# Every uniformly continuous real function has at most linear growth at infinity

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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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.