convergence in weak Lp

2.7k Views Asked by At

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.

1

There are 1 best solutions below

0
On

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.