty$. Also for $1\le p <p_0$, $||f_k||_p\to0$.
For $p_0<p$, $||f_k||_p\to\infty$. So suppose there exists $f$ such that $||f_k-f||_p\to 0$. Then by Minkowski's inequality, $||f_k||_p \le ||f_k-f||_p +||f||_p$. That means $||f||_p$ is unbounded. A contradiction.
Is my proof correct? Is there any flaw? Thanks!
EDIT: OK. I realized this shows $f_k$ doesn't converge in $L^p$ for $p>p_0$. But it's inconclusive when $p=p_0$. Any suggestion? Thanks.
Assume that $f_n\to f$ in $L^{p_0}$. Then there is $n_k\uparrow\infty$ such that $\lambda\{|f_{n_k}-f|\gt 2^{-k}\}\leqslant 4^{—k}$ (construct the sequence $(n_k,k\geqslant 1)$ by induction, using Markov's inequality and convergence in $L^{p_0}$. By Borel-Cantelli lemma, we have that $\lambda\{\limsup_k|f_{n_k}-f|\gt 2^{-k}\}=0$, hence $f_{n_k}\to f$ almost everywhere (the subsequence is common).
We prove that an arbitrary sequence of functions $(f_n)$ on a finite measure space converging in an $L^{p_0}$ to a $f$ converges also almost everywhere, up to a subsequence.
This proves that if $f_n\to f$ in $L^{p_0}$, then $f_{n_k}\to f$ a.e. for some $n_k\uparrow\infty$. But $f_n\to 0$ almost everywhere, so $f=0$. Since $\lVert f_n\rVert_{p_0}=1$, we get a contradiction.