1
$\begingroup$

I have two functions $k(t)$ and $l(t)$ in a certain closed interval $[a,b]$ both functions are continuous and differentiable in the interval. In addition we have:

  • Both functions are increasing with bounded derivatives in the interval $ A_{max} > k'(t)> A_{min} >0$ and $B_{max} > l'(t)>B_{min}>0, \forall t \in [a,b]$
  • The functions and derivatives verify that $\frac{l'(t)}{k'(t)}<\frac{l(t)}{k(t)} \ \forall t \in [a,b]$, which implies that $l(t)<k(t)\ \forall t$ (you can prove it by integrating the first inequality on $t$)
  • There is a point of the interior, $c \in (a,b)$, where both functions are zero $k(c)=l(c)=0$ (i.e. they change simultaneously their signs)

Now with this information I can affirm that the quotient $\frac{l(t)}{k(t)}$ is continuous in $[a,b]$, because the only problematic point is $\frac{l(c)}{k(c)} = \frac{0}{0}$, but according to l'Hôpital $\lim_{t\to c}\frac{l(t)}{k(t)} = \lim_{t\to c}\frac{l'(t)}{k'(t)}$ which exists because the derivatives have bounded positive values. Ok.

But the question is:

Is the quotient $\frac{l(t)}{k(t)}$ also differentiable in $[a,b]$? or equivalently Is the quotient $\frac{l(t)}{k(t)}$ differentiable in $c \in [a,b]$? If not, Is there any counter example? What would be then the minimum conditions for differentiability in $c$ (or in $[a,b]$) to be proven?

Thanks a lot, I have been working on that for some days, but no result.

$\endgroup$
  • $\begingroup$ So the quotient is not defined in the point $c$, and You want to show that its continuous extension is differentiable under these assumptions? $\endgroup$ – Peter Melech Feb 9 '16 at 21:46
  • $\begingroup$ @PeterMelech the function is not defined but the limit "exists because the derivatives have bounded positive values". Therefore we define the value of the function as the limit, of course ;) $\endgroup$ – loved.by.Jesus Feb 9 '16 at 22:46

Your Answer

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

Browse other questions tagged or ask your own question.