Prove the function $e^{-x^2}$ is uniformly contiuous on $[0,\infty)$ I have no idea how to prove this. But my teacher gave me a hint of taking cases of $\epsilon\ge 1$ and $0\lt\epsilon\lt1$ .
Please give some insight and thank you so much.
Additional information (suggestions from teacher):


*

*Note the range of the function. Also, note that the function is invertible and continuous.

*Take cases of $\epsilon\ge 1$ and $0\lt\epsilon\lt1$.

*Continuous functions on an interval of the form $[a, b]$ are uniformly continuous.

*Exploit (3) together with (1) to flush out a proof for $0\lt\epsilon\lt1$.

 A: You could do the standard proof to show that a continuous function on $[0, \infty)$ that has a limit at infinity is uniformly continuous. The idea is the following : if you go far enough, say for $x \geq A$ where $A$ is "big enough", since the function goes to $0$ at infinity, then the space between two images become as small as you want. But before, on $[0,A]$, $f$ is uniformly continuous since $[0,A]$ is a compact and you can use the corresponding $\eta$. For a more "formal" proof with a continuous function $f$ that goes to $l$ at infinity :
let $\epsilon > 0$. Since $f(x) \underset{x \to +\infty}{\longrightarrow} l $, there exists $A \geq 0$, such that $$\forall x \geq A, |f(x)- l| \leq \frac{\epsilon}{2}$$
On the other hand, $f$ is continuous and thus uniformly continuous on the compact $[0,A+1]$ : there exists $\eta_0$, such that $$\forall x,y \in [0,A+1], |x-y| \leq \eta_0 \implies |f(x)-f(y)| \leq \epsilon $$
Then, you can check that for $\eta=min(\eta_0,1)$
$$\forall x,y \in [0,\infty), |x-y| \leq \eta \implies |f(x)-f(y)| \leq \epsilon $$
because


*

*if $x$ and $y$ are in $[0,A] \subset [0,A+1]$, by the choice of $\eta$, it is true

*if $x$ and $y$ are in $(A,\infty)$ then whatever is the distance between them, you have : $$|f(x) - f(y)| = |f(x) - l + l - f(y)| \leq |f(x) - l| + |l - f(y)| \leq \frac{\epsilon}{2} + \frac{\epsilon}{2} \leq \epsilon$$

*if one is in $[0,A]$ and the other is not (let's assume $x < A < y$), since $\eta \leq 1$, you have actually $x,y \in [0,A+1]$ and hence once again by definition of $\eta$, it works.
In you case
This is the general proof, where we take into account that the function could be over or under $l$ at infinty, and is not assumed monotonous. In your special setting, you can explicit the A. Let's recall that for $f : x \mapsto e^{-x^2}$ :


*

*$f$ is decreasing from $1$ (at $0$) to $0$ (at infinty), which is its limit

*$f$ is continuous

*$f$ is invertible (let's call $f^{-1}$ its inverse)


Thus $A$ can be determined by the inverse : if we call $A(\epsilon)$ the $A$ such that $x \geq A \implies |f(x)| \leq \epsilon$, then it is when the image of $f$ reaches $\epsilon$ (after that, by monotony it will always be between $0$ and $\epsilon$), so $A(\epsilon) = f^{-1}(\epsilon)$. This is why there is a special case for $\epsilon \geq 1$, since your function is never greater than 1, the inverse $f^{-1}(\epsilon)$ is not defined for $\epsilon > 1$. Actually, you don't care because you have that for $$0 \leq x \leq y, |e^{-x^2} - e^{-y^2}| = e^{-x^2} - e^{-y^2} \leq e^{0} - lim_{y \to \infty}e^{-y^2}=1$$ and there is nothing to do (any $\eta$ will work).
You can also "improve the epsilons". We use triangle inequality for a general $f$ near its limit, but by ordering $x$ and $y$, $f$ being monotonous in our case, you can remove absolute value and it is sufficient to take $A(\epsilon)$ instead of $A(\frac{\epsilon}{2})$ when picking $A$. And now you can use your inverse as asked by the teacher. 
Concerning the $\eta_0$, since it has to do with the variations of $f$, you would have to use the MVT or derivative-related arguments to make it explicit. But it looks like you're not allowed to. Anyway, if he asks to use the fact that $f$ is uniformly continuous on a segment $[a,b]$, then it must be the argument you should use to get the $\eta$, which is not explicitely given by, so I think you can't do better and this is ok as it is.
Anyway, all of this aims at making the $A$ explicit for an "optimized" proof in this particular setting, but the general proof is also obviously working.
A: Here's the approach referred to in the comments.
A function continuous over a compact set is uniformly continuous.  While $[0,\infty)$ is not compact, we can "make it compact" by adding a point "$+\infty$" and then setting the value of $e^{-x^2}$ to be $0$ at this point.  The resulting function is continuous (since $e^{-x^2}$ is continuous over the reals, and $\lim_{x \to +\infty} e^{-x^2} = e^{-\infty^2}$ as we have defined it).  The domain is now compact, so the function is uniformly continuous.  And so the restriction of this function to $[0,\infty)$ is uniformly continuous, and we are back at our original function.
A: Hint: the derivative is $-2xe^{-x^{2}}$, it is continuous and in $0$ is $0$ and it goes to $0$ while $x\to \infty$. It is also non-positive in your domain. Then it has a minimum. Consider the absolute value of this minimum, it represent your lipschitz constant for the function.
A: For given $\varepsilon >0$ and any $r>0$, we find by continuity on compact sets a $\delta_r >0$ such that $f(x):=e^{-x^2}$ is uniformly continuous on $[0,r]$. 
For $r>0$ so large that $f(r)<\varepsilon$ and any $x,y>r$ we compute (assuming that $x<y$)
$$ |f(x)-f(y)|=f(x) \Bigl|1-\frac{f(y)}{f(x)}\Bigr|\leq f(r)<\varepsilon$$
Note that we exploited the monotonicity of the exponential function and that $y>x$. Hence $0\leq f(y)/f(x) \leq 1$.
Summarizing, we obtained uniform continuity 
$|f(x)-f(y)|< \varepsilon$ if either $|x-y|<\delta_r$ or if $x,y>r$.
What remains is to "glue" the results: So far we obtained the uniform continuity on the interval $[0,r]$, i.e. for $x,y\in[0,r]$ there holds $|f(x)-f(y)|<\varepsilon$ if $|x-y|<\delta_r$. If $x,y>r$ we obtain always $|f(x)-f(y)|< \varepsilon$. It remain to consider the case where $x\leq r<y$. This consideration will be  "the glue" to connect the results 
on the separated intervals.
We may choose $r>1$. Then the intervals $[0,r]$, $[0,2r]$ and $[r,\infty)$ overlap and the intersection is at least the interval $[r,r+1]$ (since $r>1$).
Hence, for $\varepsilon >0$ choose $r$ so large that  $f(r)<\varepsilon$. Then, choose $\delta=\min\{\delta_r,\delta_{2r},1\}$ and we infer $|f(x)-f(y)|<\varepsilon$
for any $|x-y|<\delta$
