Question on the integrability of $f''$, where $f$ is two times differentiable, verifying some conditions Let $f : [0,\infty) \rightarrow [0,\infty)$ be a two times differentiable function verifying,
\begin{align}
\exists M>0, \forall x \geq 0, f(x) \leq M \quad (1) \\
\exists \alpha>0, \forall x \geq 0,  f''(x) \geq \alpha^2 f(x) \quad (2) 
\end{align}
It is asked at some point of an exercise to show that $ \lim\limits_{x \rightarrow +\infty} f(x) =0$.

At first glance, after showing that $f' \leq 0$ using the convexity and boundedness of $f$, I wrote that, as $f$ is non-increasing and $0$ is a lower bound, then $\ell :=\lim\limits_{x \rightarrow +\infty} f(x)$ exists. By contradiction if $\ell>0$ then we have $f \geq \ell >0$, and using $(2)$ yields$$f'' \geq \alpha^2 \ell >0.$$
That is where I wanted to integrate and use the fundamental theorem of analysis, but I realized I had not much information on the regularity of $f''$. If I knew for example that $f$ was of class $\mathcal{C}^2$, then I could say that,
$$\forall x \geq 0, f'(x) \geq f'(0) + \alpha^2 \ell x \underset{x \rightarrow +\infty}{\longrightarrow} +\infty$$
giving us a contradiction.

I was therefore wondering whether the initial hypothesis on the regularity of $f$ needed to be enhanced to show that $\lim\limits_{x \rightarrow +\infty} f(x) = 0$ or not. Thus, I would like to know if there is another argument which can help proving the statement, or if it is false (which might be harder).
I know that we cannot apply the fundamental theorem of analysis to any differentiable function : Volterra's function provides a counter-example.
Any insights on this question would be greatly appreciated. Feel also free to ask for more details if needed.
 A: Here is a direct proof.
Let $\;g:[0,+\infty[\to\mathbb R\;$ be the function defined as
$\displaystyle g(x)=\int_0^x\alpha^2f(t)dt-f’(x)\qquad\forall\,x\in [0,+\infty[\;.$
Since $\;g’(x)=\alpha^2f(x)\!-\!f’’(x)\leqslant0\;,\;\forall\,x\in [0,+\infty[\;,\;$ the function $\,g(x)\,$ is non-increasing on $[0,+\infty[\;,\;$ hence ,
$\displaystyle\int_0^x\alpha^2f(t)dt-f’(x)\leqslant-f’(0)\;\;\;$ for all $\,x\in [0,+\infty[\,.$
Since $\,f(x)\,$ is non-increasing on $\,[0,+\infty[\;,\;$ we get that
$\displaystyle \alpha^2xf(x)\leqslant\!\int_0^x\!\!\!\alpha^2f(t)dt\leqslant f’(x)\!-\!f’(0)\leqslant-f’(0)$
for all $\,x\in [0,+\infty[\;\;,$
consequently ,
$0\leqslant f(x)\leqslant-\dfrac{f’(0)}{\alpha^2x}\qquad\forall\,x\in \,]0,+\infty[\;\;,$
hence , by applying Squeeze theorem ,
$\lim_\limits{x\to+\infty}f(x)=0\;.$
A: If $\ell>0$ then, forall $x\ge$ some $x_0,$ $f(x)\ge\frac\ell2$ hence by (2), $$f''(x)\ge k:=\alpha^2\frac\ell2,$$ so $x\mapsto f'(x)-kx$ is non-decreasing and
$$f'(x)\ge f'(x_0)+k(x-x_0)\to+\infty.$$
