Uniform continuity and antiderivative of $x \mapsto e^{-x^2}$ 
Show that the antiderivatives of $x \mapsto e^{-x^2}$ are uniformly continuous in $\mathbb{R}$.

So we know that for a function to be uniformly continuous there has to exists $\varepsilon$ s.t when $|x-y| < \delta$ implies $|f(x)-f(y)| < \varepsilon$.
But how do we go about this since we cannot really integrate $e^{-x^2}$and we need the antiderivative?
 A: Every function $f$ defined and differentiable over an interval, with a bounded derivative, is uniformly continuous. Indeed, by the mean value theorem, if $x\ne y$, we can say that
$$
\frac{f(x)-f(y)}{x-y}=f'(c)
$$
with $c$ between $x$ and $y$. Therefore, if $|f'(x)|\le L$ for every $x$, we get
$$
|f(x)-f(y)|\le L|x-y|
$$
Thus $f$ is not only uniformly continuous, but also Lipschitzian, which is a stronger property.
Since $0\le e^{-x^2}\le 1$, we're done.
A: Any antiderivative can be written in the form
$$
\int_0^x {e^{ - t^2 } dt}  + C,
$$
where $C$ is a constant. Thus
$$
\left| {\int_0^x {e^{ - t^2 } dt}  + C - \left( {\int_0^y {e^{ - t^2 } dt}  + C} \right)} \right| = \left| {\int_y^x {e^{ - t^2 } dt} } \right| \le \left| {\int_y^x {1dt} } \right| = \left| {x - y} \right|.
$$
Can you finish from here?
A: Let $F(x) = \int_{-\infty}^{x} e^{-t^2}dt$ be the antiderivative of $e^{-x^2}.$ 
Without loss of generalities (indeed, the problem is symmetric if we exchange $x$ and $y$), let's assume $y  > x$ and consider the following:
$$|F(y) - F(x)| = \left|\int_{-\infty}^{y} e^{-t^2}dt - \int_{-\infty}^{x} e^{-t^2}dt\right| = \left|\int_{x}^{y} e^{-t^2}dt\right| = \int_{x}^{y} e^{-t^2}dt.$$
Since $e^{-t^2} \leq 1$, then:
$$|F(y) - F(x)|  = \int_{x}^{y} e^{-t^2}dt \leq  \int_{x}^{y} dt = y-x.$$
Now, for any $\varepsilon > 0$, there exists a $\delta > 0$, such that if we choose $|y-x| < \delta = \varepsilon$, then $|F(y) - F(x)| < \varepsilon.$
Hence, the antiderivative of $e^{-x^2}$ is uniformly continuous.
A: MVT for integrals:
$f(x)=\displaystyle{\int_{a}^{x}}e^{-t^2}dt$
$|f(x)-f(y)|=|\displaystyle{\int_{y}^{x}}e^{-t^2}dt|=$
$e^{-s^2}|x-y| \le |x-y|,$
where $s \in (\min(x,y),\max(x,y))$.
