I'm given that $f:\mathbb{R} \to \mathbb{R}$ is twice differentiable and $f(0) = f'(0) =1$. Assuming that $f''(x) > f(x)$ everywhere show that $f(x) > 0$ for all $x$.
I know that $f$ and $f’$ are continuous ($f’’$ exists). Since $f(0) = f’(0)= 1$ there is some $\delta > 0$ where $f(x), f’(x) > 0$ for $-\delta \leq x \leq \delta$. I tried using the second-order Taylor approximation to extend the interval but I cannot see how to show $f(x) > 0$ for all $x < -\delta$.
A clue here is that the exponential function equals its second derivative, whereas $f’’ > f$ and, thus, $f$ has greater convexity. Hence, $g(x) = f(x)e^{-x}$ has a global minimum at $x=0$ where $g(0) = 1$.
Therefore, $g(x) \geqslant 1$, which implies $f(x) \geqslant e^x > 0$ for all $x$.
Details:
Note that $g'(x) = [f'(x) - f(x)] e^{-x} = [f'(x) - f(x)] e^{x} \cdot e^{-2x} $.
We have, since $f'' >f$,
$$\frac{d}{dx}\left([f'(x) - f(x)] e^{x}\right) = [f''(x) - f(x)]e^x > 0,$$
and $[f'(0) - f(0)]e^0 = 0$, showing that $[f'(x)-f(x)]e^x$ and, consequently, $g’(x)$ passes from negative to positive values with $x$ and $g$ has a global minimum at $x = 0$.
