# Statement(s) > [!NOTE] Statement 1 (Uniformly Continuous Real Function is Bounded by some Linear Function) > Let $f:\mathbb{R}\to \mathbb{R}$ be [[Uniformly Continuous Real Function|uniformly continuous]]. Then there exists $a,b>0$ such that for all $x\in \mathbb{R},$ $|f(x)|<a|x|+b.$ # Proof(s) **Proof of statement 1:** By definition of uniform continuity, we can choose $\delta>0$ such that for all $x,y\in \mathbb{R},$ $|x-y|<\delta \implies |f(x)-f(y)|<1.$ For all $x\neq0,$ we know $|x|\in(n\delta,(n+1)\delta\,]$ for some $n\in \mathbb{N}.$ Applying the triangle inequality, $\begin{align} |f(x)|&=|f(x)-f(n\delta)+f(n\delta)-f((n-1)\delta)+f((n-1)\delta)-\dots+f(\delta)-f(0)+f(0)| \\ &\leq n+1+|f(0)| \\ &\leq \frac{1}{\delta} |x|+|f(0)|+1 \end{align}$since $n\delta<|x|.$ $\blacksquare$ # Application(s) **Consequences**: **Examples**: # Bibliography