A question on infinite derivative.

• Jun 6th 2010, 06:42 PM
zzzhhh
A question on infinite derivative.
Definition of infinite derivative: Let $E=\text{dom }f, c\in E$, c is a limit point of E, real-valued function f is said to have a infinite derivative at c if f is continuous at c and the limit $\lim\limits_{x\to c}\frac{f(x)-f(c)}{x - c}$ is $+\infty$ or $-\infty$. We write in this case $f'(c)=+\infty$ or $f'(c)=-\infty$.
For finitely differentiable functions it is well-known that $(f+g)'=f'+g'$. But if both f and g has infinite derivative at c, what about the differentiability of the sum f+g at c? It may be still differentiable, e.g. $f=x^{1/3}$ and $g=-x^{1/3}$, f+g=0 so (f+g)'=0 at c=0 even if $f'(c)+g'(c)=(+\infty)+(-\infty)$ which is undefinable. Now I want to find an example such that $f'(c)=+\infty, g'(c)=-\infty$, but f+g is not differentiable at c, where c is a finite real number. I hope to find a function F satisfying 1)continuous at c, 2)has bounded derivative about c but except c and 3) $F'$ oscillates like $\sin\frac{1}{x}$ so that F is not differentiable at c. If I can find such F, My example can be easily constructed. But I failed to find such F, Can you help me find such F, or construct the example in another way? Thanks!
• Jun 7th 2010, 02:40 AM
Ackbeet
Try two functions that behave more or less the same way as the functions you gave, but don't combine. How about $f(x)=\sqrt[3]{x}$, and $g(x)=-\sqrt[5]{x}$? Here's an example where, I think, there's too much symmetry, and the solution involves destroying symmetry.
• Jun 7th 2010, 05:30 PM
zzzhhh
$\sqrt[3]{x}-\sqrt[5]{x}$ does not meet my requirement: it has derivative $-\infty$ at 0. We should not only destroy the symmetry, but also the destroyer is not differentiable at c. Since derivatives do not have first-kind discontinuities, I think the destroyer should have an oscillating derivative, that is, second-kind discontinuity, near c, as condition 3) in my first post indicated.
• Jun 7th 2010, 06:50 PM
Ackbeet
Wow. It is not intuitive at all that my candidate actually had a derivative at 0. I must confess I didn't actually check it. I just figured you'd have a cusp at the origin, which would not have a definable two-sided derivative. If you plot my function, it does appear to have a cusp at the origin. But it must hang to the left of the origin just enough to get that negative slope.

Anyway. You've got a tough assignment there. I would look into the construction of the everywhere continuous, nowhere differentiable function. I realize that that function, as is, does not solve your problem. But its construction might give you some ideas. You might also try cycloids which have bona fide cusps.