Let $f:[0,\infty) \to \mathbb{R}$ obey the differential inequality
$$f'' - 2\alpha f' + 2\alpha f \leq 0$$
where $0 < \alpha < 2$ is some constant. If $f(0) = 0$ and $f'(0) = 1$, can I say that $f(x) < e^x - 1$ for some $x$?


Note that the solution to the corresponding differential equation oscillates since the characteristic equation has complex roots (call this solution $g$). Thus we can certainly say $g(x) < e^x -1$ for infinitely many $x$. My first thought was to try to control $f$ by $g$ a la Gronwall's inequality. However, I was recently shown that the analogue to Gronwall for degree two differential equations doesn't hold.


Any ideas would be welcome. Also, any good references for differential inequalities that might help me solve this problem are equally welcome.



Controlling subsolutions of a second order linear ODE
Let us actually show more than requested, namely, that $f(x\_n)\le0$ for some sequence $(x\_n)$ converging to $\infty$ and all natural $n$. Moreover, the initial conditions, $f(0) = 0$ and $f'(0) = 1$, will not be used or needed. 


Indeed, suppose that, to the contrary, the statement in the first sentence of this answer is false. Then 
\begin{equation}
 \text{there is some real $x\_\*>0$ such that $f(x)>0$ for all $x>x\_\*$.} \tag{1}
\end{equation} 
Let $a:=\alpha$. 
Let $h:=f'' - 2a f' + 2a f$, so that $h\le0$. 
Let 
\begin{equation}
 r(x):=f(x)/e^{ax}. \tag{2}
\end{equation}
Then the equation $h=f'' - 2a f' + 2a f$ can be rewritten as 
\begin{equation\*}
 r''+b^2r=g,\quad\text{where}\ g(x):=h(x)/e^{ax}\le0 
\end{equation\*}
and $b:=\sqrt{(2-a)a}>0.$
In view of (1) and (2), $r>0$ on $(x\_\*,\infty)$. So, $r''=-b^2r+g<0$, so that $r$ is a strictly positive concave function on $(x\_\*,\infty)$. 
It follows that $r'\ge0$ and hence $r$ is nondecreasing on $(x\_\*,\infty)$; indeed, if $r'(x\_{\*\*})<0$ for some real $x\_{\*\*}>x\_\*$, then the concavity of $r$ implies $r(x)\le r(x\_{\*\*})+r'(x\_{\*\*})(x-x\_{\*\*})\to-\infty$ as $x\to\infty$, which contradicts the positivity of $r$. 
So, $r(x)\uparrow R$ and $r'(x)\downarrow k$ as $x\to\infty$, for some $R\in(0,\infty]$ and $k\in[0,\infty)$. 


Take now any $\rho\in(0,R)$ and let 
\begin{equation\*}
 d:=r-s,\quad\text{where }s(x):=\rho\sin bx. 
\end{equation\*}
Then $d>0$ on $[x\_{\*\*\*},\infty)$ for some real $x\_{\*\*\*}>0$. Also, $d''+b^2d=g$, so that $d''=-b^2d+g<0$ and hence $d$ is a positive concave function on $[x\_{\*\*\*},\infty)$, so that $d'(x)\downarrow \ell$ as $x\to\infty$, for some $\ell\in[0,\infty)$. 
Therefore, 
$b\rho\cos bx=s'(x)=r'(x)-d'(x)\to k-\ell$ as $x\to\infty$, which is absurd. Thus, (1) is false, and hence the statement in the first sentence of this answer is true. 


