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Let $T=[0,1]$, $$P=\{x=(x^1,x^2,…,x^n)\in\mathbb{R}^n \vert \; x^{i}\geq 0,i=1,2,…,n; x^1+x^2+…+x^n=1\}.$$ A $T$-strategy is a measurable function $y$ from $T$ to $P$. Hence $y$ is Lebesgue-integrable and we write $\int y$ for $(\int y^1(t)d\lambda,\dots,\int y^n(t)d\lambda)$, where the integration is on $T$. That is, a $T$-strategy $y\in L_1(T\times\{1,2,\dots,n\})$. Let $S$ denote the set of all $T$-strategies endowed with $L_1$ weak topology. Claim: The set $S$ is a compact space.

Questions:

  1. Does $L_1$ weak topology mean that if a sequence $\{y_{k}\}\subset L_1(T\times\{1,2,\dots,n\})$ converges weakly to $ y\in L_1(T\times\{1,2,\dots,n\})$ if $\int y_{k}^{i}(t)v(t)d\lambda\to \int y^{i}(t)v(t)d\lambda$,$\forall v\in L_{\infty}(T)$, $\forall i=1,2,…,n$?
  2. How to show the set $S$ is compact?
Shuhao Cao
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    Some quick ctrl+c/v'able reference for $\LaTeX$: http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Shuhao Cao Jun 15 '13 at 03:52

1 Answers1

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1. Yes, weak convergence means convergence of integrals against every $L^\infty$ function.

2. First, the set is closed. It's easy to see that it's closed in norm (use the relation of norm convergence and a.e. convergence). And since it's also convex, it is weakly closed as well.

It is awkward to work with the weak topology on $L^1$. Instead, use the fact that your functions are bounded by $1$. Therefore, they form a bounded set in $L^p$ for every $p$. For $1<p<\infty$, closed bounded sets in $L^p$ are weakly compact. Recall that the weak topology on $L^p$ is determined by integration against $L^q$ with $1/p+1/q=1$. Since $L^\infty$ is smaller than $L^q$ for finite $q$, it follows that the weak $L^1$ topology is weaker than the weak $L^p$ topology for $p>1$. Having fewer open sets implies having more compact sets. Thus, the set is weakly compact in $L^1$.

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  • Hi, the following links says that a banach space is reflexive if and only if its cloased unit ball is weakly compact. It seems that L^1 is not reflexive, why S a subset of unit ball is weakly compact here? http://math.stackexchange.com/questions/199478/equivalence-of-reflexive-and-weakly-compact, http://math.stackexchange.com/questions/428817/if-c-is-convex-weakly-closed-and-norm-bounded-longrightarrow-c-is-weakl – user91360 Jun 16 '15 at 17:05