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I want to know if this statement is right?

Let $(g_n)$ be a measurable sequence from $[0,T]$ to $\mathbb{R}^n$ such that $\exists \beta \in L^{\infty}_{\mathbb{R}_+}([0,T])$ where $|g_n(t)| \leq \beta(t) \forall t \in [0,T]$ then $(g_n) \in L^{\infty}_{\mathbb{R}^n}([0,T])$ and it admits a subsequence converges weakly* to some $g \in L^{\infty}_{\mathbb{R}^n}([0,T])$ satisfying $|g(t)| \leq \beta(t) \; \forall t \in [0,T]$

I try to prove it this way :

  1. since the functions $\{g_n, \forall n\}$ are measurable also bounded by $\beta$ then $\exists c >0$ such that $|g_n(t)| \leq \beta(t) \leq c$ then $g_n \in L^{\infty}_{\mathbb{R}^n}([0,T])$

2)$|g_n(t)| \leq \beta(t) \Rightarrow | \frac{g_n(t)}{\beta(t)} | \leq 1 \Rightarrow (\frac{g_n}{\beta})$ in the unit ball of $L^{\infty}$ which is weakly* compact, then we can extract a subsequence $(\frac{g_k}{\beta})$ converges to $(\frac{g}{\beta})$ in that ball where $|g(t) | \leq \beta(t) $

Please tell me if something is wrong and thank you very much.

user1000039
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Myriam
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  • The main idea looks about right to me, but for the second part you technically need to watch out for any $t$ with $\beta(t) = 0$ for the division to be well defined, but that is easy, since then $g_n(t) = 0$ anyway. – mlk Sep 05 '15 at 11:59
  • thanks for your remark, was very useful . – Myriam Sep 05 '15 at 14:28

2 Answers2

2

There's no need to divide by $\beta$. Since $||g_n||_\infty$ is bounded there is a weak* convergent subsequence; let's save typing by saying $g_n\to g$ weak*.

Now fix $\lambda>0$ and let $E$ be the set where $\beta\le\lambda$. Then $g_n\chi_E\to g\chi_E$ weak* and $||g_n\chi_E||_\infty\le\lambda$, hence $||g\chi_E||_\infty\le\lambda$. So $|g|\le\beta$ almost everywhere.

David C. Ullrich
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Yes, it is right in large, just details would need some clarification.

Define $$ h_n(t)=\left\{ \begin{array}{ll} \frac{g_n(t)}{\beta(t)}, & \beta(t)\ne 0,\\ 0, & \beta(t)=0. \end{array} \right. $$ It is measurable and $\|h_n\|_\infty\le 1$. By weak* compactness you conclude that there exists a weak* cluster point $h$ with $\|h\|_\infty\le 1$. Finally, you need to show that the (sub)sequence $g_n=h_n\beta\to g=h\beta$ weakly*: for all $f\in L^1$ $$ \int f\cdot g_n\,d\mu=\int \underbrace{f\beta}_{\in L^1}\cdot h_n\,d\mu\to \int f\beta\cdot h\,d\mu=\int f\cdot g\,d\mu $$ and $\|g\|\le \|h\beta\|\le \beta\|h\|\le\beta$.

A.Γ.
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