Prove that there exists some $c\in(-3,3)$ such that$ \ \ g(c) \cdot g''(c)<0$.

Question

$f(x)$ is a differentiable function and $g(x)$ is a double differentiable function such that $|f(x)|\leqslant 1$ and $f'(x)=g(x)$. If $$f(0)^2+g(0)^2=9$$

then prove that there exists some $c\in(-3,3)$ such that$ \ \ g(c) \cdot g''(c)<0$.

Attempt:

Let us define a function $h(x) = g(x) g~'(x)$. Then

$$h'(x) = g(x)g''(x) + \left( g'(x) \right)^2 \tag 1$$

If we prove that for some $c \in (-3,3), h~'(c) < 0,$ then $$g(c)g''(c) <0 \tag 2$$

Also, $$\left|f(0)\right| < 1 \implies f'(0) \in (-3,-2\sqrt 2 ) \cup (2\sqrt 2,3) $$

Could someone please advise me how do I move forward from here.

Thank you very much for your help in this regard.

Answer 1

(Unless I made some error, the statement actually holds with $(-3, 3)$ replaced by $(-a, a)$ for any $a > 1/\sqrt 2$.)

Without loss of generality we can assume that $$ g(0) \ge 0 \text{ and } g'(0) \ge 0 \, .$$ (Otherwise replace $f$ and $g$ by $$ f_1(x) = u f(vx) \, , g_1(x) = uvg(vx) $$ where $u = \pm 1$ and $v = \pm 1$ are chosen appropriately.)

Assume that $a > 0$ and $$ g(x) g''(x) \ge 0 \text{ for all } x \in (0, a) \, . \tag 1 $$ Define $$ h(x) = g(x)^2 \, . $$ Then $$ h(0) = 9 - f(0)^2 \ge 8 \, , \\ h'(x) = 2 g(x) g'(x) \, , \, h'(0) \ge 0 \, , \\ h''(x) = 2 g'(x)^2 + 2 g(x) g''(x) \ge 0 \, . $$ From $h'' \ge 0$ follows that $h'$ is increasing and therefore non-negative on $[0, a]$. Consequently, $h$ is increasing and therefore $h(x) \ge 8$ for all $ x \in [0, a]$.

So $f'(x) = g(x) \ge \sqrt 8$ for all $ x \in [0, a]$ and the Mean-value theorem gives $$ 2 \ge f(a) - f(0) \ge (a - 0) \, \sqrt 8 $$ and therefore $$ a \le \frac{2}{\sqrt 8} = \frac{1}{\sqrt 2} \, . $$

It follows that for any $a > 1/\sqrt 2$, $(1)$ cannot hold and $g(c)g''(c) < 0$ for some $c \in (0, a)$.

Answer 2

It is clear that $g(0)\in (-3,-2\sqrt{2}]\cup [2\sqrt{2},3)$.

Consider the interval $(0,3)$.

By $LMVT$ there exists $x=c_{1}$ where $c_{1}\in(0,3)$such that

$|g(c_{1})|=|f'(c_{1})|=\left|\frac{f(3)-f(0)}{3-0}\right|\leq \frac{|f(3)|+|f(0)|}{3}\leq\frac{1+1}{3}=\frac{2}{3}$

Similarly, on the interval $(-3,0)$ By $LMVT$ there exists $x=c_{2}$ where $c_{2}\in(-3,0)$such that

$|g(c_{2})|=|f'(c_{2})|=\left|\frac{f(0)-f(-3)}{0-(-3)}\right|\leq \frac{|f(0)|+|f(-3)|}{3}\leq\frac{1+1}{3}=\frac{2}{3}$.

WLOG,let us assume that $g(x)>0$.

Now clearly, $c_{2}<0<c_{1}$.

The values of $g(c_{2})$ and $g(c_{1})$ are less than or equal to $\frac{2}{3}$ whereas value of $g(0)\in [2\sqrt{2},3)$.

Since, $g(x)$ is twice differentiable, the graph of $y=g(x)$ must attain a point of maximum between

$c_{2}$ and $c_{1}$ and clearly at such a point of maximum ,say $x=c$ the value of $g''(c)<0$ and we are

done.