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I wonder if my reasoning works: Given $1 < p < \infty$ let $(a_n)$ be a sequence of real numbers and define $f_n(x) = a_n$ if $x \in [n,n+1]$ and $0$ otherwise. I claim that if $(a_n)$ is bounded then $f_n \rightharpoonup 0$ in $L^p$.

My idea is that $$\int f_ng = \int_n^{n+1}a_ng \leq M\int_n^{n+1}g \leq M\|g\|_q, \ \ \forall g \in L^q(\mathbb{R}).$$Now since the right side is independent of $n$ we can take the limit as $n \rightarrow \infty$ and since $g$ is arbitrary, we can make the the right side as small as we want. Thus showing that $\lim_{n\rightarrow \infty}\int_\mathbb{R}f_ng \leq \epsilon$.

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    $\begingroup$ By Jensen's inequality and dominated convergence: $|\int_{[n,n+1]}g|\leq \|\mathbf{1}_{[n,n+1]}g\|_{L^q}\to 0$. $\endgroup$ Commented Jan 2, 2024 at 17:43
  • $\begingroup$ @Snoop I would like to say that I created another post related with your answer. Thank you! $\endgroup$ Commented Jan 2, 2024 at 20:04

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Let $M=\sup|a_n|.$ If $g\in L^q$ then by the Hölder inequality we get $$ \int\limits_\mathbb{R}|f_n||g|\le |a_n|\left (\int\limits_n^{n+1}|g|^q\right )^{1/q}$$ Hence $$M^q\|g\|_q^q\ge M^q\sum_{n=1}^\infty \int\limits_n^{n+1} |g|^q\\ \ge \sum_{n=1}^\infty |a_n|^q\int\limits_n^{n+1} |g|^q\ge \sum_{n=1}^\infty\left (\int\limits_\mathbb{R}|f_n||g|\right )^q$$ The general term of the last series tends to $0$ which implies $\int f_ng\to 0.$

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  • $\begingroup$ Clean, thank you! $\endgroup$ Commented Jan 2, 2024 at 21:07
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There are two problems in your reasoning. The first one is that you integrate $g$ when $g$ does not belong to $L^1$ all the time, and this quantity may not exist. It belongs to $L^q$ where $q$ is such that $1/p + 1/q = 1$. In order to avoid this, try using Holder's inequality. Then, you cannot change the test function like that since in order to prove weak convergence, you have to prove that for any test function $g$, $\int f_n g \rightarrow 0$.

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  • $\begingroup$ That's true and I could simply use Holders inequality at the beginning. In fact, in my last inequality I used Holders. But if I understood you correctly, how would Holders help if I need to fix $g$ and make it as small as I want? $\endgroup$ Commented Jan 2, 2024 at 17:24
  • $\begingroup$ Because you do not write $\int_n^{n+1} g$ but $\int_n^{n+1} |g|^q$ instead. Then what can you say about $g$ if $\limsup \int_n^{n+1} |g|^q > 0$ ? $\endgroup$ Commented Jan 2, 2024 at 17:27
  • $\begingroup$ Where does that $|g|^q$ appears from? Because we want to show that $(g,f_n) = \int gf_n \rightarrow 0$ as $n \rightarrow \infty$. What we could do since $f_n$ is clearly on $L^p$ is use Holders, $\int f_ng \leq \|f_n\|_p\|g\|_q = a_n\|g\|_q.$ But we are back at the same. $\endgroup$ Commented Jan 2, 2024 at 18:39
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    $\begingroup$ for $g \in L^q$, you have $|\int f_n g |\leq \int |f_n g| = \int_n^{n+1} |f_n g| \leq \left( \int_n^{n+1} |f_n|^p \right)^{1/p} \left( \int_n^{n+1} |g|^q \right)^{1/q} = |a_n| \left( \int_n^{n+1} |g|^q \right)^{1/q} \leq M \left( \int_n^{n+1} g^q \right)^{1/q}$. Now, if $\limsup_n \left( \int_n^{n+1} g^q \right)^{1/q} >0$, you have $g \notin L^q$. Hence, $\limsup_n \left( \int_n^{n+1} |g|^q \right)^{1/q} =0$ and thus $\int f_n g \rightarrow 0$. $\endgroup$ Commented Jan 2, 2024 at 18:51
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    $\begingroup$ The $\leq M \dots$ is simply a weird formatting of the website. It is part of the big series of inequalities before. $\endgroup$ Commented Jan 2, 2024 at 20:37

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