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Let $f(x)=e^x \sin x$. Prove that the function is not uniformly continuous in the interval $[0,\infty)$.

What I tried:

Let $x=2\pi k, y=2\pi k + \frac{\pi}{2}$. I have to prove that there is some $\varepsilon>0$ such that $|f(y)-f(x)|\geq \varepsilon$ where $|y-x|<\delta$.

$$|f(y)-f(x)|=\left|e^{2\pi k+\frac{\pi}{2}}\cdot \sin\left(2\pi k+\frac{\pi}{2}\right) - e^{2\pi k}\cdot \sin(2\pi k )\right|.$$

I know that $\sin (2\pi k)=0, \sin\left(2\pi k + \frac{\pi}{2}\right)=1$, therefore: $$|f(y)-f(x)|=|e^{2\pi k+\frac{\pi}{2}}\cdot 1 - e^{2\pi k}\cdot 0|=|e^{2\pi k+\frac{\pi}{2}}|.$$

Now it seems like all I have to do is select some $\varepsilon>0$, for instance $\varepsilon=1$ and I'm finished.

Am I?

Is my solution correct?

DMcMor
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Alan
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1 Answers1

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Not quite. You need to prove that for some $\epsilon$, no matter how small you make $\delta$, you can find $x, y$ such that $|y - x| < \delta$ and $|f(y) - f(x)| \geq \epsilon$. One approach that could work: note that for $\delta > 0$ small enough, $\sin (2\pi k + \delta) > \delta/2$, so let's choose, say, $x = 2\pi k$ for some very large integer $k$ and $y = 2\pi k + \delta/2$. Obviously $|y - x| < \delta$, and $$|f(y) - f(x)| = \sin (\delta/2) e^{2\pi k + \delta/2} > \frac{\delta}{4} e^{2\pi k + \delta/2} $$ which may be made arbitrarily large, no matter how small $\delta$ is, by choosing $k$ large enough.

A reminder: in logical symbols, the criterion for uniform continuity on an interval $I$ is $$(\forall \epsilon > 0) (\exists \delta > 0) (\forall x \in I)(\forall y \in I) (|y-x| < \delta \implies |f(y) - f(x)| < \epsilon)$$ (unlike ordinary continuity, where $\forall x$ and $\exists \delta$ are reversed); the negation of this, by de Morgan's laws, is $$(\exists \epsilon > 0) (\forall \delta > 0) (\exists x, y \in I) (|y - x| < \delta \wedge |f(y) - f(x)| \geq \epsilon)$$ (recall that $A \implies B$ is equivalent to $(\neg A) \vee B$).

Connor Harris
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  • Thanks! how do we know for sure that $$|f(2\pi k + \delta) - f(2\pi k)| > \frac{\delta}{2} e^{2\pi k + \delta} $$? – Alan Apr 11 '17 at 13:35
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    $f(2\pi k) = 0$ and $f(2\pi k + \delta) = \sin(\delta) e^{2\pi k + \delta}$, and $\sin(\delta) > \delta/2$ for $0 < \delta \ll 1$. – Connor Harris Apr 11 '17 at 13:36
  • Is there a way of solving this using derivative? – Alan Apr 11 '17 at 13:52
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    @Alan it is sufficient to know that the derivative is unbounded. – Umberto P. Apr 11 '17 at 13:55
  • See here: http://math.stackexchange.com/questions/291166/prove-bounded-derivative-if-and-only-if-uniform-continuity – Connor Harris Apr 11 '17 at 13:56
  • @UmbertoP. if the deriviative is unbounded then the function is not uniformly continuous? I know the opposite is true - if the deriviative is bounded, the function is uniformly continuous. – Alan Apr 11 '17 at 14:03
  • In general, no. For this specific function, yes. – Umberto P. Apr 11 '17 at 14:04
  • @UmbertoP. thank you. However, I couldn't quite understand why this specific case proves that. Could you elaborate? Thanks! – Alan Apr 11 '17 at 14:08
  • @ConnorHarris the only issue I have with your answer is that when I looked up in my book, the definition is that $|y-x|<\delta$ where in your answer $|y-x|\leq \delta$. Can I still use it? Thanks! – Alan Apr 11 '17 at 14:45
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    @Alan The difference between strict and non-strict inequality is immaterial, but I'll edit my answer above to adjust to your version. – Connor Harris Apr 11 '17 at 15:03
  • Is there any short method to solve the problem?. If this question is asked some entrance examination, How would I do? –  Oct 09 '17 at 03:27