Let $v$ be a real-valued function defined on any Polish space $\mathbb{X}$ such that $v \geq 1$. The Polish space is a separable and completely metrizable space, which is commonly used to avoid measurability concerns.

Denote by $B_v(\mathbb{X})$ the space of real-valued functions $f$ on $\mathbb{X}$ such that $f(x)/v(x)$ is bounded as $x$ ranges over $\mathbb{X}$, and consider the weighted sup-norm on $B_v(\mathbb{X})$: $$ \|f \|_v = \sup_{x \in \mathbb{X}} \frac{|f(x)|}{v(x)} < \infty.$$

Evidently, endowed with the finite weighted sup-norm, $B_v(\mathbb{X})$ is a Banach space.

However, I'm wondering that is $B_v(\mathbb{X})$ a Banach lattice? If so, how to prove this?

My attempt: I was thinking to transform this weighted sup-norm $ \|f \|_v = \sup_{x \in \mathbb{X}} \frac{|f(x)|}{v(x)} = \| \frac{f}{v}\|_{\infty}$ to a common supremum norm. I know that when $v(x) \equiv 1$, the vector lattice $B_1(\mathbb{X})$ of all bounded real-valued functions on $\mathbb{X}$ is a Banach lattice, but I got stuck in a more general weight function $v: \mathbb{X} \to [1, \infty)$.

I also thought that if the weight function $v: \mathbb{X} \to [1, \infty)$ is continuous, then $B_v(\mathbb{X})$ and $B_1(\mathbb{X})$ are homeomorphic. As a result, $B_1(\mathbb{X})$ is a Banach lattice, so is $B_v(\mathbb{X})$. I'm not sure whether this brief proof is correct, it is just my rough idea.

Any suggestions or idea is much appreciated!


1 Answer 1


Its straightforward that $B_u(X)$ is a Riesz space under the pointwise order and for $f\in B_u(X)$ its modulus $|f|$ is given by $|f|(x)=|f(x)|$ for every $x\in X$. Now let $f, g\in B_u(X)$ such that $|f|\leq |g|$, that is $|f(x)|\leq |g(x)|$ for every $x$ in $X$. Since $u$ is positive, $\frac{|f(x)|}{u(x)} \leq \frac{|g(x)|}{u(x)}$ for every $x\in X,$ which implies that $\|f\|_u\leq \|g\|_u$.


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