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I would like to show that $$f(n) > g(n)$$ for all $n$ within a certain range.

If I can show that both $f(n)$ and $g(n)$ are strictly increasing with $n$, and that both are strictly concave, and that $f(n) > g(n)$ at both the lower and upper bounds of $n$, is that sufficient?

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1 Answer 1

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No. Consider, for example, $f(x)=1+12x-x^2$ and $g(x)=20x-10x^2$ between $0$ and $1$.

Plotted by Wolfram Alpha.

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Thanks, Henning! I am trying to understand your answer. And I am trying to figure out how to interpret the parametric version. Is it helpful if I add the restrictions that both f and g can only range from 0 to 1? –  Angada Oct 6 '11 at 21:38
    
@Angada, the parametric plot is just there because I'm not sure how to make Wolfram Alpha not display it. It's the ordinary two-function plot that matters. –  Henning Makholm Oct 6 '11 at 21:43
    
Hmmm... so I guess it is required to show that $f$ can never equal $g$? –  Angada Oct 6 '11 at 21:44
1  
@Angada: yes, showing $f(x)\ne g(x)$ always would work. But if you can do that, you don't need concavity or monotonicity -- just continuity and a single point where $f(x)>g(x)$. –  Henning Makholm Oct 6 '11 at 21:50
2  
Since concavity is itself defined by inequalities, at least those inequalities in the definition can be be proved by showing a function is concave. Your problem here is that what you're really dealing with is the difference between the two functions -- and knowing that $f$ and $g$ are both concave does not allow you to conclude anything about their difference. The difference is the sum of the concave $f$ and convex $-g$, and the sum of concave and convex can be anything. –  Henning Makholm Oct 6 '11 at 22:05

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