Let $1\leq p < \infty$ and consider a sequence $(x_{n})_{n}\subseteq L^{p}[0,1]$. Show the equivalence of:

$1.$ $x_{n} \xrightarrow{ w} 0$

$2.$ $\sup\limits_{n \in \mathbb N} \vert \vert x_{n}\vert\vert_{p}<\infty $ and $\int_{A}x_{n}(t)dt\xrightarrow{n \to \infty} 0$ for any borel sets $A$ on $[0,1]$.

for $1. \Rightarrow 2.$

note that for any $\ell \in (L^{p})^{*}$ we have $\ell(x_{n})\xrightarrow{n \to \infty} 0$ and hence:

$\sup\limits_{n \in \mathbb N}\vert \ell(x_{n})\vert<\infty$ for any $\ell \in (L^{p})^{*}$ but now for $1<p < \infty$ we know that $L^{p}$ is reflexive, so we can write: $\sup\limits_{n \in \mathbb N} \vert \vert x_{n}\vert\vert_{p}=\sup\limits_{n \in \mathbb N} \vert \vert Jx_{n}\vert\vert_{*}<\infty$ by the uniform boundedness principle and the fact that $\sup\limits_{n \in \mathbb N} \vert \ell (x_{n})\vert= \sup\limits_{n \in \mathbb N}\vert Jx_{n}(\ell)\vert$ by reflexiveness. But we have now only shown this for $1< p < \infty$, since $L^{1}$ is not reflexive. How do we show this when $p=1$?

For "$\int_{A}x_{n}(t)dt\xrightarrow{n \to \infty} 0$ for any borel sets $A$ on $[0,1]$", consider any $1 \in L^{q}[0,1]$, note that because of Riesz there has to exist a $\ell \in (L^{p})^{*}$ so that $\ell(x)=\int 1 x(t)dt $ for any $x\in L^{q}$. Then it is clear that $\ell(x_{n})=\int_{0}^{1}x_{n}(t)dt\xrightarrow{n \to \infty}0$ by assumption. But how can I show that $\int_{A}x_{n}(t)dt\xrightarrow{n \to \infty}0$ for any Borel set $A$? Particularly since I have no assumption on whether the $(x_{n})_{n}$ are positive functions?

for $2. \Rightarrow 1.$

note that for any $\ell \in (L^{p})^{*}$ we can find a unique $y \in L^{q}$ where $\frac{1}{p}+\frac{1}{q}=1$ so that:

Note that $\vert \ell(x_{n})\vert =\vert\int_{0}^{1}y(t)x_{n}(t)dt\vert\leq \sup\limits_{n \in \mathbb N} \vert \vert x_{n}\vert\vert_{p}\vert\vert y\vert\vert_{q}<\infty$ and can I state from $\int_{A}x_{n}(t)dt\xrightarrow{n \to \infty} 0$ for any borel sets $A$ that $x_{n} \to 0$ almost everywhere, but I am not sure as the assumption of positivity of $x_{n}$ is once again missing. If I could use this and dominated convergence, then

$\lim\limits_{n\to \infty}\ell(x_{n})=\lim\limits_{n\to \infty}\int_{0}^{1}y(t)x_{n}(t)dt=0$ and we are done.

  • 1
    $\begingroup$ The same question was asked here so you should take a look. I'm not marking this as a duplicate of that question because I think you have questions about your particular attempt which aren't covered there. $\endgroup$ – Rhys Steele Jul 21 '19 at 10:49
  • The statement "a weakly convergent sequence $x_n$ is bounded in norm" is true in every Banach space $X$ and does not need reflexivity. Even when $X$ is not reflexive, you can still identify your vectors $x_n$ with bounded linear functionals on $X^*$ in the usual way. Then the uniform boundedness principle lets you conclude that $\sup_n \|x_n\|_{X^{**}} < \infty$. But by the Hahn-Banach theorem you can show that $\|x\|_X = \|x\|_{X^{**}}$ for $x \in X$, i.e. the natural map of $X$ into $X^{**}$ is an isometry. So your argument that $\sup_n \|x_n\|_{L^p} < \infty$ also works when $p=1$.

  • Hint: for any Borel set $A$, the indicator function $1_A$ is in $L^q([0,1])$. You can write $\int_A x(t)\,dt = \int_0^1 1_A(t) x(t)\,dt$ and use Hölder's inequality.

  • If 2 holds, then show that $\int_0^1 x(t) y(t) \,dt \to 0$ for all simple functions $y$. Now take advantage of the fact that simple functions are dense in $L^q$. There's a little trick with the triangle inequality, and your assumption that $\sup_n \|x_n\|_{L^p} < \infty$ will be essential.

  • $\begingroup$ On your hint: for what do we need to use Hölder? Surely the fact that $x_{n} \xrightarrow{w} 0$ shows that indeed $0=\lim\limits_{n\to \infty} \int x_{n}(t) 1_{A}(t)dt = \lim\limits_{n\to \infty} \int_{A} x_{n}(t)dt$ for any Borel set $A$ $\endgroup$ – SABOY Jul 21 '19 at 19:20
  • $\begingroup$ @SABOY: Hölder's inequality is used to verify that $x \mapsto \int x(t) 1_A(t)\,d$ is a continuous linear functional on $L^p$, which you need to be able to take advantage of the weak convergence. $\endgroup$ – Nate Eldredge Jul 22 '19 at 1:16
  • $\begingroup$ I don't understand. Surely the fact that for any $\ell \in (L^{p})^{*}$, where $1 \leq p < \infty$ we know that by the Riesz representation there exists a unique $g \in L^{q}$ so that $\ell (f) = \int fg $, so $x \mapsto \int x(t)1_{A}(t)dt$ is bounded by definition? I struggle to see why we need Hölder. $\endgroup$ – SABOY Jul 22 '19 at 8:52
  • $\begingroup$ @SABOY: Maybe you already know the fact that "for any $g \in L^q$, the functional $\ell(f) = \int fg$ is a bounded linear functional on $L^p$", but at some point it had to be proved, and Hölder's inequality is the most convenient proof I know. Please note carefully that the fact I wrote down is not the same as the one you wrote down, but is in some sense its converse. [...] $\endgroup$ – Nate Eldredge Jul 22 '19 at 14:50
  • $\begingroup$ @SABOY: As you've written your fact, we cannot apply it to $\ell(f) = \int f 1_A$ until we know that $\ell \in (L^p)^*$, which is to say that we must first show that $\ell$ is a bounded linear functional. And once we have shown that, we do not actually need the fact you wrote down. $\endgroup$ – Nate Eldredge Jul 22 '19 at 14:50

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