Let $f:(0,1)\to (0,1)$ be a continuously differentiable function.Then we can conclude that

  • $g=\frac{1}{f}$ is a continuous function on $(0,1)$.
  • $g=\frac{1}{f}$ is a continuously differentiable function on $(0,1)$.
  • $g=\frac{1}{f}$ is a uniformly continuous function on $(0,1)$.
  • $h$ defined by $h(x)=x(1-x)f(x);x\in (0,1)$ is uniformly continuous.

My effort: Since $f(x)\neq 0\forall x$ so $\frac{1}{f}$is a continuously differentiable function. Hence 1 and 2 are correct.

For 3 I have chosen $f(x)=x$

For 4 ;We know that $h$ will be uniformly continuous over $(0,1)$ iff it can be extended continuously over $[0,1]$.Moreover $h$ is a differentiable function so if I can show that $h^{'}$ i.e. if I can show that $f^{'}$is bounded then it will be uniformly continuous.

But how can I conclude that $f^{'}$ is bounded?Please help me out.

  • 2
    $\begingroup$ The first three of your questions are identical. For the last, look at functions like $\frac 1{x^2}$. $\endgroup$ – lulu Aug 8 '16 at 11:55
  • $\begingroup$ I have done required edits;your example does not work $\endgroup$ – Learnmore Aug 8 '16 at 12:01
  • $\begingroup$ Quite right as to bad example. Ignored condition on range. $\endgroup$ – lulu Aug 8 '16 at 12:24

Since $f$ is bounded, the factor $x(1-x)$ forces the continuity in $x=0$ and $x=1$. A continuous function on a compact set is automatically uniformly continuous.

  • $\begingroup$ Thank you very much for the details $\endgroup$ – Learnmore Aug 8 '16 at 12:41

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.