If $f$ bounded and $f''>f$ then $f$ decreases exponentially Let $f$ be a function of class $C^2$ from $ \mathbb{R^{+,*}}$ to $\mathbb{R^{+, *}}$ (where  $\mathbb{R^{+,*}}$ denotes the set of nonnegative reals) such that $f''>f$ and $f$ is bounded above. Show that $f \le f(0)e^{-x}$.
 A: $f'$ is increasing because $f''(x) > f(x) \ge 0$. Then $f'$ must be
negative because  $f$ would be unbounded otherwise.
First define $g(x) = e^{-x} (f'(x) + f(x))$. Then
$$
 g'(x) = e^{-x}(f''(x) - f(x)) > 0
$$
so that $g$ is increasing. It follows that for $ x < a$
$$
  e^{-x} (f'(x) + f(x)) \le e^{-a} (f'(a) + f(a)) \le e^{-a}M 
$$
where $M$ is an upper bound for $f$. Letting $a  \to \infty$ we get
that the left-hand side is $\le 0$, i.e.
$$
  f'(x) + f(x) \le 0 \text{ for all } x \ge 0 \, .
$$
Finally define $h(x) = e^x f(x)$. Then
$$
 h'(x) = e^x(f'(x) + f(x)) \le 0
$$
so that $h$ is decreasing. It follows that
$$
  e^x f(x) = h(x) \le h(0) = f(0) \, .
$$
A: Let $0\le a \le x$. It follows that
$$
f''(x)-f'(x)>-f'(x)+f(x)
\\
\frac{d}{dx}(e^{-x}f'(x))>-\frac{d}{dx}(e^{-x}f(x))
\\
e^{-x}f'(x)-e^{-a}f'(a)>-e^{-x}f(x)+e^{-a}f(a)
\\
\frac{d}{dx}(e^{x}f(x))>e^{2x-a}(f'(a)+f(a))
\\
e^{x}f(x)-e^{a}f(a)>\frac12e^{-a}(e^{2x}-1)(f'(a)+f(a))
\\
f(x)>e^{a-x}f(a)+\frac12(e^{x-a}-e^{-x-a})(f'(a)+f(a))
$$
As the left side is bounded, so is the right side. This can only be if
$$
\frac12e^{x-a}(f'(a)+f(a))\le0,
$$
and this must hold for all $a>0$.
Thus $\frac{d}{dx}(e^xf(x))\le0$ which directly leads to the claim.
