Assuming $f:\mathbb R\to\mathbb R $ be an uniform continuous function, how to prove $$\exists a,b\in \mathbb R^+ \quad \text{such that}\quad |f(x)|\le a|x|+b.$$
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4$\begingroup$ think about $|f(x) - f(0)|\le C|x-0|$ from the definition of uniform continuous. $\endgroup$Yimin– Yimin2013-02-15 19:54:33 +00:00Commented Feb 15, 2013 at 19:54
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$\begingroup$ See math.stackexchange.com/questions/14735/… $\endgroup$Jonas Meyer– Jonas Meyer2013-02-15 20:22:36 +00:00Commented Feb 15, 2013 at 20:22
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1 Answer
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By definition, there is a $\delta > 0$, such that $$|f(x) - f(y)| < 1$$ whenever $|x - y| < \delta$. Consequently, $$|f(0) - f(y)| < k$$ whenever $|y| < k \delta$ by the triangle inequality. Another application of the triangle inequality gives $$|f(y)| \le |f(0) - f(y)| + |f(0)| < k + |f(0)|$$ for each such $y$.
Assuming $\delta > 0$ (otherwise there is not much to prove), we can set $$m(z) = \min\{k \in \mathbb N \colon |z| < k \delta\}$$ for arbitrary $z \ne 0$. We then have $m(z) \le |z|/\delta + 1$ and $$|f(z)| < m(z) + |f(0)| \le |z|/\delta + 1 + |f(0)|$$ which finishes the proof.
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1$\begingroup$ :your attempt is nice but why minimime of this set{k : |z| < k $\delta$}exist $\endgroup$M.H– M.H2013-02-16 03:22:23 +00:00Commented Feb 16, 2013 at 3:22
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1$\begingroup$ Can we agree that the set is non-empty? The set is bounded from below (by 0), so it has a minimum. (I just made an edit to make clear that the minimum is taken over the non-negative integers). $\endgroup$anonymous– anonymous2013-02-16 11:17:01 +00:00Commented Feb 16, 2013 at 11:17
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1$\begingroup$ In which book I can find this? $\endgroup$Ricci Ten– Ricci Ten2024-12-06 01:20:00 +00:00Commented Dec 6, 2024 at 1:20