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Given that $e^{{-x}^2}$ is a continuous function: $ℝ → ℝ$; and $\lim_{x→\infty}e^{{-x}^2}=0$ and $\lim_{x→-\infty}e^{{-x}^2}=0$, prove that $e^{{-x}^2}$ is uniformly continuous on $ℝ$.

So I know the limit definition is as follows: $\lim_{x→\infty}f(x)=A$ means: given $\epsilon>0$, $\exists R\inℝ$ depending on $\epsilon$ such that if $x>R$ then $|f(x)-A|<\epsilon$.

Is it sufficient, then, to say that since $f(x)=e^{{-x}^2}$ is a continuous function with limits as $x$ approaches $-\infty$ and $\infty$ then $f(x)$ is uniformly continuous? Help writing a full, complete proof would be awesome. Thanks.

Error 404
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  • The first step would be to write down the definition of uniform continuity. – mathematician Nov 14 '16 at 09:59
  • Addition to your question: Generally if $f$ is continuous on $\Bbb R$ with $\lim_{x\to \infty} f$ and $\lim_{x\to -\infty}f$ exists and finite, then $f$ would be uniformly continuous on $\Bbb R$. – Nick Nov 14 '16 at 10:03
  • With a bit of searching, you can probably find several posts about uniform continuity of this function on this site" http://math.stackexchange.com/questions/772458/prove-the-function-e-x2-is-uniformly-contiuous-on-0-infty http://math.stackexchange.com/questions/759004/uniform-continuity-proof-fx-e-x2 http://math.stackexchange.com/questions/2013318/prove-uniform-continuity-of-e-x2-on-r http://math.stackexchange.com/questions/1150454/sketch-the-proof-of-e-x2-being-uniformly-continuous-proof-is-given – Martin Sleziak Nov 14 '16 at 13:33
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    There are four questions in the last few hours from you with the same title. Please work a little harder to make your titles better reflect the question, so folks here know which ones they might want to help you with. – Ethan Bolker Nov 14 '16 at 13:40

3 Answers3

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Hint. Since $f$ is continuous it is uniformly continuous on compact set. Hence, if $f(x)$ is NOT uniformly continuous on $[0,\infty)$, by considering the proper negation of the definition of uniform continuity, there exist two sequences $x_n\to+\infty$ and $y_n\to+\infty$ such that $x_n-y_n\to 0$ and $f(x_n)-f(y_n)\not\to 0$.

Is this property in contradiction with the fact that $\lim_{x\to\infty}f(x) = L\in\mathbb{R}$?

Note that by using the same argument you can show that if $f$ is continuous in $[0,+\infty)$ and it has a linear asymptote $mx+q$ at $+\infty$ (possibly oblique) then $f$ is uniformly continuous in $[0,+\infty)$.

Robert Z
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You already have the existance of an $R>0$ s.t. $|f(x)| < \frac{\varepsilon}{2}$ $(-\infty,-R)\cup (R,\infty)$. So you are done for this area due to $|f(x) - f(y)| \le |f(x)| + |f(y)| < \varepsilon$

Still need to be shown it's uniformly continuous on $[-R,R]$, but what do you know about continuous functions on compact intervals?

Gono
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You can first show that if $f$ is a differentiable and if $|f'|$ is bounded, then $f$ is uniformly continuous. Then show that $e^{-x^2}$ satisfies the conditions. This will be a nice exercise.

Alternatively, you can use that continuous functions on compact domains are uniformly continuous.

Mathematician 42
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