Prove uniform continuity of $e^{-x^2}$ on $\Bbb R$.

Question

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.

Answer 1

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?

Answer 2

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 $[0,+\infty)$.

Answer 3

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.