Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is uniformly continuous. Show there exists a constant $K > 0$ such that $$|f(x)| \leq K(1+|x|), \forall x \in \mathbb{R}.$$

My attempt:

Since $f$ is uniformly continuous, for all $\epsilon > 0$ there exists a $\delta > 0$ such that for all $x,y$ with $|x-y| < \delta, |f(x) - f(y)| < \epsilon.$

In particular, letting $y=0$, we have that $|x| < \delta$ implies $|f(x) - f(0)| < \epsilon.$

So, $|f(x)| < \epsilon - |f(0)| \leq \epsilon + \epsilon|x| = \epsilon(1 + |x|) \leq K(1+|x|)$, for $K> \epsilon$.

Is this right? It seems a little weird to me, but I'm not sure why.


Select some $ϵ>0$, for instance $ϵ=1$, and get the $δ>0$ from the uniform continuity.

Set $x_k=kδ/2$, then $|f(x_{k+1})-f(x_k)|<ϵ$. Now iterate this to get $$|f(x_k)|\le|f(0)|+|k|ϵ=|f(0)|+\frac{2ϵ}δ|x_k|.$$

Now add some wiggle room $ϵ$ for the function values in between the sample points (as $|x-x_k|\leδ/2<δ$ for some $k$) and you are done.

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