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Let $f(x)$ and $g(x)$ be two increasing and differentiable functions from $\Bbb{R}$ to $\Bbb{R}$. If $f'(x)>g'(x)$ $\forall x$ then there exists an interval $[a,\infty)$ for some real number $a$ for which $f(x)>g(x)$.

This theorem is false, as shown by many counter examples. But how do I modify the conditions so that the theorem becomes true.?

curious
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  • Without thinking too hard, so I have no idea if this is true, but possibly also requiring $f''(x) >= 0$ would allow the rest of the theorem to hold. Because I can only think of an oblique asymptote from below as a counterexample, which means negative curvature. – Justin Jan 13 '15 at 17:16
  • Please avoid editing your question so that it invalidates existing answers. If the answers to your question suggest another, new question, please ask it separately. You can always link from the new question to the earlier one for context. – Ilmari Karonen Jan 13 '15 at 17:28
  • @IlmariKaronen My initial question did include this but since the issue was not addressed therefore I changed the title of the question. – curious Jan 13 '15 at 17:35
  • @Quincunx: I don't think $f''(x) \ge 0$ is sufficient; consider $g(x)=e^x$, $f(x)=e^x-e^{-x}$. (This is only convex for $x\ge0$, but that's easy enough to fix.) – Ilmari Karonen Jan 13 '15 at 17:43
  • @IlmariKaronen Oh yes, I forgot to consider if $g''(x)>0$. So then $f''(x)>g''(x)$ should be enough, (possibly with $\ge$ instead of $>$) – Justin Jan 13 '15 at 18:20

3 Answers3

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No, this isn't true. $g$ could be a line, for instance, and $f$ could increase to it as an oblique asymptote. You need $f'>g'+\epsilon$ for some positive $\epsilon$, in which case the fundamental theorem of calculus will get you the result right away.

Kevin Carlson
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You could reformulate your proposition as follows, by considering $h = f - g$: If $h$ is a differentiable real function such that $h'(x) > 0$ for all $x$ then necessarily there exists a $c$ such that $h(c) > 0$.

Note that if $h(c)>0$, since $h' >0$ it will be true that $h(x) > h(c) >0$ for all $x > c$, so that element of your proposition has been captured as well.

When you write it this way I think you can see the improbability of this claim. Does a counterexample come to mind?

Although it is not the least assumptions, this proposition would be correct if the function is assumed to be both increasing and concave up - since a concave up function sits above its tangent lines, and the tangent lines have positive slope, the function must eventually be positive.

e: I'd like to include Kevin's condition as well - if $h'(x)$ is not merely positive but $h'(x) > \epsilon$ for some $\epsilon > 0$, then the proposition holds as well.

Jason Knapp
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Let $h(x)=f(x)-g(x)$ .
$h'(x)>0$ so $h(x)$ is strictly increasing.
Let $x>a$ . We know that $$h(x)>h(a)\Rightarrow f(x)-g(x)>f(a)-g(a)\Rightarrow f(x)>g(x)+f(a)-g(a)$$ If $f(a)-g(a)\geq 0$ we could conclude that $ f(x)>g(x)$ for every $x$ greater than $a$.
So, a sufficient condition is that there is an $a$ for which $f(a)\geq g(a)$ holds.

Konstantinos Gaitanas
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