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?

  • $\begingroup$ So, $f$ is bounded convex function. Now follow, math.stackexchange.com/questions/518091/… $\endgroup$
    – User
    Jun 20 '20 at 15:23
  • $\begingroup$ 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? $\endgroup$
    – user762914
    Jun 20 '20 at 15:25
  • $\begingroup$ math.stackexchange.com/questions/513887/… $\endgroup$
    – User
    Jun 20 '20 at 15:29
  • $\begingroup$ 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. $\endgroup$ Jun 20 '20 at 15:53
  • $\begingroup$ The non negativity of the second derivative implies convexity in any interval, bounded or unbounded. $\endgroup$
    – Angelo
    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.

  • $\begingroup$ Great solution!! Absolutely spot on!! Thank you very much. $\endgroup$ 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.


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