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Is it true that a twice continuously differentiable bounded function from R to R with non negative second derivative for all x in R is necessarily a constant? If not give a counter example. The given function is convex throughout R since it’s second derivative is non negative. And its boundedness geometrically implies it is a constant. How should I rigorously prove the result? Or is my geometric intuition wrong? Help me please.

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    $\begingroup$ More general result: A bounded convex function on $\mathbb R$ is constant. $\endgroup$
    – zhw.
    Feb 11 '20 at 17:06
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Under these assumptions, by Taylor's theorem there exists $\xi$ between $x$ and $y$ such that

$$f(x) = f(y) + f'(y)(x-y) + \frac{1}{2} f''(\xi)(x-y)^2\geqslant f(y) + f'(y)(x-y)$$

Assume that $f$ is not constant. Then either $f'(y) > 0$ or $f'(y) < 0$ for some $y \in \mathbb{R}$.

If $f'(y) > 0$ the inequality above gives $f(x) \to +\infty$ as $x \to +\infty$. If $f'(y) < 0$ then the inequality above gives $f(x) \to +\infty$ as $x \to -\infty$. This contradicts the hypothesis that $f$ is bounded. Therefore $f'(y) = 0$ for every $y$ and $f$ is constant.

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  • $\begingroup$ Thank you for the insight. $\endgroup$ Feb 11 '20 at 14:39
  • $\begingroup$ @LawrenceMano: You're welcome. $\endgroup$
    – RRL
    Feb 11 '20 at 14:40
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$f$ is a convex function. If $x<y$ and $N>y$ then we can write $y=tx+(1-t)N$ where $t=\frac {N-y} {N-x}$ and so $f(y) \leq tf(x)+(1-t)f(N)$. Letting $N \to \infty$ in this yields $f(x) \leq f(y)$. A similar argument using the points $-N <x <y$ gives the reverse inequality. Hence $f(x)=f(y)$.

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  • $\begingroup$ The value of t when substituted does not give y. Can you please explain that part. Other things are clear to me because as N tends to infinity, t goes to one and hence we get the inequality given. $\endgroup$ Feb 11 '20 at 14:11
  • $\begingroup$ @LawrenceMano There was a typo. I should have written $t=\frac {N-y} {N-x}$. You just have to solve for $t$ from $y=tx+(1-t)y$. $\endgroup$ Feb 11 '20 at 23:16
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If $f''\ge 0$ then $f'$ is increasing. Because if $x<y$ and $f'(x)>f'(y)$ then by the MVT there exists $z\in (x,y)$ such that $0>\frac {f'(y)-f'(x)}{y-x}=(f')'(z)=f''(z)\ge 0,$ which is absurd.

If $f$ is differentiable and not constant then $f'$ is not everywhere $0.$ For if $f(x)\ne f(y)$ then by the MVT there exists $z$ between $x$ and $y$ with $0\ne \frac {f(y)-f(x)}{y-x}=f'(z).$

Therefore:

If $f''\ge 0$ and $f'(z)>0$ then for $x>z$ we have $$f(x)=f(z)+\int_z^xf'(t)dt\ge f(z)+\int_z^xf'(z)dt=f(z)+(x-z)f'(z)$$ which ( for a fixed $z$) is unbounded above as $x\to \infty.$

If $f''\ge 0$ and $f'(z)<0$ then for $x<z$ we have $$f(x)=f(z)+\int_z^x f'(t)dt=f(z)+\int_x^z(-f'(t))dt\ge f(z)+\int_x^z(-f'(z))dt=$$ $$= f(z)+(z-x)(-f'(z))$$ which (for a fixed $z$) is unbounded above as $x\to -\infty.$

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