# Bounded function whose second derivative is non negative

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.

• More general result: A bounded convex function on $\mathbb R$ is constant.
– zhw.
Feb 11 '20 at 17:06

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.

• Thank you for the insight. Feb 11 '20 at 14:39
• @LawrenceMano: You're welcome.
– RRL
Feb 11 '20 at 14:40

$$f$$ is a convex function. If $$x and $$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 gives the reverse inequality. Hence $$f(x)=f(y)$$.

• 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. Feb 11 '20 at 14:11
• @LawrenceMano There was a typo. I should have written $t=\frac {N-y} {N-x}$. You just have to solve for $t$ from $y=tx+(1-t)y$. Feb 11 '20 at 23:16

If $$f''\ge 0$$ then $$f'$$ is increasing. Because if $$x and $$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 $$x we have $$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.$$