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Let $v$ be a real-valued function defined on any Polish space $\mathbb{X}$ such that $v \geq 1$.

Denote by $B(\mathbb{X})$ 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 $B(\mathbb{X})$: $$ \|f \|_v = \sup_{x \in \mathbb{X}} \frac{|f(x)|}{v(x)} < \infty.$$

The positive cone $B(\mathbb{X})_+$ of this Banach space $( B(\mathbb{X}), \| \cdot \|_v)$ is defined as $$B(\mathbb{X})_+ = \{ f \in B(\mathbb{X}) \colon f \geq 0\} .$$

I'm wondering that how to prove that this positive cone $B(\mathbb{X})_+$ has non-empty interior?

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 $v(x) \equiv 1$, $B(\mathbb{X})_+$ has non-empty interior, but I got stuck in a more general $v: \mathbb{X} \to [1, \infty)$.

Any suggestions or idea is much appreciated!

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Since $v(x)\geq 1$ for all $x\in\mathbb X$ and $\|v\|_v=1$, we get $v\in B(\mathbb X)_+$.

Claim: $v$ is in the interior of $B(\mathbb X)_+$.

Proof:

Let be $\varepsilon\in(0,1)$. Suppose there exists $f\in B(\mathbb X)$ such that $\|f-v\|_v<\varepsilon$ and there exists $x_0\in\mathbb X$ such that $f(x_0)<0$. Then $$|f(x_0)-v(x_0)|=v(x_0)-f(x_0)>v(x_0)$$and we conclude$$\|f-v\|_v\geq \frac{|f(x_0)-v(x_0)|}{v(x_0)}>\frac{v(x_0)}{v(x_0)}=1>\varepsilon.$$This is a contracation and therefore$$\{f\in B(\mathbb X)~:~\|f-v\|_v<\varepsilon\}\subset B(\mathbb X)_+$$and the interior of $B(\mathbb X)_+$ is not empty.

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