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Problem:

"Show that if $f:\left[a,+\infty\right] \rightarrow \mathbb R$ is continuous and if $\lim_{x \rightarrow + \infty} f(x)$ exists, then $f$ is uniformly continuous".

I have a question about this problem.

My partial solution:

$\lim_{x\rightarrow +\infty}f(x) = L \Rightarrow \forall \varepsilon>0, \exists M>0$ such that $x \geq M \Rightarrow |f(x)-L|<\frac{\varepsilon}{2}$.

Given $x_1$,$x_2$ $\in$$[M, \infty)$$\Rightarrow |f(x_1)-L|<\frac{\varepsilon}{2}$ and $|f(x_2)-L|<\frac{\varepsilon}{2}$ $\Rightarrow |f(x_1)-L|<\frac{\varepsilon}{2}$ and $|L-f(x_2)|<\frac{\varepsilon}{2}$ $\Rightarrow |f(x_1)-L+L-f(x_2)|\leq |f(x_1)-L|+|L-f(x_2)|< \varepsilon$ $\Rightarrow |f(x_1) - f(x_2)|<\varepsilon$ $\Rightarrow$ $f$ is uniformly continuous on $[M, +\infty)$.

How can I extend this result towards $f$ uniformly continuous on $[a, +\infty]$?

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3  
Hint: $f$ is uniformly continuous on $[a,M+1]$. –  David Mitra Jun 14 '13 at 18:45
    
See this post for details. –  David Mitra Jun 14 '13 at 18:51
    
$f$ is continuous on a compact set $[a,\ M+1]$ hence $\ldots$ –  M. Strochyk Jun 14 '13 at 19:09
    
The function $\sin( x^2)$ shows the hypothesis that $\lim\limits_{x\to\infty} f(x)$ exists is essential. –  Pedro Tamaroff Jun 15 '13 at 22:29

1 Answer 1

Here's an outline of one approach:

$\ \ \ 1)$ Let $\epsilon>0$.

$\ \ \ 2)$ Choose $N$ so that $|f(x)-L|<\epsilon/3$ whenever $x\ge N$.

$\ \ \ 3)$ Argue that $f$ is uniformly continuous on $[a,N]$.

$\ \ \ 4)$ Choose $ \delta>0$ so that $|f(x)-f(y)|<\epsilon/3$ whenever $|x-y|<\delta $ and $x\in[a,N]$, $y\in[a,N]$.

$\ \ \ 5)$ Show that in fact $|f(x)-f(y)|<\epsilon $ whenever $|x-y|<\delta $ and $x\in[a,\infty)$, $y\in[a,\infty)$

$\ \ \ \ \ \ \ $(consider three cases depending on the relationship between $x$, $y$, and $N$).


A slightly more elegant approach:

$\ \ \ 1)$ Let $\epsilon>0$.

$\ \ \ 2)$ Choose $N$ so that $|f(x)-L|<\epsilon/2$ whenever $x\ge N$.

$\ \ \ 3)$ Argue that $f$ is uniformly continuous on $[a,N+1]$.

$\ \ \ 4)$ Choose $ 1>\delta>0$ so that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta $, $x\in[a,N+1]$, and

$\ \ \ \ \ \ $$y\in[a,N+1]$.

$\ \ \ 5)$ Show that in fact $|f(x)-f(y)|<\epsilon $ whenever $|x-y|<\delta $ and $x\in[a,\infty)$, $y\in[a,\infty)$

$\ \ \ \ \ \ \ $(consider two cases depending on whether one (or both) of $x$, $y$, exceeds $N+1$).

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