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Have this in one of my finals practice questions:

Given $f(x) = e^{\sin x}$, prove $f(x)$ is uniformly continuous on $\mathbb R$.

The direction I'm thinking of is choosing $x_1 = 2\pi k + \delta$, and $x_2 = 2\pi k + \frac{\delta}2$.

so I need to prove that for all $\varepsilon > 0$, there exists $\delta > 0$ such that $|x_1-x_2|< \delta$ gives $|f(x_1)-f(x_2)|< \varepsilon$.

So, my choice of $x_1$ and $x_2$ gives $|x_1-x_2| = \delta/2 < \delta$,

and $|f(x_1)-f(x_2)| = |e^{\sin(2\pi k + \delta)} - e^{\sin(2\pi k+ \delta/2)}|$

This stage is a little dodgy for me.. any help on what to do next would be great.

Joe
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user2970357
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3 Answers3

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Since $|e^{\sin(x)}|\leq e$, we have

$$|e^{\sin(x)}-e^{\sin(y)}|= |e^{\sin(x)}||e^{[\sin(y)-\sin(x)]}-1|\leq e|e^{[\sin(y)-\sin(x)]}-1|.$$

Since $e^x$ is continuous at $x=0$, for any $\epsilon >0$ there exists $\delta(\epsilon) >0 $ independent of $x$, such that if $|x|<\delta(\epsilon)$ then $|e^x-1|< \epsilon.$

Consequently, if

$$|\sin(y)-\sin(x)|<\delta_0=\delta(\epsilon/e)$$

then

$$|e^{[\sin(y)-\sin(x)]}-1|< \epsilon/e.$$

But

$$|\sin(y)-\sin(x)|=2\left|\sin\frac{(y-x)}{2}\right|\left|\cos\frac{(y+x)}{2}\right|\leq2\left|\sin\frac{(y-x)}{2}\right|\leq|y-x|.$$

Therefore, if $|y-x| < \delta_0$ then $|e^{\sin(x)}-e^{\sin(y)}|<\epsilon$ for all $x,y \in \mathbb{R}$. Since $\delta_0$ does not depend on $x$ and $y$, the function is uniformly continuous.

RRL
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  • Guys I got what you are saying, but I need some input on what is wrong with the method I came up with, just in case, so I wont repeat it. What I did was: let delta = d. I chose x1 = 2piK+d, and x2= 2pik -d/2. so for |x1-x2| I get 0.5d Which is indeed < d as required. now, I want to show that for all eps>0 and for this delta such that |x1-x2| < d, |f(x1)-f(x2)| < eps. so I get |e^sin(2pik+d) - e^sin(2pik+d/2)| = |e^sin(d)-e^sin(d/2)| <= |e^d - e^(d/2)|, now select epsilon = e^d and i get |e^d - e^(d/2)| < epsilon. can i define epsilon in terms of delta anyway? Thanks for your input! – – user2970357 Jun 19 '14 at 20:23
  • The multiple of $2\pi$ is unecessary -- it cancels when you take $x_1-x_2$ You are just picking pairs that differ by some $\delta/2$, say $x_2$ and $x_1 = x_1 + \delta/2.$ We know for each $x_2$ we can find a $\delta(\epsilon,x_2)$ that makes $|f(x_2)-f(x_1)| < \epsilon$ The hard part is to find a $\delta$ independent of $x_2$. My proof shows how to find it. It is $\min(-\ln(1-\epsilon/e),\ln(1+\epsilon/e)$. – RRL Jun 19 '14 at 20:55
  • The key is to show $|e^{\sin x_1}-e^{\sin x_2}|= |e^{\sin x_1}||e^{(\sin x_2- \sin x_1)}-1|$ and to bound the first factor. Then you find the independent $\delta$ as the the one used to show $e^x$ is continuous at a fixed point -- $0$. – RRL Jun 19 '14 at 21:01
  • RRL, yea your proof is very nice, just want to make it clear.. by showing (in my proof), that i can find x1 and x2 such that |x1-x2| < delta such that |fx1-fx2| < eps, does not prove anything?. i mean, I understand that i only show this for x1,x2 that are delta/2 apart, but still.. and by the way, is it legit defining epsilon = e^delta ? (after noticing that value will make fx1-fx2 < eps). thanks for your time! – user2970357 Jun 19 '14 at 21:18
  • Yes because $e$ is just a constant. Since $e^x$ is continuous at $0$, there exists $\delta$ such that $|e^x-1|<\hat{\epsilon}$ for ANY $\hat{\epsilon}>0$ when $|x|< \delta$. Now find the $\delta$ for $\hat{\epsilon} = \epsilon/e$ – RRL Jun 19 '14 at 21:26
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One approach would be to say that, because the function is periodic with period $2 \pi$, a modulus of continuity at $x \in [0,2\pi]$ (that is, the $\delta=\delta(x,\varepsilon)$ so that $|x-y| < \delta \Rightarrow |f(x)-f(y)|<\varepsilon$) also works at $x+2k\pi$ for any $k \in \mathbb{Z}$. Then you could use the usual result about uniform continuity on compact sets.

Ian
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Note that $$ |F'(x)| = |e^{\sin(x)}\cos(x)| \leq e $$ So try using the Mean-Value Theorem.

Prahlad Vaidyanathan
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  • yea i also figured this part just but looking at the derivative I see its bounded by e , so its a lipchitz function hence uniformly cont. but i would like a formal delta-epsilon proof without using a derivative – user2970357 Jun 19 '14 at 16:08
  • @user2970357 You can work it out formally this way by noting $|f(x)-f(y)| \leq \int_x^y |f'(x)| dx \leq e(y-x)$. – Ian Jun 19 '14 at 17:31
  • Guys I got what you are saying, but I need some input on what is wrong with the method I came up with, just in case, so I wont repeat it. What I did was: let delta = d. I chose x1 = 2piK+d, and x2= 2pik -d/2. so for |x1-x2| I get 0.5d Which is indeed < d as required. now, I want to show that for all eps>0 and for this delta such that |x1-x2| < d, |f(x1)-f(x2)| < eps. so I get |e^sin(2pik+d) - e^sin(2pik+d/2)| = |e^sin(d)-e^sin(d/2)| <= |e^d - e^(d/2)|, now select epsilon = e^d and i get |e^d - e^(d/2)| < epsilon. can i define epsilon in terms of delta anyway? Thanks for your input! – user2970357 Jun 19 '14 at 20:21
  • $\delta$ is small, so you're not covering all possibilities for $x_1$ by doing what you're describing, you're only covering a small neighborhood of each multiple of $2 \pi$. In notation like yours, you would let $x_1 = 2 \pi k + x$ for an arbitrary $x \in [0,2 \pi)$, then find a $\delta(x_1,\varepsilon)$, then show that $\delta$ takes on a minimum value as $x_1$ varies. The key difference is that $x_1$ needs to be arbitrary, whereas in your setup it is not. – Ian Jun 20 '14 at 03:17