Prove the second derivative of a function has a zero in an interval given constraints

Question

I'm wondering how to express this question in a form computable by Mathematica:

• Assume $f(x), f^\prime (x),$ and $f^{\prime \prime} (x)$ exist and are continuous for all $x$.
• $|f(x) - \sin x| \leq 1/3$ for all $x$.

Show that $f^{\prime \prime} (x)$ has at least one zero in the open interval $(0, 2 \pi)$.

My (incorrect) attempt is:

\!\(
\*SubscriptBox[\(\[ForAll]\), \(x\)]\(\((Abs[
    f[x]\  - \ Sin[x]\  <= \ 1/3])\)\ \(
\*SubscriptBox[\(\[Exists]\), \(x\  \[Element] \((0, 2\ \[Pi])\)\)]D[
     f[x], {x, 2}] == 0\)\)\)

---

Extended response to 64494's comment, below:

The above is Mathematica. Consider

• "Automated vector space proofs using Mathematica," Aaron E. Naiman, ACM Communications in Computer Algebra 56(1):1--13, issue 219, March 2022

Clearly (by its title and peer-reviewed publication) dealing with Mathematica. In that paper the author uses Mathematica to prove (or disprove) theorems about vector spaces.

A sample:

Reduce[!( *SubscriptBox[([ForAll]), (s)]( *SubscriptBox[([Exists]), (t)]\ s\ t\ < \ 0)), Reals]

The Mathematica statement can be read: "For all $s$, there exists a $t$ such that $s$ times $t$ is negative."

(This statement is of course FALSE, given the case $s=0$, and Mathematica computes precisely that answer.)

This is precisely the kind of automated reasoning I'd like to apply to the problem posed above.

Here's another example:

Resolve[ForAll[x, a x^2 + b x + c > 0]]

as listed here.

Mathematica is rather powerful for automated reasoning about equations and relations, but I had difficulty applying it to the problem stated above. I have a feeling somebody can solve it.

Clear?

Answer 1

The following shows a +/- band of 1/3 around Sin[x]

Plot[{Sin[x], Sin[x] + 1/3, Sin[x] - 1/3}, {x, 0, 2 Pi}]

In the interval {1,2] f must somewhere have a negative second derivative. Similar, it must have a positive second derivative somewhere in [4,5]. As the second derivative is continuous, it must have a zero somewheer in between.

Answer 2

I appreciate the purpose of this question. Is to automate proof. @DanielHuber answer is almost a visual proof or at least motivates proof.

I have upvoted @DanielHuber

I apologize for the tardy nature of the post and of course for errors.

The inequalities, repeated use of mean value theorem and final intermediate value theorem could, perhaps,be mimicked in Mathematica but I am uncertain how to “automate reasoning”.