Q5 exam 2020

Hello,

I don't understand why \( -log(x \) is not smooth, as is twice differentiable on \( \mathbb{R}^+ \) with \( f''(x) = \frac{1}{x^2} > 0 \forall x \in \mathbb{R}^+ \) Using theorem 1.9 along with lemma 6.1 (easy direction of lemma 2.4) shouldnt we get that f is smooth?

Thank you for any help!

Cheers

Yann

Top comment

A function is smooth if the absolute value of its second derivative is upper bounded by some positive constant. Since x is arbitrary small, it can reach arbitrary large values.

I'm not a teacher.

But the reason is :

Intuitively : look at the left side of the logarithm, it goes up very fast (you can't bound it)
Formaly : There is no upper bound for \(\dfrac 1 {x^2}\) (think of x going to 0)

Page 1 of 1

Add comment

Post as Anonymous Dont send out notification