proving that $f(x)=\sin(e^x)$ is not evenly continuous

Question

how do you prove that a given function, $f(x)=\sin(e^x)$ is not evenly continuous(I think that it's called uniformly continuous also).

I don't even know how to approach it or how to show that it is not continuous and where. please explain to me how to approach it or how to prove it.

wish I could explain which steps I used, but I honestly don't know how to solve this one.

thank you very much, hope you can help and explain to me so i'll know how to approach this kind of problems in the future.

Answer 1

It can be done with sequences. Take $x_n=\ln(\frac{\pi (n+1)} {2} )$ and $y_n=\ln(\frac{\pi n} {2}) $. We have that $ |x_n - y_n|\to 0 $ if $n \to \infty$ but: \begin{align} |\sin(e^{x_n}) - \sin(e^{y_n} ) | = 1 \end{align}

Conclude.

Answer 2

You are going to use the definition of continuity, but you need to show that your choice of $\delta$ will depend on $x$. If a function is uniformly continuous on $\mathbb{R}$, then the choice of $\delta$ is the same for every $x$.

In other words, If a function is just continuous, you still will be able to find a $\delta$ for every $x$, it just will be a different $\delta$.

Consider $|sin(e^{ln(\pi)}-sin(e^{ln(2 \pi})|< \epsilon$ and $|ln(\pi)- ln(2 \pi)|$

We obtain $|0|<\epsilon$ and $|\pi|< \delta$. Choose $\delta=\pi + \epsilon$. However, for $|sin(e^{ln(\pi/2)}-sin(e^{ln( \pi)})|< \epsilon$ we get $|1|<\epsilon$ and $|\pi- \pi/2|< \delta$ , which requires a different $\delta$. Pick $\delta=\pi/2 + \epsilon$.

Keep in mind also if a function's derivative is bounded, then the function is Lipschitz continuous, which implies uniform continuity.

Answer 3

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\eps}{\varepsilon}$Here's a visual strategy: Plot your function $f:\Reals \to \Reals$. Now, fix a positive number $\delta$ (arbitrarily small), and imagine sliding an interval of length $\delta$ along the number line. Take the part of the graph $y = f(x)$ lying over your interval and project it to the $y$-axis. As the $\delta$-interval slides across the domain, the "shadow" on the $y$-axis will change position and length. The question you're trying to answer is:

If $\eps > 0$ is arbitrary, can we ensure the lengths of the shadows on the $y$-axis are no longer than $\eps$ just by taking sufficiently short intervals in the domain?

If yes, then $f$ is uniformly continuous: For every $\eps > 0$, there exists a $\delta > 0$ such that if $|x - y| \leq \delta$, then $|f(x) - f(y)| \leq \eps$.

If no, then $f$ is not uniformly continuous: There is an $\eps > 0$ such that for every $\delta > 0$, there are points $x$ and $y$ in the domain of $f$ such that $|x - y| \leq \delta$ but $|f(x) - f(y)| > \eps$.

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In the case of $f(x) = \sin(e^{x})$, it should be clear that the graph $u = e^{x}$ is "arbitrarily steep". This suggests we can find points $x$ and $y$ that are "close", but such that for some integer $n$, we have $e^{x} = 2n\pi - \frac{\pi}{2}$ and $e^{y} = 2n\pi + \frac{\pi}{2}$, in which case $$ |f(x) - f(y)| = |\sin(e^{x}) - \sin(e^{y})| = |\sin(-\pi/2) - \sin(\pi/2)| = |-1 - 1| = 2. $$ In the animation, note how the image interval "pegs" on $[-1, 1]$ as the domain interval nears the right edge of the frame. The point here is, no matter how small $\delta > 0$ is chosen, there is a $\delta$-interval sufficiently far to the right (possibly outside the animation frame) on which "$f$ goes through at least half a complete oscillation". The corresponding image has length $2$, which is not arbitrarily small.