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S is closed, unbounded set on $\mathbb{R}$, and $f$ is bounded, continuous function on S. Suppose $\lim_{x\rightarrow\infty}f(x) \lim_{x\rightarrow-\infty}f(x)$ exists, Show that $f$ is uniformly continuous on S

Martin Sleziak
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H. Inigo
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    what are your thoughts on this? – Kushal Bhuyan Nov 27 '15 at 02:00
  • Some related posts: http://math.stackexchange.com/questions/252387/show-f-is-bounded-on-a-infty-if-continuous-there-and-lim-limits-x-to-i, http://math.stackexchange.com/questions/75491/how-does-the-existence-of-a-limit-imply-that-a-function-is-uniformly-continuous, http://math.stackexchange.com/questions/346292/uniform-continuity-on-0-infty – Martin Sleziak Nov 27 '15 at 11:14

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Let $A=\lim_{x\to-\infty}f(x)$, $B=\lim_{x\to\infty}f(x)$, and let $\varepsilon>0$. Then there exist $x_0<y_0$ in $S$ such that $|f(x)-A|<\varepsilon/2$ if $x\leq x_0$ and $|f(y)-B|<\varepsilon/2$ if $y\geq y_0$ in $S$. Now, since $[x_0,y_0]$ is compact in $\mathbb{R}$, there exists $\delta>0$ such that $|f(x)-f(y)|<\varepsilon/2$ if $|x-y|<\delta$ in $S$. The conclusion follows easily.

Ángel Valencia
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