# $x_{n} \xrightarrow{w} 0 \iff \sup\limits_{n \in \mathbb N}\vert \vert x_{n} \vert \vert_{L^{p}} < \infty$

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 $$1 we know that $$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.

• 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. – Rhys Steele Jul 21 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.
• 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 $A$ – SABOY Jul 21 at 19:20
• @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. – Nate Eldredge Jul 22 at 1:16
• 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. – SABOY Jul 22 at 8:52
• @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. [...] – Nate Eldredge Jul 22 at 14:50
• @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. – Nate Eldredge Jul 22 at 14:50