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Is $\{ f \in C[a,b] | |f(x)| \leq 2+f(x)^2 \}$ a complete metric space with the metric $\rho_{\infty}$ (the sup norm metric)?

So far, I know that continuous functions on a finite interval are complete in the sup norm. So, $|f_n(x) - f_m(x)| \leq \epsilon \implies f_n(x) \rightarrow f(x) \in C[a,b]$, but does $f(x)$ necessarily belong to the above subset?

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    $\begingroup$ Is there a function $f \in C([a,b])$ that doesn't satisfy $|f(x)| \le 2 + f(x)^2$? $\endgroup$ – Umberto P. Apr 6 '17 at 18:05
  • $\begingroup$ I can't think of any, but I worry that sometimes there's an exception that is not too obvious $\endgroup$ – bashmike Apr 6 '17 at 18:12
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    $\begingroup$ Well, either $|f(x)| \le 2$, or $|f(x)| > 2$ in which case $|f(x)| < f(x)^2$. $\endgroup$ – Umberto P. Apr 6 '17 at 18:14
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    $\begingroup$ All real numbers $u$ satisfy $|u|\leq 2+u^2$. Therefore your space is just $C\bigl([a,b]\bigr)$. $\endgroup$ – Christian Blatter Apr 6 '17 at 18:26
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Hint : Show that $\{f \in C[a,b]:|f(x)| \le 2+|f(x)|^2\}$ is closed in $(C[a,b],||.||_{\infty})$

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