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?

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.

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)

## 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

A function is smooth if the absolute value of its second derivative is

upper boundedby 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)

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