We know there exists some functions, such that Weierstrass one, that are continuous everywhere on $[0 ; 1]$ and nowhere differentiable. Their expression (and a little bit of work) would yield a proof. However, I was wondering if we can just prove the existence of such a function, without finding it. I searched on the web and found nothing. Would some of you have a reference or a proof?

Thanks in advance!

  • $\begingroup$ It's been a while, but I think this is a case where Baire category theorems come into play. $\endgroup$ – Cameron Williams Jan 16 at 21:11
  • $\begingroup$ Thank you I'll check that. $\endgroup$ – Cauchy is my master Jan 16 at 21:17
  • $\begingroup$ There is an example using the Baire Category Theorem in "A Primer of Real Functions" by Ralph Boas. $\endgroup$ – awkward Jan 17 at 15:32

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