A question about increasing convex functions

[frozenfish]

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!

 

[daon2]

How about \(\displaystyle f(x) = x^2-4\), \(\displaystyle x=3, y=2, z=3\).

 

[frozenfish]

How about \(\displaystyle f(x) = x^2-4\), \(\displaystyle x=3, y=2, z=3\).

Thanks a lot! However, I have realized that another condition must also hold, that is \(\displaystyle f(x) > 0, \forall x\). I will try to find a counter-example for this case as well.

 

[daon2]

Thanks a lot! However, I have realized that another condition must also hold, that is \(\displaystyle f(x) > 0, \forall x\). I will try to find a counter-example for this case as well.

I have a feeling you'll want keep imposing more conditions, but:

\(\displaystyle f(x)=\frac{1}{2}x^2, 0<x<\infty\) with \(\displaystyle x=z=1\), \(\displaystyle y=4/3\) (since your post wants y>1).

 

[frozenfish]

I have a feeling you'll want keep imposing more conditions, but:

\(\displaystyle f(x)=\frac{1}{2}x^2, 0<x<\infty\) with \(\displaystyle x=z=1\), \(\displaystyle y=4/3\) (since your post wants y>1).

Sorry, you are right about imposing more conditions! It may be useful to also have \(\displaystyle f(x) > x\), that is, the function should give something larger than its argument. I do appreciate your help so far, thank you! I will try to figure this out on my own as well.

 

[daon2]

You're killin me lol

Let \(\displaystyle f(x) = \frac{1}{2}(x+1)^2\), then \(\displaystyle f(x)>x\) for all \(\displaystyle x\).

If \(\displaystyle x=22, y=z=\sqrt{22}\) (could even be a smidgen bigger), then

\(\displaystyle f(x) = 264.5 > f(y)f(z) \approx 262.13\)

 

[frozenfish]

Sorry! Your example is great, thank you very much! It does seem there is no property than can be said about \(\displaystyle f(x)\) and \(\displaystyle f(y)\), \(\displaystyle f(z)\) at the moment.