I've seen this statement so I'm not wanting to argue it, but am thus curious as to where my logic falls short.
Folland 6.2 problem
If $1<p<\infty, \;f_n\rightarrow f$ weakly in $l^p(A)$ iff sup$_n||f_n||_p<\infty$ and $f_n\rightarrow f$ pointwise.
The classic example for $\{f_n\}\nrightarrow f $ pointwise but $\{f_n\}\rightarrow f$ in $L^p$ strongly is $\{f_n\}=\{\chi_{[0,.5]},\chi_{[0,.25]},\chi_{[.25,.5]},\chi_{[0,.125]},\chi_{[.125,.25]},…\}$ and so on in such a fashion w/ indicator functions of length $\frac{1}{2}$ to a power.
It is easy to show that this sequence does not converge pointwise but does converge in $L^p$ strongly for $1\le p<\infty$. If the sequence converges strongly then it must also converge weakly, so this sequence has to converge weakly but not pointwise inside $L^p\text{ for }1<p<\infty.$
Obviously my logic has to be wrong but where. Thanks for the help.
Best Answer
The question was answered in the comments: Folland's statement is for $\ell^p$, the example is for $L^p$. I'll add the reason for the difference. The underlying measure space of $\ell^p$ is discrete, with each point having positive measure. This allows us to test weak convergence against characteristic functions of singletons. By the definition of weak convergence, the integrals $\int f_n\chi_{\{p\}}$ must converge, which implies pointwise convergence of $f_n$.
For $L^p$, the underlying measure space contains no atoms, and the above reasoning does not apply.