L^∞(Ω×Ω)是否存在更便捷的稠密子空间？

Great question! Let's unpack this by contrasting with the Stone-Weierstrass result you mentioned and breaking down the different topological contexts for $L^\infty$, since that's the key distinction here.

First: Recap the Gap Between $C(\Omega\times\Omega)$ and $L^\infty(\Omega\times\Omega)$ #

In $C(\Omega\times\Omega)$ (with the uniform norm), the Stone-Weierstrass theorem gives us a clean, structured dense subspace: all finite linear combinations of products a(x)b(y) where $a,b \in C(\Omega)$. But $L^\infty(\Omega\times\Omega)$ operates very differently—it's equipped with the essential supremum norm, and we also need to consider the weak* topology (since $L^\infty$ is the dual space of $L^1$), which changes what "dense" means.

Now: Dense Subspaces for $L^\infty(\Omega\times\Omega)$ #

Let's split this into two cases based on the topology you're working with:

1. Norm Topology (Essential Supremum) #

• The simple functions (finite linear combinations of indicator functions $\chi_A(x,y)$ for measurable $A \subset \Omega\times\Omega$) are indeed the standard dense subspace here. For any $f \in L^\infty(\Omega\times\Omega)$ and $\varepsilon > 0$, you can always find a simple function that approximates $f$ to within $\varepsilon$ in the essential supremum norm—this is a fundamental result in measure theory.
• Unfortunately, the product-based subspace (finite combinations of a(x)b(y) with $a,b \in L^\infty(\Omega)$) is not dense in the norm topology. For example, take $\Omega = (0,1)$ and consider the indicator function of the diagonal: $f(x,y) = \chi_{x=y}(x,y)$. There's no way to approximate this function arbitrarily well with product functions in the essential supremum norm—any product-based combination behaves "separably" in $x$ and $y$, while the diagonal function has a tight coupling between the two variables that can't be captured uniformly.

2. Weak* Topology (Dual to $L^1(\Omega\times\Omega)$) #

If you're working with the weak* topology (extremely common in functional analysis when dealing with $L^\infty$), we get a nice, Stone-Weierstrass-style dense subspace:

• The subspace spanned by product functions a(x)b(y) (where $a,b \in L^\infty(\Omega)$) is weak dense* in $L^\infty(\Omega\times\Omega)$.

Why is this true? By the Hahn-Banach theorem, a subspace is weak* dense if it separates points in the predual space (here, $L^1(\Omega\times\Omega)$). Suppose $g \in L^1(\Omega\times\Omega)$ satisfies $\int_{\Omega\times\Omega} a(x)b(y)g(x,y) dxdy = 0$ for all $a,b \in L^\infty(\Omega)$. Then for any $a \in L^\infty(\Omega)$, $\int_\Omega a(x) \left(\int_\Omega b(y)g(x,y) dy\right) dx = 0$; choosing $b$ as indicator functions shows that $\int_\Omega g(x,y) dy = 0$ almost everywhere, which implies $g = 0$ almost everywhere. Since the product subspace separates all non-zero elements of $L^1$, it's weak* dense in $L^\infty$.

Final Takeaway #

• If you need density in the norm topology, simple functions (indicator combinations) are the most straightforward dense subspace—there's no simpler structured subspace that works here.
• If you're okay with weak density*, then the product-based subspace $\text{span}{a(x)b(y) \mid a,b \in L^\infty(\Omega)}$ is a far more convenient, Stone-Weierstrass-style dense subspace, avoiding the need to work with arbitrary measurable set indicators.

内容的提问来源于stack exchange，提问作者user254433