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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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    $u$ is a $C^2$ bounded function.2012-11-26
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    You're asking for an oxymoron: you say u is *bounded* and positive, then you ask whether u is unbounded...! Please do check **carefully* what you want to ask.2012-11-26
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    Im sorry, this is a part of a demonstration. The argue was by contradiction and I lost here. I just want to show that $u$ is unbounded.2012-11-26
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    @WillJagy Nonnegative reals. Please, don't shout.2012-11-26
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    @MichaelGreinecker, thank you. Then my answer is fine. The question would begin to be sensible if he dropped the $\mathbb R^+$ in favor of $\mathbb R.$2012-11-26
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    @WillJagy Why is the question not sensible?2012-11-26
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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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    or $$u(x) = \frac{1}{x+1}$$2012-11-26
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    @GEdgar: Those were my first thoughts, too. Will's is unbounded in $\Bbb R^+$, and yours is also unbounded (though not in $\Bbb R^+$), so neither is really a counterexample. (Both can be fixed to give one, of course.)2012-11-26
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    The OP never mentions any domain but $\mathbb R^+$.2012-11-26
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    Fair point, @GEdgar.2012-11-26
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    Im sorry, my question was incomplet. I edited and is better to compreend.2012-11-26
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    @CameronBuie, what does $\mathbb R^+$ mean? I'm not sure I've ever used the notation.2012-11-26
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    @WillJagy: It often indicates the set of positive or nonnegative reals.2012-11-26
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    @WillJagy In my case, if $u'<0$ I cannot take $x sufficiently large. What can I do in this case? I know that $u(0)=0$ and $u>0$ in $\mathbf{R}^+$. This implies that $u'>0$?2012-11-26
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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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    I forgot to say that $u$ is bounded in $\mathbf{R}^+$ and $u''>0$ in the whole $\mathbf{R}^+$! Im sorry.2012-11-26
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    The approach I described above works just fine for that. First, figure out what $g''(x)$ is--which will be some polynomial times $g(x)$, in this case--then use the fact that $g(x)$ is a positive function to conclude that $g''(x)>0$ if and only if $\frac{g''(x)}{g(x)}>0$. Figuring out the latter will show you that $g''(x)>0$ for all $x$ in $(\alpha,\infty)$ for some $\alpha>0$, at which point $u(x):=g(x+\alpha)$ will be a bounded function and $u''$ will be positive on $\Bbb R^+$.2012-11-26
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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 $x>0$.