# a function that has an asymptote is uniformly continuous

• Jun 28th 2011, 11:45 AM
ymar
a function that has an asymptote is uniformly continuous
Let $f:[0,\infty]\longrightarrow \mathbb{R}$ be a continuous function and $\lim_{x\rightarrow\infty}(f(x)-(ax+b))=0$. Show that $f$ is uniformly continuous.

I'd like to use this charactetization of uniform continuity. A function $f$ is uniformly continuous iff for all sequences $x_n,\,y_n$ if $|x_n-y_n|\rightarrow 0$, then also $|f(x_n)-f(y_n)|\rightarrow 0$.

Suppose $|x_n-y_n|\rightarrow 0$.
If $x_n$ is a bounded sequence, then $y_n$ must be bounded too, and so if $|x_n-y_n|\rightarrow 0$, then also $|f(x_n)-f(y_n)|\rightarrow 0$, because $f$ is continuous and therefore uniformly continuous on a compact set. If, on the other hand, $x_n\rightarrow\infty$, then also $y_n\rightarrow\infty$ and we have

$|f(x_n)-f(y_n)|\leq |f(x_n)-(ax_n+b)|+|f(y_n)-(ay_n+b)|+|(ax_n+b)-(ay_n+b)|\rightarrow 0$

But what if $x_n$ is neither bounded nor convergent to infinity?
• Jun 28th 2011, 01:35 PM
hatsoff
Re: a function that has an asymptote is uniformly continuous
Quote:

Originally Posted by ymar
Let $f:[0,\infty]\longrightarrow \mathbb{R}$ be a continuous function and $\lim_{x\rightarrow\infty}(f(x)-(ax+b))=0$. Show that $f$ is uniformly continuous.

I'd like to use this charactetization of uniform continuity. A function $f$ is uniformly continuous iff for all sequences $x_n,\,y_n$ if $|x_n-y_n|\rightarrow 0$, then also $|f(x_n)-f(y_n)|\rightarrow 0$.

Suppose $|x_n-y_n|\rightarrow 0$.
If $x_n$ is a bounded sequence, then $y_n$ must be bounded too, and so if $|x_n-y_n|\rightarrow 0$, then also $|f(x_n)-f(y_n)|\rightarrow 0$, because $f$ is continuous and therefore uniformly continuous on a compact set. If, on the other hand, $x_n\rightarrow\infty$, then also $y_n\rightarrow\infty$ and we have

$|f(x_n)-f(y_n)|\leq |f(x_n)-(ax_n+b)|+|f(y_n)-(ay_n+b)|+|(ax_n+b)-(ay_n+b)|\rightarrow 0$

But what if $x_n$ is neither bounded nor convergent to infinity?

Let $g=f-(ax+b)$ and $\epsilon>0$. Then there is $M>0$ such that if $x,y\geq M$ then $|g(x)|,|g(y)|\leq\epsilon/3$. Notice also that $f$ is uniformly continuous on $[0,M+\delta_1]$ (by Heine-Cantor), where $\delta_1=\epsilon/3|a|$. So there is $\delta_2>0$ such that if $|x-y|<\delta_2$ for $x,y\in[0,M+\delta_1]$ then $|f(x)-f(y)|<\epsilon$. Now let $\delta=\min\{\delta_1,\delta_2\}$, and suppose $|x-y|<\delta$. If $x,y\geq M$ then $|f(x)-f(y)|<\epsilon$ (details omitted). Otherwise $M>x$ or $M>y$, giving us $x,y\in[0,M+\delta_1]$ and hence $|f(x)-f(y)|<\epsilon$. So in both cases we have $|x-y|<\delta$ implying $|f(x)-f(y)|<\epsilon$.