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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):

enter image description here

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

1 Answer 1

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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.

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  • $\begingroup$ 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 $\lvert f'(x) \rvert < 1$. $\endgroup$
    – V.S.e.H.
    Oct 4, 2020 at 18:04
  • $\begingroup$ Thank you for picking that up. I stuffed up the definitions of $\ \eta_1\ $, $\ \eta_2\ $, which I think I have now fixed. $\endgroup$ Oct 4, 2020 at 18:21
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    $\begingroup$ 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 $3/2 \le g(x_0) =f(x_0)^3< 1$. $\endgroup$
    – Martin R
    Oct 4, 2020 at 19:26
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    $\begingroup$ Thank you for the tip. "Simplify the proof slightly" is a nicely considerate litotes. The simplification seems to me to be pretty substantial. $\endgroup$ Oct 4, 2020 at 22:59

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