# Prove that the function is uniformly continuous

Let $$f(x)$$ be a continuous function in $$[0,\infty)$$

there are $$a,b \in \mathbb{R}$$ such that $$\lim_{x\to\infty} [f(x) - (ax +b)] =0$$

prove that $$f(x)$$ is uniformly continuous in $$[0,\infty)$$

how i started:

using that function limit definition: let $$\epsilon >0$$ there is a $$M>$$ such that for every $$x>M, |f(x) - (ax +b)|<\epsilon$$ in the interval $$[0,M]$$ the function is uniformly continuous (by weierstrass theorem).

this is there part i got stuck in, i know that f(x) "Converges" with the $$(ax+b)$$, but i cant find a $$\delta$$ that will prove what i need

• Sloppy outline: Let $\epsilon>0$. If you choose $M$ so that $|f(x)-(ax+b)|<\epsilon$ whenever $x>M$, then using the triangle inequality, you can show, for $x,y>M$, that $|f(x)-f(y)|<3\epsilon$ with an additional constraint on $x$ and $y$ (namely $|a||x-y|<\epsilon$). Now find a $0<\delta<1$ that "works" for $x,y\in[0,M+1]$. Then put the pieces together. – David Mitra Jan 23 '14 at 20:05

As $g(x)=ax+b$ is uniformly continuous on $\mathbb R$ and the sum of uniformly contnuous functions is also a uniformly continuous function, we can assume that $$\lim_{x\to \infty} f(x)=0.$$ Let $\varepsilon>0$ and $M>0$, such that whenever $x\ge M$, then $|f(x)|<\varepsilon/3$. As $f$ is uniformly continuous in $[0,M]$, then there exists a $\delta>0$, such that for $x,y\in [0,M]$ $$|x-y|<\delta\quad\Longrightarrow\quad |f(x)-f(y)|<\frac{\varepsilon}{3}.$$ Now let $x,y\in[0,\infty)$ with $|x-y|<\delta$.

Case I. $x,y\in [0,M]$, then clearly $|f(x)-f(y)|<\varepsilon/3<\varepsilon$.

Case II. $x,y>M$, then $|f(x)|, |f(y)|<\varepsilon/3$ and hence $|f(x)-f(y)|<2\varepsilon/3<\varepsilon$.

Case III. $x<M<y$. Then $$|f(x)-f(y)|\le |f(x)-f(M)|+|f(M)-f(y)|\le \varepsilon/3+2\varepsilon/3=\varepsilon.$$

• $$\lim_{x\to \infty} f(x)=0.$$ this isn't necessarily true... – guynaa Jan 23 '14 at 20:13
• @yiorgos S.Smyrlis very nice argument. complete one. – GA316 Jan 23 '14 at 20:22
• @guynaa: What do you not agree with? – Yiorgos S. Smyrlis Jan 23 '14 at 20:33
• $f(x)$ can easily go to infinity if $a$ is not $0$ – guynaa Jan 23 '14 at 22:50
• @guynaa: That's not a problem. $g(x)=ax+b$ is uniformly continuous for every $a\in\mathbb R$, since $|g(x)-g(y)|=|a||x-y|$. – Yiorgos S. Smyrlis Jan 24 '14 at 6:43