I would like to sort out the relations for strong/weak convergences for $L^p(X)$ mainly between $[p=1; p>1]$ and $[\mu(X) <\infty ; \mu(X) = \infty]$
For the purpose of strong/weak convergence, what is the intuitive difference between $L^1$ and $L^p: 1<p<\infty$?
- When the domain has finite measure, the implication is clear since $L^p\subset L^1$.
- For $m(X) = \infty$, I always think of the example $\chi_{[n,n+1]}$, it converges weakly to 0 in $L^p$ but not in $L^1$, since the value of $g\in {L^p}^*$ is essentially small outside of $[-N,N]$ which would make $\lim \int g\chi_{[n,n+1]} = 0$, but $\int 1\chi_{[n,n+1]} = 1$ for $1\in L^\infty$.
If $f_n \rightarrow f$ a.e. and $f_n \rightharpoonup f$ in $L^p: 1\leq p<\infty$, under what condition would these imply $f_n \rightarrow f$ strongly.
For example, let $f_n\geq 0$, $f_n\rightarrow 0$ a.e. and $f_n \rightharpoonup 0$ in $L^1$, then take $1\in L^\infty$, we have $$\lim_n \int f_n = 0$$ which means $||f_n||_1 \rightarrow 0$. Does this imply $f_n \rightarrow 0 $ strongly in $L^1$ even without the assumption of $f_n$ pointwise a.e. convergence to $0$. If this were true, then we could say more about the case when $\mu(X) <\infty$.
When $\mu(X) <\infty$ I know strong/weak convergence in $L^p: 1<p<\infty$ implies strong/weak convergence in $L^1$. Can we say anything more about the chase when $\mu(X) = \infty$?
Thank you very much!
To partially address 2.:
In $L_1(X,\mu)$ with $\mu(X)<\infty$, a weakly convergent sequence is uniformly integrable. This follows, for instance, from the Dunford-Pettis theorem. See Theorem 3 from this paper (which is certainly worthy of careful study) for the statement and proof of the Dunford-Pettis theorem.
So, if $(f_n)$ is weakly convergent and a.e. pointwise convergent in $L_1(X,\mu)$ with $\mu(X)<\infty$, it then follows from Vitali's Convergence Theorem that $(f_n)$ is norm convergent.
If $f_n\ge 0$ and $(f_n)$ weakly convergent to $0$ in $L_1$, then, as you show, $(f_n)$ is norm convergent to $0$. Without the positivity condition, this is false. For example, the Rademacher functions converge weakly to $0$ in $L_1[0,1]$, but not in norm to $0$.
For $X$ of infinite measure, I'm not sure whether or not weak and pointwise convergence in $L_1(X)$ implies norm convergence.
For $1<p<\infty$, the sequence $(f_n)$ with $f_n=n^{1/p}\chi_{[0,1/n]}$ converges pointwise a.e. and weakly to $0$ in $L_p[0,1]$, but each $f_n$ has norm $1$.
On the other hand, see this post for a sufficient condition that a weakly convergent sequence in $L_p$, $1<p<\infty$, be norm convergent.