$0<\frac {1}{\sqrt{2+f(x)}+\sqrt{2+g(x)}}<1$ in $C(\mathbb{R}_+)$

Asked Jun 16 '21 at 09:17

Active Jun 16 '21 at 09:17

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So let $C(\mathbb{R}_+)$ be all the continuous functions in $[0,\infty)$,

I have the metric space

$(C(\mathbb{R}_+),d) \to (C(\mathbb{R}_+),d)$

I have taken $f,g \in (C(\mathbb{R}_+),d)$

And i want to prove that $0<\frac {1}{\sqrt{2+f(x)}+\sqrt{2+g(x)}}<1$

Question : If not mistaken $f,g \in (C(\mathbb{R}_+),d)$ means that f,g are function with their starting points at $[0,\infty)$ not having a clue about their image/mapping, therefore is there a way to prove the thing I want?

asked Jun 16 '21 at 09:17

stefanos diamantis

$f \in C(\Bbb R_+)$ includes that $f: \Bbb R_+ \to \Bbb R_+$. Does that help to show what you need? – Gono Jun 16 '21 at 09:21
@Gono that means that every $f:(0,\infty) \to (0,\infty)$ correct? – stefanos diamantis Jun 16 '21 at 09:23
I would assume $f:[0,\infty) \to [0,\infty)$ but this depends on your definition of $\Bbb R_+$ – Gono Jun 16 '21 at 10:02

$0<\frac {1}{\sqrt{2+f(x)}+\sqrt{2+g(x)}}<1$ in $C(\mathbb{R}_+)$

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