Wednesday, October 24, 2018

real analysis - Is this a Darboux function?



Let $f(x)=x$ if $0\leq x\leq 1$ and $f(x)=x-\frac{1}{2}$ if $1

Questions


  • Am I right? Is $f$ a Darboux function?

  • If yes, then why do authors give complicated examples of Darboux functions like $g(x)=\sin\left(\frac{1}{x}\right)$ if $x>0$ and $g(0)=0$ or $h(x)=\left(x^2\sin\left(\frac{1}{x}\right)\right)'$ which are hard to draw near $x=0$?. Even worse when they mention Conway base 13 function.
  • Answer



    It doesn't satisfy the intermediate value theorem :



    $$f(\frac{9}{10}) = \frac{9}{10}$$



    $$f(\frac{11}{10}) = \frac{6}{10}$$




    But there is no $x \in [\frac{9}{10}, \frac{11}{10}]$ such that $f(x) = \frac{8}{10}$


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