$f \in C^2(\mathbb R)$ , $(f(x))^2 \le 1$ ; $(f'(x))^2+(f''(x))^2 \le 1 $ ; then is $(f(x))^2+(f'(x))^2 \le 1 $? 
Let $f \in C^2(\mathbb R)$ be such that $$(f(x))^2 \le 1 ;
 (f'(x))^2+(f''(x))^2 \le 1 , \forall x \in \mathbb R$$  Then is it
  true that $(f(x))^2+(f'(x))^2 \le 1 , \forall x \in \mathbb R$ ?

I haven't gotten anywhere with this problem . Please help . Thanks in advance 
 A: Let $g(x) = f^2(x) + f'^2(x)$. We want to show that $g$ is bounded by $1$ when $f^2(x) \leq 1$ and $f'^2(x) + f''^2(x) \leq 1$. 
The extremal value(s) of $g$ are found at points $x_*$ where $g'(x_*)  = 2f'(x_*)(f(x_*) + f''(x_*)) = 0$. If $f'(x_*) = 0$ then $g(x_*) = f^2(x_*) \leq 1$ and if $f(x_*) = -f''(x_*)$ we have $g(x_*) = f'^2(x_*) + f''^2(x_*) \leq 1$.  $g$ is therefore bounded by $1$ at any local maximum point.
If $g$ has an infinite number of maximum points $\{x_n\}_{n=1}^\infty$ and $x_n$ is unbounded in both directions then it follows that $g$ has to be bounded by $1$ for all $\mathbb{R}$.
Now assume (as suggested in the comments by Eric Thoma) there exist a $x_*$ such that $g(x_*) > 1$. We can wlog assume $g’(x) > 0$ for all $x>x_*$ as otherwise there exist a point where $g’(x) = 0$ and $g(x) > 1$ contradicting the result above (if $g'(x) < 0$ then the same argument can be applied to the function $g(-x)$ for which $g' > 0$). Since $g$ is bounded and strictly increasing for $x>x_*$ the limit $\lim_{x\to\infty} g(x)$ exists and satisfy
$$\lim_{x\to\infty} g(x) = \lim_{x\to\infty} f^2(x) + f’^2(x) = a^2 > 1$$
Since $g'(x) > 0$ and $g$ is bounded we must also have 
$$\lim_{x\to\infty} g'(x) = \lim_{x\to\infty}f'(x)(f(x) + f''(x)) = 0$$
Since $f,f',f''$ are bounded we must either have $ \lim_{x\to\infty}f'(x) = 0$ or $\lim_{x\to\infty}f(x) + f''(x) = 0$. If $\lim_{x\to\infty}f'(x) = 0$ then $a^2 = \lim_{x\to\infty} f^2(x) \leq 1$ gives us a contradiction and if $\lim_{x\to\infty} f(x) + f''(x) = 0$ then
$$a^2 = \lim_{x\to\infty} f^2(x) + f’^2(x) = \lim_{x\to\infty} f''^2(x) + f’^2(x) \leq 1$$
also gives us a contradition. It follows that $g$ is bounded by $1$ for all $x$.
