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I am trying to prove or disprove $f(x) = e^{x^{2}}$ and $g(x) = e^{-x^{2}}$ is uniformly continuous for $x \in (- \infty, \infty)$.

My book says a function is uniformly continuous on an interval $I$ if for every number $\epsilon \gt 0$ there exists a $\delta = \delta(\epsilon)$ such that $\left\{ x, x_0 \in I \text{ and } |x - x_0| \lt \delta \right\} \Rightarrow |f(x) - f(x_0)| \lt \epsilon$. So apparently all I need to do in order to prove uniform continuity is to find a $\delta$ that is independent of $x_0$. I have seen a lot of similiar examples but the choice of $\delta$ always seem rather arbitrary to me.

I have some stuff that I am allowed to refer to as well:

$\large{\text{Theorem 1}}:$ If $f$ is continuous on the closed, finite interval $\left[a, b\right]$, then $f$ is uniformly continuous on that interval.

$\large{\text{Satz 1}}:$ Assume $f$ is differentiable and that the derivative is bounded on the interval $I$. Then $f$ is uniformly continuous on $I$.

$\large{\text{Satz 2}}:$ If for every $h \gt 0$ is such that $|f(x + h) - f(x)|$ is bounded on $I$, then $f$ is not uniformly continuous on $I$.

Thankful for any help I can get, completely stuck with this. I think it's rather hard to see how to choose the $\delta$'s in particular.

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First, we show that $e^{x^2}$ is not uniformly continuous. If we suppose $|x-y| = \delta$, we may choose $x$ and $y$ sufficiently large such that $|x+y| \geq \frac{1}{\delta}$, in which case $|x^2 - y^2| \geq 1$. Then, assuming for convenience $x >y$, $$|e^{x^2} - e^{y^2}| = e^{x^2} (1 - e^{y^2 - x^2}) \geq e^{x^2} \big( 1 - e^{-1} \big)$$ And we see that the above difference can be made as large as we please, no matter which $\delta$ is given. Thus this function is not uniformly continuous.

For the second function, simply note that $e^{-x^2}$ has derivative bounded by $2/e$. Thus by the mean value theorem, choosing $\delta = e \cdot \epsilon /2$, $|x-y| < \delta$ implies $$|e^{-x^2} - e^{-y^2}| \leq \frac{2}{e} \cdot |x-y| < \epsilon$$ A fun generalization for the second problem: suppose $f : \mathbb{R} \to \mathbb{R}$ is continuous and $\displaystyle \lim_{|x| \to \infty} f(x) = 0$. Prove that $f$ is uniformly continuous on $\mathbb{R}$.

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  • $\begingroup$ Thank you. But isn't it supposed to be $\left| x - y\right| \lt \delta$, how can you pick $\left|x - y\right| = \delta$ ? $\endgroup$ – Amoz Feb 23 '18 at 14:28
  • $\begingroup$ I'm just calling that distance $\delta$, since it is an arbitrary positive number. $\endgroup$ – Rellek Feb 23 '18 at 18:02
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Assume $f$ was uniformly continuous. Applying the definition for $\epsilon = 1$, we find that there is a $\delta > 0$ such that the function $$F(x) := |e^{(x+\delta)^2} - e^{x^2}| < 1 \quad \forall x \in (-\infty,\infty).$$ By applying the mean value theorem, we find $$F(n) = |e^{(n+\delta)^2} - e^{n^2}| = 2\delta \xi_n e^{\xi_n^2} $$ Where $\xi_n \in [n,n+\delta]$. This shows that $\lim_{n\to\infty} F(n) = \infty$, a contradiction.

EDIT: We have $g'(x) = -2xe^{-x^2}$. By the mean value theorem, we find $$|g(x) - g(y)| \leq |g'(\xi)||x-y|$$ for some $\xi \in \mathbb{R}$. Series Expansion shows that $ 1 + x^2 \leq e^{x^2}$, which implies $\frac{1}{1+x^2}\geq e^{-x^2}$, which in turn implies $|g'(x)|\leq 1$ for all $x\in\mathbb{R}$. Therefore we have $$|g(x)-g(y)|\leq |x-y|.$$ More generally, a differentiable function with bounded derivative is Lipschitz continuous and therefore uniformly continuous.

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  • $\begingroup$ Thanks. I saw now that I had a bad typo, it was meant to be $e^{-x^{2}}$ not $e^{x^{-2}}$. If you wanna edit your answer I'd be very thankful :D $\endgroup$ – Amoz Feb 23 '18 at 14:24
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Hint: take the derivative of $e^{x^2}$ and show that it exists at every point, and that it is bounded on any interval (which isn't difficult).

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  • $\begingroup$ You could show that it is bounded on bounded intervals. Which isn't helpful. $\endgroup$ – user159517 Feb 19 '18 at 11:30

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