Prove that $f_n \rightharpoonup f$ iff $\sup_n ||f_n||_{L^{p}} < \infty$ in $p \in (1, \infty), f_n,f \in L^p (0,1)$

Question

Let be $p \in (1, \infty), f_n,f \in L^p (0,1)$ .

a) Prove that $f_n \rightharpoonup f$ iff $sup_n ||f_n||_{L^{p}} < \infty$

and $\lim_{n \rightarrow \infty} \int_{0}^{x}f_n(y)dy = \int_{0}^{x}f(y)dy $

b) Is still true without the hypothesis $\sup_n ||f_n||_{L^{p}} < \infty$ ?

My ideas:

The hypothesis of the sup mean uniform continuity. For the weak convergence, I know that $\int f_ng \rightarrow \int fg$ for $g \in C_c^\infty$, $g \in L^\infty$. The weak convergence could be interpreted as $L^1$ norm, so we have:

$||f_ng||_{L^1} \leq||f_n||_{L^p}\cdot ||g||_{L^q}$ . Since the definition of $g$, the norm is ok (I suppose, is it ok?), so we have to check the other one.

$||f_n||_{L^p} = (\int_{0}^{1}|f_n|^p dx)^{p^{-1}} = (\int_{R}^{}X_{(0,1)}f_n^p dx)^{p^{-1}}$

This could be interpreted as $||X_{(0,1)}f_n^p ||_{L^1}$? Here I don't know how to continue; maybe this has something to do with the Banach Steinhouse theorem (the uniform limitedness principle). Simple functions notion could be used in this?

Another idea is to prove it by absurd, supposing that $\sup_n ||f_n||_{L^{p}} \not< \infty$

For the point b) I think that a good functions could be like $s(x) = (n \sin(nx))^n$, but I need to check if the hypothesis is needed (I suspect not).

I'm asking some tips for this exercise, because I'm stuck, thanks

{---------------- EDIT -----------------}

After the suggestions, here my tentative for point a)

$\lim_{n \rightarrow \infty} \int_{0}^{x}f_n(y)g(y)dy = > \int_{0}^{x}f(y)g(y)dy $

I assume $g:= \cup_i X_i = (0,x)$ where $x \in (0,1)$ where the $i$ are open disjointed set ($X_i \cap X_j$, $i \not = j$)

$\int_{0}^{1}f_n(y)\cup_i X_idy =\cup_i \int_{0}^{1}f_n(y) X_idy$ now I consider a single set, $X_k \in \cup_i X_i$. So we have:

$\int_{0}^{1}f_n(y)X_kdy$ this is a $L^1$ norm, so I can use Holder

$||f_n X_k|| \leq ||f_n||_{L^{p}} \cdot||X_k||_{L^{q}}$. $f_n$ converge in $L^p$ by assumption.

Now $\sup\{ \int_{0}^{1} f_n dy \}< \infty$ because if I split the function in the interval above, it converge in every interval so basically it'll be bounded and it'll have a finite sup.

Something is missing for this point?

Answer 1

a) It is assumed that the convergence $$ \tag{*}\lim_{n\to +\infty}\int_{(0,1)}f_n(y)g(y)dy=\int_{(0,1)}f(y)g(y)dy $$ for all function $g$ of the form $g\colon y\mapsto \mathbf 1_{(0,x)}(y)$, $x\in (0,1)$. Extend (*) to the case where $g$ is the indicator function of a disjoint union of open sets, then when $g=\mathbf 1_O$, where $O$ is an open set. And finally when $g$ is a simple function. Then approximate a function $g\in L^{p/(p-1)}$ in this space by a simple function and use the boundedness assumption.

b) It is known that a weakly convergence sequence is bounded. Here the question is rather whether the validity of (*) for all function $g$ of the form $g\colon y\mapsto \mathbf 1_{(0,x)}(y)$, $x\in (0,1)$ guarantee boundedness in $L^p$ of $(f_n)$. Look at functions $f_n$ of the form $c_n\mathbf 1_{(0,\delta_n)}$.