Bound on $|f(x)|^2 + |f'(x)|^2$ Let $f\in C^2(\mathbb{R})$ be a twice differentiable function satisfying
$$|f(x)|^2\le  a$$ and $$|f'(x)|^2 + |f''(x)|^2\le  b$$ for all real $x$,
where $a$ and $b$ are positive constants.
Prove that
$|f(x)|^2 + |f'(x)|^2\le  \max(a, b)$ for all real $x$.
 A: To find the supremum of $|f(x)|^2+|f'(x)|^2$, note that
$$
\begin{align}
|f(x)|^2+|f'(x)|^2
&\le|f(x)|^2+\left(|f'(x)|^2+|f''(x)|^2\right)\\
&\le a+b\tag{1}
\end{align}
$$
so $|f(x)|^2+|f'(x)|^2$ is bounded. Furthermore,
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\left(|f(x)|^2+|f'(x)|^2\right)
&=2f(x)f'(x)+2f'(x)f''(x)\\
&=2f'(x)\left(f(x)+f''(x)\right)\tag{2}
\end{align}
$$
1. If $|f(x)|^2+|f'(x)|^2$ reaches a maximum, either $f'(x)=0$ or $f(x)=-f''(x)$.
$\quad$a. If $f'(x)=0$, then $|f(x)|^2+|f'(x)|^2=|f(x)|^2\le a$.
$\quad$b. If $f(x)=-f''(x)$, then $|f(x)|^2+|f'(x)|^2=|f'(x)|^2+|f''(x)|^2\le b$.
2. If $|f(x)|^2+|f'(x)|^2$ doesn't attain its maximum, then either
$\quad$a. there is a sequence of increasing local maxima as $x\to\infty$ or $x\to-\infty$, or
$\quad$b. eventually, it monotonically increases to its maximum as $x\to\infty$ or $x\to-\infty$.
In case 2a, apply 1. to each local maximum.
In case 2b, the mean value theorem says that there is a sequence $x_n\to\infty$ or $x_n\to-\infty$ so that $\frac{\mathrm{d}}{\mathrm{d}x}\left(|f(x_n)|^2+|f'(x_n)|^2\right)\to0$ as $|f(x)|^2+|f'(x)|^2$ increases to its supremum. Thus, by $(2)$, eventually $\left|f'(x_n)(f(x_n)+f''(x_n))\right|\le\epsilon^2$. Then either $\left|f'(x_n)\right|\le\epsilon$ or $\left|f(x_n)+f''(x_n)\right|\le\epsilon$.
$\quad$c. If $|f'(x_n)|\le\epsilon$, then $|f(x_n)|^2+|f'(x_n)|^2\le a+\epsilon^2$.
$\quad$d. If $|f(x_n)+f''(x_n)|\le\epsilon$, then
$$
\begin{align}
|f(x_n)|^2+|f'(x_n)|^2
&\le|f'(x_n)|^2+|f''(x_n)|^2+|f(x_n)+f''(x_n)|\,|f(x_n)-f''(x_n)|\\
&\le b+(\sqrt{a}+\sqrt{b})\epsilon
\end{align}
$$
Since $\epsilon$ was arbitrary, we are done.
