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If $u$ is a positive function such that $u''>0$ in the whole $\mathbf{R}^+$ then $u$ is unbounded?

In fact, I know that if $u''>0$ then $u$ is strictly convex. I think that implies $u$ is coercive. I want to prove it.

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    @MichaelGreinecker, only because it has easy examples as in $e^{-x}.$ Perhaps a better word is nontrivial. If the domain were switched to the whole line, somewhere the derivative is either positive or negative, and in the appropriate direction growth is at least linear, as in $ -x + \sqrt {1 + x^2}. $ All of which you know, so...2012-11-26

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$ u(x) = e^{-x} {}{}{}{}{}{}{}{} $

EDIT: if you actually meant the entire real line $\mathbb R,$ then any $C^2$ function $u(x)$ really is unbounded. Proof: as $u'' > 0,$ we know that $u'$ cannot always be $0.$ as a result, it is nonzero at some $x=a.$ If $u'(a) > 0,$ then for $x > a$ we have $u(x) > u(a) + (x-a) u'(a),$ which is unbounded. If, instead, $u'(a) < 0,$ then for $x < a$ we have $u(x) > u(a) + (x-a) u'(a),$ which is unbounded as $(x-a)$ is negative. Both of these are the finite Taylor theorem. Examples with minimal growth include $ x + \sqrt{1 + x^2} $ and $ -x + \sqrt{1 + x^2} $

Note that $C^2$ is not required, it suffices that the second derivative always exist and is always positive. Taylor's with remainder.

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    Finally, game over. Since that $u(0)=0$ and u>0, we have that u'>0 in a neighborhood of $0$. If u''>0 then $u'$ is strictly increase, so u'>0 in the whole $\mathbf{R}^+$. We use your argument and finish the question. Thank you.2012-11-26
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Consider $g(x)=e^{-x^2}$. Observe that this is a bounded, positive, everywhere differentiable function, and that for sufficiently large $x$, we have $g''(x)>0$. Thus, an appropriate translation of $g(x)$--in particular, some function of the form $g(x+\alpha)$ for some sufficiently large $\alpha$--will give you a counterexample function $u$.

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    **Note**: I assumed (perhaps incorrectly) that by $\mathbf{R}^+$ you meant the positive reals. If you instead meant the nonnegative reals, then instead of finding some \alpha>0 so that g''(x)>0 for all $x$ in $(\alpha,\infty)$, we'll need to find some \alpha>0 so that g''(x)>0 for all $x$ in $[\alpha,\infty)$. From there, we'll proceed as previously described. If you mean something else altogether by $\mathbf{R}^+$, please specify your intended definition, and I'll alter my answer accordingly.2012-11-26
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Consider the function $f(x):={1\over 1+x}\qquad(x>0)\ .$ Then $0 and $f''(x)>0$ for all $x>0$.