# Twice continuously differentiable bounded functions with non negative second derivative

Let $$f: \mathbb R \rightarrow\mathbb R$$ be twice continuously differentiable. Suppose further that $$f$$ is bounded and $$f’’(x)\geq 0$$ for every $$x \in \mathbb R$$. Then prove that $$f$$ is infinitely differentiable.

Except constant functions I am not able to get any function satisfying the hypothesis. The exponential function satisfies all properties but is unbounded. $$\arctan$$ satisfies all properties except that its second derivative is not nonnegative in $$\mathbb R$$. Are there non-constant functions satisfying the hypothesis? How do we prove the result?

• So, $f$ is bounded convex function. Now follow, math.stackexchange.com/questions/518091/…
– User
Jun 20 '20 at 15:23
• The claim is that all bounded and convex functions on $\mathbb{R}$ that are twice differentiable are also infinitely differentiable. That doesn't sound like it is true? If you start with a non-decreasing $f'''$ that has kinks and integrate back up to $f$, you'll construct a twice continuously differentiable function with a non-differentiable third derivative, but no apparent contradictions?
– user762914
Jun 20 '20 at 15:25
• math.stackexchange.com/questions/513887/…
– User
Jun 20 '20 at 15:29
• The non negativity of the second derivative implies convexity only in a bounded interval. Refer Baby Rudin. Here the interval is the whole Real line which is unbounded. Jun 20 '20 at 15:53
• The non negativity of the second derivative implies convexity in any interval, bounded or unbounded. Jun 20 '20 at 16:35

It is sufficient to prove that $$f$$ is constant in $$]-\infty,+\infty[$$.

If it was not constant there would exists $$x_0\in \mathbb{R}$$ such that $$f’(x_0)\ne0$$. There are only two possibilities: $$f’(x_0)>0$$ or $$f’(x_0)<0$$.

Without loss of generality we can suppose that $$f’(x_0)>0$$.

The hypothesis $$f’’(x)\ge0$$ for all $$x\in \mathbb{R}$$ implies that $$f’(x)$$ is a monotone nondecreasing function, so $$f’(x)\ge f’(x_0)>0$$ for all $$x\in ]x_0,+\infty[$$.

Moreover for all $$x \in ]x_0, +\infty[$$ we can apply Langrange Theorem to the interval $$[x_0,x]$$, so there exists $$c \in ]x_0,x[$$ such that $$f(x)-f(x_0)=f’(c)(x-x_0)\ge f’(x_0)(x-x_0)$$.

It means that $$f(x)\ge f(x_0)+f’(x_0)(x-x_0)$$ for all $$x \in ]x_0, +\infty[$$. So $$f$$ is not bounded from above in $$]x_0, +\infty[$$, but it is absurd because it contradicts one of the hypothesis.

Hence it is impossible that $$f$$ is not constant in $$]-\infty,+\infty[$$.

It means $$f$$ is constant in $$]-\infty,+\infty[$$, so it is infinitely differentiable.

• Great solution!! Absolutely spot on!! Thank you very much. Jun 20 '20 at 20:30

As you concluded, the only continuous bounded convex functions on $$\mathbb R$$ are constant functions. One characterization of convex functions is that their graphs lie above their tangent lines: $$f(x)\ge f(a)+f'(a)(x-a)$$ for all $$x$$ and $$a$$. If $$f$$ is bounded, so if $$x\mapsto f(a)+f'(a)(x-a)$$, so $$f'(a)=0$$, and so on.