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Let $f:mathbb{R}tomathbb{R}$ be a function, two times
differentiable with $left|f(x)right|leq1,forall xinmathbb{R}$
and $f^2(0)+left(f'(0)right)^2=4$. Prove that there exists at least
one $x_0inmathbb{R}$, such that $f(x_0)+f''(x_0)=0$.

I have tried the usual Rolle method by $e^x$, but didn't go far... Any help available?

Let's define $g(x) = f(x)^2 + f'(x)^2$. We have:

• $g'(x) = 2f'(x)(f(x)+f''(x))$


• $g(0) = 4$

We will find a local maximum of $g$, noted $x_0$, which verifies $g(x_0) ge 4$. We shall distinguish two options:

• $g(0) = 4 ge g(x) forall x in mathbb{R}$, in which case $0$ is the local maximum that we are trying to find.




• There exists $y in mathbb{R} setminus {0}$ with $g(y) > 4$. Let's supose that $y > 0$ (otherwise apply the following reasoning to $-g$) and try to find a number $x' > y$ which verifies $g(x') = 4$. If there isn't such a number, then we would have $g(x) > 4$ for every $x > y$. Thus, $left| f'(x) right| > sqrt 3$ for every $x > y$. Using the mean value theorem on $x > max{2,y}$ we get a contradiction:

$$ 2 ge left| f(x) - f(0)right| = left| f'(xi_x) x right| > 2 sqrt 3 > 2 $$

As a consequence, we can take $x' > y$ with $g(x') = 4$. The function $g$ has a global maximum in $[0, x']$ because it is continuous. Let $x_0$ be that global maximum. Then, $g(x_0) ge g(y) > 4$ and, consequently, $x_0 ne 0, x'$. Hence, $x_0$ is a local maximum of $g$ with $g(x_0) ge 4$.

Finally, since $x_0$ is a local extrema we have:

$$ 0 = g'(x_0) = 2f'(x_0)(f(x_0)+f''(x_0)) $$

Furthermore, $f'(x_0)^2 ge 4 - f(x_0)^2 ge 3 > 0$ and, thus, $f(x_0)+f''(x_0) = 0$.

Try to explore the extrema of
$$
g(x)=f(x)^2+f'(x)^2implies g'(x)=2f'(x)(f(x)+f''(x)).
$$