Where $a$ and $b$ can be any real numbers, including fractional or negative. But $x$ is positive.
Asked
Active
Viewed 82 times
1
-
1No. Take $a=b=-1$. – Parcly Taxel Apr 20 '17 at 00:26
-
@ParclyTaxel what do you mean? $1/x$ is convex on $(0,\infty)$. – user251257 Apr 20 '17 at 01:28
1 Answers
2
Hint: $\log(e^{x^a+x^b})$ simplifies to $x^a+x^b$, which is twice differentiable, so it's convex iff its second derivative is nonnegative over its domain (in your example the domain is $x>0$). This should lead to a counterexample. For simplicity try $a=b$.
grand_chat
- 40,909