f(x)+f(y)<f(x+y)

Nov 2007
15
0
Let's assume that f:[0,∞[ -> R is differentiable twice and f(0)=0 and f''(x)>0, when x>0.

Show that f(x)+f(y)<f(x+y) when x,y>0.


It's really important that I get this done, so any help is highly appreciated!!
 

Opalg

MHF Hall of Honor
Aug 2007
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Leeds, UK
Let's assume that f:[0,∞[ -> R is differentiable twice and f(0)=0 and f''(x)>0, when x>0.

Show that f(x)+f(y)<f(x+y) when x,y>0.
For fixed y, let \(\displaystyle g(x) = f(x+y)-f(x)-f(y)\). Then \(\displaystyle g(0) = 0\). Show that \(\displaystyle g'(x)>0\) for all x>0, and deduce that \(\displaystyle g(x)>0\) for all x>0.
 
Nov 2007
15
0
Thanks Opalg for the hint. Is f a (strictly) convex function because f''(x)>0? Or is it not because f''(x)=0 when x=0 (because f(0)=0)?

I'm still pretty confused with this.
 

Opalg

MHF Hall of Honor
Aug 2007
4,039
2,789
Leeds, UK
Thanks Opalg for the hint. Is f a (strictly) convex function because f''(x)>0? Or is it not because f''(x)=0 when x=0 (because f(0)=0)?
The only reason that you need the condition f''(x)>0 (for x>0) is that it implies that f'(x) is a strictly increasing function, so that f'(x+y) > f'(x) when y>0.

The value of f''(x) when x=0 is not relevant. The fact that f''(x) > 0 when x>0 does in fact imply that f is strictly convex.
 
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