1
$\begingroup$

Let $0 < a < 1$; let $b$ be an odd integer; let $ab > 1 + \frac{3\pi}{2}$; let $f: x \mapsto \sum_{n \geq 0}a^{n}\cos (b^{n}\pi x): \Bbb{R} \to \Bbb{R}$. Then $f$ is everywhere continuous and nowhere differentiable.

On page 39 of Counterexamples in Analysis (Dover edition), the authors write the Weierstrass function $f$ as $f(x) = \sum_{n \geq 0}b^{n}\cos (a^{n} \pi x)$; then I do not see how it can converge (of this "typo" series)...

If this is really an obvious typo, then there should already be someone that pointed it out; but so far I did not find any source doing this, so I ask here to clarify this point.

$\endgroup$
2
  • $\begingroup$ It's weird. In the index it's stated that there's an errata, but it's not included somehow. $\endgroup$
    – user99914
    Nov 1 '15 at 5:47
  • $\begingroup$ @JohnMa Thank you for feedback. Maybe I should contact the authors (if they are still alive. :)) $\endgroup$
    – Megadeth
    Nov 1 '15 at 5:52
1
$\begingroup$

The typo is the constraints on $a$ and $b$. In the book it says that $b$ is an odd integer, it should actually be $a$. And $b$ should be $0<b<1$. That, or the $a$ and $b$ should be switched in the series.

To see this converges, just use the Weierstrass M-test. $|a^n\cos(b^n \pi x)|\leq a^n$ and $\sum a^n$ converges.

$\endgroup$
8
  • $\begingroup$ Thanks for feedback. The remedy I guess is like yours: $a$ should be put where $b$ is and $b$ should be put where $a$ is. $\endgroup$
    – Megadeth
    Nov 1 '15 at 5:50
  • $\begingroup$ @GudsonChou If you look at en.wikipedia.org/wiki/Weierstrass_function#Construction you'll see they have the same definition with $a$ and $b$ switched. $\endgroup$
    – user223391
    Nov 1 '15 at 5:51
  • $\begingroup$ Oh, right, thanks for that. Now I tend to maintain more that it is simply a typo. $\endgroup$
    – Megadeth
    Nov 1 '15 at 5:54
  • $\begingroup$ @GudsonChou Actually the proof it converges is really straightforward. Just use Weierstrass M-test. $\endgroup$
    – user223391
    Nov 1 '15 at 5:58
  • $\begingroup$ Sorry? Yes, it is clear that it converges; but this is not my question :) $\endgroup$
    – Megadeth
    Nov 1 '15 at 5:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.