Let $f:[a,b]→\mathbb{R}$ be regulated and non-negative. Prove that $g:[a,b]→\mathbb{R}$ defined by $g(x)=\sqrt{f(x)}$ is regulated.
A function $f:[a,b]\to\Bbb R$ is a regulated function if $\forall$ $\varepsilon>0$ $\exists$ a step function $\varphi:[a,b]\to\Bbb R$ such that $\Vert f-\varphi\Vert<\varepsilon$.
I've tried to use the definition of a regulated function but haven't been able to make any progress. Is there a way of using the fact that a linear combination of regulated functions is regulated? Or am I not even close?
@Arthur Is this attempt at all correct or at least along the lines of what you mean:
We have a step function $\varphi_f$ for $f$ and $\varepsilon_f$. Let $\sqrt{\varphi_f}=\varphi_g$.
We know $\Vert f-\varphi_f\Vert<\varepsilon_f \implies \Vert f-\varphi_g^2\Vert<\varepsilon_f$
$\implies \Vert (\sqrt{f}+\varphi_g)(\sqrt{f}-\varphi_g)\Vert<\varepsilon_f$
$\implies \Vert (g+\varphi_g)\Vert \Vert(g-\varphi_g)\Vert<\varepsilon_f $
$\implies \Vert(g-\varphi_g)\Vert < (\varepsilon_f)/\Vert (g+\varphi_g)\Vert $
So letting $\varepsilon_g = (\varepsilon_f)/\Vert (g+\varphi_g)\Vert $ proves that $\Vert(g-\varphi_g)\Vert < \varepsilon_g$ and hence $g(x)$ is regulated.
Best Answer
Use the fact that $f([a,b]) \subset [0,R]$ for some $R > 0$ and that $[0,R] \to \Bbb{R}, x \mapsto \sqrt{x}$ is uniformly continuous.
Finally, you should use that if $\varphi$ is a (non-negative) step function, then so is $\sqrt{\varphi}$.
Of course, you will have to show that the step function $\varphi$ approximating $f$ can be taken to be nonnegative, but this is easy.