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Given continuous and Periodic function, How can I prove that it is Uniformly continuous?

Thank you!

1 Answers1

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Observe that it is uniformly continuous when restricted to a closed interval whose length is a period of your function (or twice that, to simplify things a little bit), and then use periodicity to extend to the whole of $\mathbb R$.

Mariano Suárez-Álvarez
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  • This is a hint... It is really not hard to do what you want. If you extend your question with details about what you have already tried or done, then it would probably be easier to help you. – Mariano Suárez-Álvarez Mar 03 '11 at 15:25
  • Thank you for the comment.. I tried to use cantor theroem on [0,T] interval, and play a little bit with the definition using delta and epsilon, for any random x1 and x2 on R, Didn't manage to complete the proof. –  Mar 03 '11 at 15:31
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    @Nir: I would really suggest that you play more with it before someone posts a complete solution... – Mariano Suárez-Álvarez Mar 03 '11 at 15:48
  • :) o.k I'm going to play. –  Mar 03 '11 at 15:54
  • game is over.. Can't do that. –  Mar 03 '11 at 18:41
  • The problem can only be due to the boundaries of your interval, i.e., to show that for $x_1=T-\alpha \delta$ and $x_2= T+ (1-\alpha) \delta$ with $0< \alpha <1$ it follows that $f(x_1) - f(x_2) < \epsilon$. Am I correct? Did you consider Mariano Suárez-Alvarez comment to use the interval [0,2T] for the Heine-Cantor theorem? Did you use that $f(x) = f(x+T)$? – Fabian Mar 03 '11 at 20:24
  • Another hint: There is a mighty theorem saying that a function which is continuous on some compact set $X$ is automatically uniformly continuous on $X$. – Christian Blatter Mar 03 '11 at 21:26