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I am trying to show that $f(x)=e^{-x}$ is uniformly continuous on $[0,\infty)$ and not having much success. I'm attempting to use a modified version of the following result (which I found on Math Stack Exchange here: https://math.stackexchange.com/questions/172988/please-prove-uniform-continuity ), that if $f$ is continuous on $[a,\infty)$ and the limit of $f(x)=L$ as $x$ approaches infinity, then $f$ is uniformly continuous on $[a,\infty)$.

This is the work I've done so far. I really don't know if it's valid reasoning... The first problem I'm encountering is that the book has not yet defined functional limits, so I can't really use the proof directly. We have, however defined sequential limits and the notion of a decreasing function. If I define a sequence $x_n=n-1$ for $n$$\in$$\mathbb{N}$, then as $n$ goes to infinity, $f(x_n)\rightarrow0$. For $\epsilon/2$, $\exists$N$\in$$\mathbb{N}$ s.th for all n $\geq$ N, |$f(x_n)|$ <$\epsilon/2$. Then, since $f(x)$ is a decreasing function, this inequality holds for all $x$$\geq$$x_N$. Since $\mathcal{S}$=$[0,N+1]$ is closed and bounded, and hence a compact set in $\mathbb{R}$, and $f(x)=e^{-x}$ is continuous on $\mathcal{S}$, then by the Uniform Continuity Theorem, $f$ is uniformly continuous on $\mathcal{S}$.

So, if $x,y$ $\in$$\mathcal{S}$,$\forall$$\epsilon>0,$ $\exists$ $\delta>0$ s.th |$x-y$|<$\delta$ then |$f(x)-f(y)$|<$\epsilon$. Then (really not sure about this part), if $x, y$ $\in$$(N+1,\infty)$, for any $x$ and $y$ in this set, we have $|f(x)-f(y)|=|f(x)+(-f(y))|$$\leq|f(x)|+|f(y)|<\epsilon/2+\epsilon/2=\epsilon$. So $f$ is uniformly continuous on $(N+1,\infty)$. Provided this part is correct, I am having trouble with the case where $x$$\in[0,N+1]$ and $y$ $\in(N+1,\infty)$. Provided it's not correct, I'm having trouble on that part as well. Provided none of this is correct, well... I don't know.

Any help would be appreciated. Is this an appropriate way to approach the problem? Are there other ways?

Community
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mrmingus
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    Forget about limits. What can you say about the set of values of the derivative $f'(x) = - e^{-x}$ for $x \in [0,+\infty)$? – Lee Mosher Apr 25 '14 at 15:57
  • We have not yet defined the derivative. We're using Elements of Real Analysis by Bartle. – mrmingus Apr 25 '14 at 16:00

2 Answers2

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For any $x>0$ and $y>0$ we have $$ \left\vert \mathrm{e}^{-x} - \mathrm{e}^{-y} \right\vert = \left\vert \mathrm{e}^{-\min(x,y)} \left(1 - \mathrm{e}^{-|y-x|}\right) \right\vert = \mathrm{e}^{-\min(x,y)} \left(1 - \mathrm{e}^{-|y-x|}\right) \leqslant \mathrm{e}^{-\min(x,y)} |y-x| $$ Hence for all $x>0, y>0$ such that $|x-y| < \delta$ $$ \left\vert \mathrm{e}^{-x} - \mathrm{e}^{-y} \right\vert < \delta $$ which verifies the direct definition of uniform continuity.

Sasha
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  • Any reason for the downvote? Is the argument wrong? not what the OP is asking? – Sasha Apr 25 '14 at 16:19
  • I like the trick with $\max(x,y)$ :), so I upvoted. – Caran-d'Ache Apr 25 '14 at 16:20
  • @Caran-d'Ache Thanks for your vote. – Sasha Apr 25 '14 at 16:22
  • To just expand Sasha's answer, recall that $$\max(x,y) = \frac{x +y + |x - y|}{2}$$ – IAmNoOne Apr 25 '14 at 16:41
  • @Sasha I don't understand the first equality. If I substitute in the formula given by Nameless, it doesn't work out. Can you please explain? – user137090 Apr 27 '14 at 11:30
  • @user137090 Consider first the case of $x<y$. Then $\exp(-x) - \exp(-y) = \exp(-x) \left(1-\exp(-(y-x)\right)$. Now consider $y<x$. By the way, I had an error in the formula. It should be $\min$, instead of $\max$. – Sasha Apr 27 '14 at 12:11
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    @user137090 More formally:$$ | \exp(-x) - \exp(-y) | = | \exp(-\max(x,y)) - \exp(-\min(x,y)) | = \exp(-\min(x,y)) | 1 - \exp(-\left(\max(x,y)-\min(x,y)\right)) | = \exp(-\min(x,y)) | 1 - \exp(-|y-x|) |$$ – Sasha Apr 27 '14 at 12:14
  • Hello there. I don’t understand why does $1-e^{-|y-x|}\leq |y-x|$ – EMS Dec 08 '22 at 16:27
  • Please see https://math.stackexchange.com/a/504671/11069 – Sasha Dec 10 '22 at 00:21
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Proof 1

$f'(x)$ is bounded on $[0,\infty)$: $|f'(x)|\leq M $ for some $M\geq0$

Lagrange mean value theorem

$$|f(x)-f(y)|\leq M |x-y|$$

$f$ is uniformly continuous on $[0,\infty)$

Proof 2

If you want to use the fact that the limit of $f(x)=L$ as x approaches infinity, then $f$ is uniformly continuous on $[a,\infty)$

1). $\lim f(x)=L $, so, $\forall \epsilon \gt0$, there is $T \gt 0$ such that

$$|f(x)-f(y)|\lt \epsilon \qquad x,y\gt T$$

2). $f$ is uniformly continuous on $[0,T+1]$, that is, there is $\delta_1 \gt 0$ such that $x,y\in [0,T+1] ,|x-y|\lt \delta_1$

$$|f(x)-f(y)|\lt \epsilon $$

3). Let $\delta=\min\{1,\delta_1\}$. If $x,y\geq0, |x-y|\lt \delta $, then $x,y\in [T,\infty) $ or $x,y\in [0,T+1]$. From 1) and 2), we have

$$|f(x)-f(y)|\lt \epsilon $$

so $f$ is uniformly continuous on $[0,\infty)$

ziang chen
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  • This is good. I like proof 2. I'll keep proof 1 in mind for future use once we've defined the derivative. Thank you. – mrmingus Apr 25 '14 at 16:43