I was solving the next exercise:
Let $f:\mathbb{R}\to\mathbb{R}$ be an uniformly continuous function. Prove that there exist $a,b\in\mathbb{R}^{+}$ such that, for all $x\in\mathbb{R}$, $|f(x)|\leq a|x|+b$.
I found the proof here and is beautiful but I can't understand the first step, i.e., the part where the proof says that if $|y|< k\delta$, with $k\in\mathbb{N}$, by the triangle inequality, then $|f(0)-f(y)|<k$. I tried to prove using the next argument: $$|f(0)-f(y)|=\left|f(0)-f\left(\frac{y}{k+1}\right)+f\left(\frac{y}{k+1}\right)-f\left(\frac{y}{k+2}\right)+f\left(\frac{y}{k+2}\right)+\cdots-f\left(\frac{y}{2k}\right)+f\left(\frac{y}{2k}\right)-f(y)\right|$$Clearly $\left|\frac{y}{k+1}\right|<\left|\frac{y}{k}\right|<\delta$. Moreover, if $n\in\{1,\dots,k-1 \}$ $$\left|\frac{y}{k+n}-\frac{y}{k+n+1} \right|=\left|\frac{y}{(k+n)(k+n+1)} \right|<\left|\frac{y}{k}\right|<\delta$$If we apply triangle inequality a lot of times we obtain $$|f(0)-f(y)|\leq \left|f(0)-f\left(\frac{y}{k+1}\right) \right|+\left| f\left(\frac{y}{k+1}\right)-f\left(\frac{y}{k+2}\right)\right|+\cdots+\left|f\left(\frac{y}{2k}\right)-f(y) \right|$$All terms are less thant $\delta$ except the last one. In fact, I can't bound that difference. Not in a direct way. What I am missing? I really aprecciate any help. Thanks.
Your idea is in the right direction, you just need the correct representation. Try writing:
$$|f(0)-f(y)| = \\ \left| \left(f(0)-f\left(\frac y{k}\right)\right) +\left(f\left(\frac y{k}\right) - f\left(\frac {2y}{k}\right)\right) +\left(f\left(\frac {2y}{k}\right) - f\left(\frac {3y}{k}\right)\right) +\\ \dots +\left(f\left(\frac {(k-1)y}{k}\right) - f\left(y\right)\right) \right|$$
In other words, a telescoping sum:
$$|f(0)-f(y)| = \left|\sum_{n=0}^{k-1} f\left(\frac n k y\right) - f\left(\frac{n+1}k y\right)\right|$$