Originally Posted by

**Opalg** Definition of uniform continuity: $\displaystyle \forall\epsilon>0\;\exists\delta>0$ such that $\displaystyle |x-y|<\delta\;\Rightarrow\;|f(x)-f(y)|<\varepsilon$. Apply this with $\displaystyle \varepsilon=1$ to get a $\displaystyle \delta>0$ with $\displaystyle |x-y|<\delta\;\Rightarrow\;|f(x)-f(y)|<1$. In particular, with y=0, the triangle inequality gives $\displaystyle |x|<\delta\;\Rightarrow\;|f(x)|<1 + |f(0)|$.

Suppose in addition that $\displaystyle |x|>\delta/2$. Then $\displaystyle \delta/2<|x|<\delta\;\Rightarrow\;|f(x)|<1 + |f(0)|< \tfrac2{\delta}|x| + |f(0)|$.

Next (assuming from now on that $\displaystyle \delta/2<|x|<\delta$), $\displaystyle |2x-x|<\delta$ and so $\displaystyle |f(2x)-f(x)|<1$. By the triangle inequality again, $\displaystyle |f(2x)|<2+|f(0)|<\tfrac2{\delta}|2x| + |f(0)|$.

That should suggest an inductive argument: $\displaystyle |f(nx)|<\tfrac2{\delta}|nx| + |f(0)|$.

But multiples of the interval $\displaystyle [-\delta/2,\delta]$ cover all the positive reals except for the interval $\displaystyle [0,\delta/2]$. So take $\displaystyle a=\tfrac2{\delta}$ and $\displaystyle b=1+|f(0)|$, and you'll have $\displaystyle |f(x)|\leqslant a|x|+b$ for all real x.