4

I have to prove that $e^x \sin x$ is not uniformly continuous on $[0,\infty)$ using the mean value theorem. I proved that $\sin x + \cos x \geq 1$ for all $x\in [2 \pi k, 2\pi k + \pi/2], k\in\mathbb{N}$, and I see that $f'(x)=e^x (\sin x + \cos x)\geq e^x$, but I just can't think of way proving this with MVT. Any suggestions?

user249733
  • 43

2 Answers2

4

If $x,y \in [2 \pi k,2 \pi k+\pi/2]$, then the mean value theorem and the inequalities you've written imply that $|f(x)-f(y)| \geq e^{2 \pi k}|x-y|$. So if you pick $\varepsilon=1$ and let $\delta>0$ be arbitrary, then you can choose $k$ large enough that $e^{2 \pi k} \delta > 1$. This line of thinking should give you your result.

Ian
  • 101,645
  • @user249733 Given $k$ with $e^{2 \pi k} \delta>1$, let $x=2 \pi k$. Then there is a $y$ with $|x-y|<\delta$ and $|f(x)-f(y)| \geq 1/2$. – Ian Jun 21 '15 at 20:37
  • @user249733 I did one sloppy thing. I should have said to find $k$ with $e^{2 \pi k} \frac{\delta}{2} > 1$. Then $|f(2 \pi k)-f(2 \pi k + \delta/2)|>1$ and $\delta/2 < \delta$. This annoying factor of two floating around is because we need $|x-y|<\delta$ (strictly), whereas what I was initially doing would only have worked with $|x-y|=\delta$. – Ian Jun 22 '15 at 11:09
  • @user249733 Similarly, you technically have to split into the case $\delta>\pi/2$ and $\delta \leq \pi/2$ (because this is the range where you have the derivative inequality). But the same $x,y$ that you choose for $\delta$ of, say, $\pi/4$ will work for $\delta>\pi/2$. In other words, only $\delta$ in a small neighborhood of zero really matter at all. – Ian Jun 22 '15 at 11:19
  • @user249733 Yes, you should ceil that expression. – Ian Jun 22 '15 at 13:17
  • @user249733 That's the same thing; there's still something piecewise (in your case, the $\min$ function) that separates the situation with "large" $\delta$ from the situation with "small" $\delta$. – Ian Jun 22 '15 at 13:39
0

Let $f(x) = e^x \sin x$ then $f(2\pi k) = 0$ and $f(2\pi (k + 0.5)) = e^{2\pi (k+0.5)} $. Then using the mean value theorem, there is a point $c \in 2\pi k + [0, \frac{\pi}{2}]$ where the derivative is the slope of this line:

$$ f'(c) = \frac{e^{2\pi (k+0.5)} - 0}{2\pi (k+ 0.5) - 2\pi k} = \frac{e^{2\pi (k+0.5)} }{\pi k}$$

Since $f$ has a derivative at $c$ it is continuous there.

cactus314
  • 24,438