# Thread:

1. ## Uniform Continuity

Let f(x) be continous on the real number line. Suppose that lim(x-> infinity) f(x) = 0 and lim(x-> negative infinity) f(x) = 0. Show that f(x) is unformly continous or else give a counterexample.

2. Originally Posted by seams192
Let f(x) be continuous on the whole real number line. Suppose that lim(x-> infinity) f(x) = 0 and lim(x-> negative infinity) f(x) = 0. Either show that f(x) is uniformly continuous or else give a counterexample.

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A perfect example of this would be a function like 1/(x^2+1) which is uniformly continuous. I would have difficulty proving it in the general case, but am also blind to any counterexamples that could exist.

Any thoughts?
More generally, if $\lim_{x\to\infty}f(x)=L_1,\lim_{x\to-\infty}f(x)=L_2<\infty$ and $f$ is continuous then it's uniformly continuous.

Start like this, for some $\varepsilon>0$ choose $A>0$ such that $|f(x)-L_1|<\frac{\varepsilon}{2},\text{ }x\in[A,\infty)$ AND $|f(x)-L_2|<\frac{\varepsilon}{2},\text{ }x\in(-\infty,-A]$.

You clearly have then that $|f(x_1)-f(x_2)|<\varepsilon,x_1,x_2\in\mathbb{R}-[-A,A]$ and you surely have that $f$ is unif. cont. on $[-A,A]$. See if you piece the riece together.

3. Why does it follow that it is uniformly continuous on the interval [-M,M]? I can see the intervals (-inf,-M] and [M,inf) , since the delta neighborhood of x values need not depend on x, only epsilon as the slope varys less and less towards the ends of the domain.

4. Originally Posted by seams192
Why does it follow that it is uniformly continuous on the interval [-M,M]?
Any function that is continuous on a compact set of real numbers is uniformly continuous. $[-M,M]$ is compact.

5. Originally Posted by seams192
Why does it follow that it is uniformly continuous on the interval [-M,M]? I can see the intervals (-inf,-M] and [M,inf) , since the delta neighborhood of x values need not depend on x, only epsilon as the slope varys less and less towards the ends of the domain.
The specific case we need of the theorem says that

If $f:E\to\mathbb{R}$ is continuous with $E\subseteq\mathbb{R}$ compact then $f$ is in fact uniformly continuous
Are you aware of this theorem?

6. Ohh right. The Uniform Continuity theorem. Thanks!
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