TITLE: Proving the positivity of a twice-differentiable real-valued function
QUESTION [9 upvotes]: This is a problem from Berkeley prelim exams, Spring '99 

Suppose that $ f $ is a twice differentiable real-valued function on
  $\mathbb{R}$ such that $ f(0) = 0 $, $ f'(0) > 0 $, and $ f''(x) \geq f(x) $ for all $ x \geq 0 $. Prove that
  $ f(x) > 0 $ for all $ x > 0 $.

Are there general techniques to solve problems like this?

REPLY [2 votes]: Aryabhata posted an objection to Lierre's and Christian Blatter's answers. This is one way to fix the problem ... which still requires the mean value theorem. 

The expression 
$$ f''(t) - f(t) = e^t\frac{d}{dt}\left( e^{-2t} \frac{d}{dt} e^t f\right) $$
By the mean value theorem, if $u$ is differentiable with derivative $v$, then $u(b) - u(a) \geq (b-a) \inf_{[a,b]} v$; similarly $u(b) - u(a) \leq (b-a) \sup_{[a,b]} v$. 
Therefore define $S_L(y,0;v)$ to be the lower Darboux sum of $v$ between $0$ and $y$, and similarly $S_U(y,0;v)$ to be the upper Darboux sum. We have that $S_L(y,0;y') \leq u(y) - u(0) \leq S_U(y,0;u')$ whenever $u$ is differentiable.  
Using that the range of the exponential function is positive, we have that then if $f'' - f = g$ pointwise:
$$ f(x) - f(0) \geq \mathcal{G}_L(x,0;g,f') := e^{-x} S_L\left[ x,0; s \mapsto e^{2s} S_L\left( s,0; e^{-\cdot} g(\cdot)\right) + f'(0)\right] $$
Similarly defining $\mathcal{G}_U(x,0;g,f')$ using $S_U$ instead of $S_L$, we have that also
$$ f(x) - f(0) \leq \mathcal{G}_U(x,0;g,f')~.$$
Using that Darboux sums are monotone (meaning that if $u \leq v$ then $S_L(y,0;u) \leq S_L(y,0;v)$ and similarly for the upper sum), we get the following result: if $f$ and $h$ are twice differentiable functions such that $f'' - f \geq h'' - h$ pointwise everywhere with $f'(0) \geq h'(0)$, then we have
$$ f(x) - f(0) \geq \mathcal{G}_L(x,0;h'' -h,h') $$
and
$$ h(x) - h(0) \leq \mathcal{G}_U(x,0;f'' -f,f') $$
In particular, if $h$ is twice continuously differentiable (or simply if $h'' - h$ is Riemann integrable), then the Darboux sums converge to the Riemann integral and which using the fundamental theorem of calculus we get that necessarily 
$$ f(x) - f(0) \geq h(x) - h(0)~. $$

The fact about comparing upper and lower Darboux sums against the "primitive" is standard. The key to being able to apply this result uses the fact that the linear differential equation can be formally solved by integrating factors, which is related to the fact that the ODE has signed Green's function. (In the study of linear partial differential equations, the fact that the Green's function has a sign is closely connected to the existence of some form of maximum principle. )