# Want help with proving a calculus theory

Let $$f(x)$$ have a second derivative on the closed interval $$[-2,2]$$. If $$\left| f(x) \right| \le 1$$ and $$\frac{1}{2} (f^{\prime}(0))^2+f(0)^3>\frac{3}{2}$$ when $$-2\le x\le2$$, now I need to prove that there must be a point $$x_{0}$$ on the interval $$(-2,2)$$ such that $$f^{\prime \prime}\left(x_{0}\right)+3f\left(x_{0}\right)^2=0$$.

(Series[1/2 (f'[x])^2 + f[x]^3, {x, 0, 1}]) // FullSimplify


The above method does not reveal the nature of the problem and solve it cleverly. I want to use a more generic and heuristic method to verify the conclusion of this abstract function problem. What can I do to solve this problem?

The source of this problem (张宇高等数学18讲):

• Are you sure of the conclusion ? Is it not $f''(x_0)+3f(x_0)^2=0$ ? Sep 6, 2020 at 10:35
• @TheSilverDoe I'm sorry, I have updated the question, if you can, please give us a clever way. Sep 6, 2020 at 11:08
• Is the inequality at 0 or for every $x$ in $-2\le x \le 2$? Sep 6, 2020 at 11:14
• @Miguel That inequality holds only if x = 0. Sep 6, 2020 at 22:40

First note that $$\ \displaystyle\frac{f'(0)^2}{2}+f(0)^3>\frac{3}{2}\$$ and $$\ |f(x)|\le1\$$ for $$\ x\in[-2,2]\$$ implies that $$|f'(0)|>1\ .$$ By the mean-value theorem, there must exist $$\ \xi_1\in(-2,0)\$$ and $$\ \xi_2\in(0,2)\$$ such that $$\ \frac{f(0)-f(-2)}{2}=f'\left(\xi_1\right)\$$ and $$\ \frac{f(2)-f(0)}{2}=f'(\xi_2)\$$. It follows that \begin{align} \left|f'\left(\xi_1\right)\right|&\le\frac{|f(-2)|+|f(0)|}{2}\\ &\le1\ ,\\ \left|f'\left(\xi_2\right)\right|&\le\frac{|f(0)|+|f(2)|}{2}\\ &\le1\ \text{, and}\\ \frac{f'(\xi_i)^2}{2}+ f(\xi_i)^3&\le\frac{1}{2}+1\\ &\le\frac{3}{2}\ \text{ for }\ i=1,2\ . \end{align} Thus, since $$\ \displaystyle\frac{f'(x)^2}{2}+f(x)^3\$$ is differentiable on $$\ (\xi_1,\xi_2)\$$, continuous on $$\ [\xi_1,\xi_2]\$$, $$\ \displaystyle\frac{3}{2}\ \ge \frac{f'(\xi_1)^2}{2}+f(\xi_1)^3\$$, $$\ \displaystyle\frac{3}{2}\ \ge \frac{f'(\xi_2)^2}{2}+f(\xi_2)^3\$$, and $$\ \displaystyle\frac{f'(0)^2}{2}+f(0)^3>\frac{3}{2}\$$, it must achieve its supremum over the interval $$\ [\xi_1,\xi_2] \$$ at some point $$\ x_0\in(\xi_1,\xi_2)\$$ in the interval's interior, where its derivative must vanish: \begin{align} 0&=f'\left(x_0\right)f''(x_0)+3f(x_0)^2f'(x_0)\\ &=f'(x_0)\left(f''(x_0)+3f(x_0)^2\right)\ . \end{align} But since $$\ \frac{f'\left(x_0\right)^2}{2}>\frac{3}{2}-f\left(x_0\right)^3\ge\frac{1}{2}\$$, it follows that $$f''(x_0)+3f(x_0)^2=0\ .$$ Acknowldgement: This proof incorporates a simplification suggested by Martin R in the comments.
• How do you conclude that $\lvert f'(x) \rvert \geq 1$ for all $x\in[\eta_1,\eta_2]$? There clearly can be multiple points on $[\eta_1,\eta_2]$ such that $\lvert f'(x) \rvert < 1$. Oct 4, 2020 at 18:04
• Thank you for picking that up. I stuffed up the definitions of $\ \eta_1\$, $\ \eta_2\$, which I think I have now fixed. Oct 4, 2020 at 18:21
• I think one can simplify the proof slightly, you don't need $\eta_1$ and $\eta_2$. It suffices that $g=f'^2/2 +f^3$ satisfies $g(0) > 3/2$ and $g(\xi_{1, 2}) \le 3/2$. At the point where $g$ attains its maximum one can exclude that $f'(x_0) = 0$ because that would imply $3/2 \le g(x_0) =f(x_0)^3< 1$. Oct 4, 2020 at 19:26