frozenfish
New member
- Joined
- Sep 16, 2013
- Messages
- 4
Hello!
I was wondering whether there is a useful property that can be said about a general convex function \(\displaystyle f\) given some relations. I've already tried to search engine about properties of functions in general, but I didn't come across anything that would answer my question.
Okay, so what we know is this. We have three positive real numbers, \(\displaystyle x \geq 1, y > 1, z \geq 1\) and we have that \(\displaystyle x \leq y \cdot z\) (\(\displaystyle x\) and \(\displaystyle z\) can be integers if that's more helpful, but I doubt it). Now, for a general (increasing) convex function \(\displaystyle f\), we have that \(\displaystyle f(x) \leq f(y \cdot z)\). My question is, is there a property that says \(\displaystyle f(x) \leq f(y) \cdot f(z)\)? (Note that "\(\displaystyle \cdot\)" is simply multiplication.) I could not find a counterexample for this, but I could easily find counterexamples to show that \(\displaystyle f(x) > y \cdot f(z)\).
Thanks for your help!
I was wondering whether there is a useful property that can be said about a general convex function \(\displaystyle f\) given some relations. I've already tried to search engine about properties of functions in general, but I didn't come across anything that would answer my question.
Okay, so what we know is this. We have three positive real numbers, \(\displaystyle x \geq 1, y > 1, z \geq 1\) and we have that \(\displaystyle x \leq y \cdot z\) (\(\displaystyle x\) and \(\displaystyle z\) can be integers if that's more helpful, but I doubt it). Now, for a general (increasing) convex function \(\displaystyle f\), we have that \(\displaystyle f(x) \leq f(y \cdot z)\). My question is, is there a property that says \(\displaystyle f(x) \leq f(y) \cdot f(z)\)? (Note that "\(\displaystyle \cdot\)" is simply multiplication.) I could not find a counterexample for this, but I could easily find counterexamples to show that \(\displaystyle f(x) > y \cdot f(z)\).
Thanks for your help!
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