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I have $\Omega \subseteq \mathbb{R}^n$ such that $m(\Omega) < \infty$ (Lebesgue measure) and $1 \le p < \infty$, and I want to prove that if $C \subseteq L^p (\Omega)$ is closed in the weak topology $\sigma(L^p, L^q)$ of $L^p$, with $1/p + 1/q =1$, then $C \cap L^\infty$ is closed in the weak-* topology $\sigma(L^\infty, L^1)$ of $L^\infty$.

Frankly speaking, I don't know how to start. My first thought was on this theorem:

Let X be a separable Banach space. A convex set $Z \subseteq X^*$ is weak-* closed iff it is weak-* sequentially closed.

but $C$ is not convex... It seems that the only (?) way is to work directly with the topology.

I need only an hint.

Thank you in advance.

gangrene
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    You have an error in the cited theorem, it should read $Z \subset X^*$. Since $L^1$ is separable, you could apply this theorem. However, your $C$ is not convex. – gerw May 05 '14 at 06:40
  • You're right, I have been careless, and I'm going to correct. Thank you! I have to find another way to proceed... – gangrene May 05 '14 at 12:10

2 Answers2

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Hint: $L^\infty \subset L^p$, and $L^q \subset L^1$, since $m(\Omega) < \infty$. Thus

$$L^\infty \setminus (C\cap L^\infty) \subset L^p\setminus C.$$

Ponder the relations between $\sigma(L^\infty,L^1)$ and $\sigma(L^\infty, L^q)$.

The abstract argument is:

$L^q \subset L^1$ means the weak* topology $\sigma(L^\infty,L^1)$ on $L^\infty$ is finer than the topology $\sigma(L^\infty,L^q)$ induced on $L^\infty$ by $L^q$. That means the inclusion

$$\iota \colon (L^\infty,\sigma(L^\infty,L^1)) \hookrightarrow (L^p,\sigma(L^p,L^q))$$

is continuous. But then $C\cap L^\infty = \iota^{-1}(C)$ is $\sigma(L^\infty,L^1)$-closed, since $C$ is $\sigma(L^p,L^q)$-closed.

The more concrete argument using $L^\infty \setminus (C\cap L^\infty) \subset L^p\setminus C$ is:

Let $f \in L^\infty \setminus (C\cap L^\infty)$. Since $f\in L^p\setminus C$, and $C$ is weakly closed in $L^p$, there is a weak neighbourhood

$$V = V(g_1,\dotsc,g_n;\varepsilon) = \left\{ h \in L^p : \left\lvert \int_\Omega g_i(x)(h(x)-f(x))\,dx\right\rvert < \varepsilon \text{ for } 1 \leqslant i \leqslant n\right\}$$

of $f$ in $L^p$ with $V\cap C = \varnothing$. But since all $g_i \in L^1$,

$$V\cap L^\infty = W(g_1,\dotsc,g_n;\varepsilon) = \left\{ h\in L^\infty : \left\lvert \int_\Omega g_i(x)(h(x)-f(x))\,dx\right\rvert < \varepsilon \text{ for } 1 \leqslant i \leqslant n \right\}$$

is a weak* neighbourhood of $f$ in $L^\infty$ that doesn't intersect $C\cap L^\infty$.

Daniel Fischer
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  • The relations between $\sigma(L^\infty, L^1)$ and $\sigma(L^p, L^q)$, right? Anyway I haven't still got the point, but I'm thinking about it... Have I to recall the Riesz representation theorem (to clarify the situation)? Or is simply a topologic argument?

    Thank you very much for your time!

     – gangrene May 05 '14 at 18:34
  • Looking at the relation between $\sigma(L^\infty,L^1)$ and $\sigma(L^p,L^q)$ also is a way to see it. You don't need any representation theorem, it's pure topology. – Daniel Fischer May 05 '14 at 18:38
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Sometimes such arguments are easier to understand using nets.

Let $\{f_\alpha\}$ be a net in $C \cap L^\infty$, and suppose $f_\alpha \to f$ weakly in $\sigma(L^\infty, L^1)$. This means that for every $g \in L^1$, we have $\int f_\alpha g \to \int f g$. But $L^q \subset L^1$ on a space of finite measure, so a fortiori we have $\int f_\alpha g \to \int fg$ for every $g \in L^q$. Thus $f_\alpha \to f$ in $\sigma(L^p, L^q)$, and since $C$ is closed this topology, $f \in C$. Thus $f \in C \cap L^\infty$ and hence this set is closed.

Nate Eldredge
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