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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?

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    $\begingroup$ You have made it difficult to understand the question by putting Professors comment inbetween two conditions. Please edit the question. $\endgroup$ – Kavi Rama Murthy Sep 21 '19 at 12:25
  • $\begingroup$ Done! Hope it's more clear, let me know $\endgroup$ – Alessar Sep 21 '19 at 12:27
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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 $c_n\mathbf 1_{(0,\delta_n)}$.

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  • $\begingroup$ I'll let you know, thanks for this tip. One thing, how will I ever consider a space like $L^{p/{p-1}}$? I can't figure how to find/reach it $\endgroup$ – Alessar Sep 23 '19 at 5:46
  • $\begingroup$ It is the dual space of $L^p$: we represent a bounded linear functional by an element of $L^{p/(p-1)}$. $\endgroup$ – Davide Giraudo Sep 23 '19 at 6:39
  • $\begingroup$ Ok; the boundedness assumption is the point ii) of this I suppose math.stackexchange.com/questions/85009/…. Why I have to check the dual space? I can't use $L^p$? Pardon me, for the trivial questions, I m learning for my exam alone, I can t follow courses in university $\endgroup$ – Alessar Sep 23 '19 at 7:57
  • $\begingroup$ Ahhh I understand now! $f$ is by definition in Lp, so in the dual space, using Holder (because the weak condition is substantially a L1 norm), the indicator function is in the space Lq, that is the dual of Lp $\endgroup$ – Alessar Sep 23 '19 at 8:51
  • $\begingroup$ added my solution $\endgroup$ – Alessar Sep 23 '19 at 9:16

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