Let $(f_{n})_{n} \subseteq L^{2}(\mathbb R, \lambda)$ where $f_{n}(x)=1_{[n,n+1]}(x)$. Investigate:
$1. \lambda-$a.e. convergence
$2. $ weak convergence
$3.$ strong convergence
My ideas:
$1.$ for any $ x \in \mathbb R$ $\exists N \in \mathbb N$ namely where $N \geq x$ so that $f_{n}(x)=0$ for all $n \geq N$, thus $(f_{n})_{n}$ converges pointwise and further will then converge $\lambda-$a.e. to $0$
$2.$ Let $\epsilon > 0$. Since $L^{2}$ is a Hilbert space, we can identify the dual of $L^{2}$ as $L^{2}$ itself. And rather than taking $\ell \in (L^{2})^{*}$, we will take $g \in L^{2}$. But now do I show convergence according to the $L^{2}-$norm? I assume $f=0$ is a good candidate. $\vert \vert f_{n}g-fg\vert \vert_{2}=\vert \vert (f_{n}-0)g\vert \vert_{2}=\vert \vert f_{n}g\vert \vert_{2}=\int_{[n,n+1]}g^{2}d\lambda$ but since $g \in L^{2}$ we can choose an $ N \in \mathbb N$ so that $\int_{\vert x \vert > N}g^2 d\lambda<\epsilon$. Now $\forall n \geq N$: $\vert \vert f_{n}g\vert \vert_{2}=\int_{[n,n+1]}g^{2}d\lambda \leq \int_{\vert x \vert > N}g^2 d\lambda<\epsilon\Rightarrow f_{n}g\xrightarrow{L^{2}}fg \Rightarrow f_{n} \xrightarrow{w} f$
$3.$ $\lambda-a.e.$ convergence does not imply strong convergence. How do we find an ideal candidate for strong convergence? I'd assume it'd be $0$ but I am not sure how to go about it. I'd say $f_{n} \leq f_{n+1}$ for all $n \in \mathbb N$ and $\lim_{n \to \infty} f_{n}(x)=0$ for any $x \in \mathbb R$. Thus by monotone convergence theorem: $\lim\limits_{n \to \infty}\int_{\mathbb R}\vert f_{n}-0\vert^{2} d\lambda=\lim\limits_{n \to \infty}\int_{\mathbb R}\vert f_{n}\vert^{2} d\lambda=\int_{\mathbb R}\lim\limits_{n \to \infty}\vert f_{n}\vert^{2} d\lambda=0$
Any corrections?
My questions:
I am interested in the interplay between the above three notions of convergence:
I know that strong convergence (in this case $L^{2}$ convergence) implies weak convergence but has no relationship (in terms of implications) to $\lambda-$a.e.. But in general, is the pointwise limit of a sequence of functions a good choice for convergence of $(f_{n})_{n}$ in $L^{2}$ ? And more importantly, is the pointwise limit the only possible limit in $L^{2}$. In other words, can I have $f_{n}(x) \xrightarrow{n\to \infty} f(x)$ for all $x \in \mathbb R$ but there is a $g \neq f$ so that $f_{n} \xrightarrow{ L^{2}} g$. Does the converse also hold?
If $f_n \to f$ in $L^{2}$ then there is a subsequence which converges to $f$ almost everywhere. Hence, if we also know that $f_n \to g$ a.e. the $f=g$ a.e. This essentially answers all your questions: since $\|f_n\|_2$ does not tend to $0$ in the given example it follows that strong convergenec fails.