Suppose that we have a function $f:[0,T]\to \mathbb R$ in $C^2$, such that for some $C>0,$ $f'' + Cf \geq 0,$ with $f(0)=f'(0)=0.$ This is frequently seen in many places. Is it possible for us to prove that $f\geq 0$ for all $x>0$, or at least, $f(x)\geq 0$ for $x<\delta(C)$ for some constant $\delta(C)$ which depends only on $C$? Or, at least give a lower bound function for $f$?
This seems to be very easy, but I have got little clue.
Let's rewrite the inequality $f''+Cf\geq 0$ as $$ f''(x)+Cf(x)=g(x), \tag{1} $$ where $g(x)\geq 0$. It's easy to verify that $$ f(x)=\int_0^x \frac{\sin k(x-x')}{k}\,g(x')\,dx'\qquad(k=\sqrt{C}) \tag{2} $$ is the solution to $(1)$ with the initial conditions $f(0)=f'(0)=0$. Since $g$ is nonnegative and $\sin k(x-x')\geq 0$ for $0\leq x'\leq x \leq \frac{\pi}{k}$, it follows from $(2)$ that $f(x)\geq 0$ for $x\leq \frac{\pi}{k}$.