Equivalence of weighted supremum norm

Question

We define (for a measurable space $\Omega$, a bounded and measurable function $\phi$) a weighted supremum norm: $$|| \phi ||_\beta := \sup\limits_{x\in\Omega}\, \frac{|\phi(x)|}{1+\beta V(x)}$$ where $V:\Omega\to[0,\infty)$ a function. I am trying to show that $||\,\cdot\,||_\beta$ and $||\,\cdot\,||_1 =||\,\cdot\,||$ are equivalent for any $\beta>0$, because I need $$||\,\cdot\,||_\beta \le C\,\,||\,\cdot\,||\quad \text{for a $C>0$} $$ at some point. Is it generally true that these norms are equivalent or do I need a different approach?

Answer 1

It suffices to show that for any $\beta$, there is a constant $C=C\left(\beta\right)$ such that for any non-negative $y$, $$\frac{1+y}{1+\beta y}\leqslant C.$$ Since $$\frac{1+y}{1+\beta y}=\frac 1\beta\frac{\beta-1+1+\beta y}{1+\beta y} =\frac{\beta-1}{\beta\left(1+\beta y\right)}+\frac 1{\beta}, $$ we have for any non-negative $y$,
$$\frac{1+y}{1+\beta y}\leqslant \frac{\left|\beta-1\right|+1}{\beta}.$$