Prove that $f(x) \leq f(0)e^{-x}$ function $f : [0,\infty) \to [0,\infty)$ satisfies next two conditions.
a) $$\lim_{x \to \infty} f(x)=0$$
b) $f''(x)$ exists, is continuous and $$f''(x) \geq f(x)$$ for all $x$ in domain.
Then, prove that $$f(x) \leq f(0)e^{-x}$$ for $\forall x \in [0,\infty)$.
I don't know how to prove it...
 A: First, we prove that there has an increasing sequence $(\xi_n)_{n} \subset [0, \infty)$ such that $\xi_n \to \infty$ and $f'(\xi_n)\leqslant 0$ for all $n$. Indeed, let $x_0=0$, since $\lim_{x\to \infty} f(x)=0$ and $f(x)\geqslant 0$ on $\mathbb R$,  we can find an $x_1>\max\{1, x_0\}$ such that $0\leqslant f(x_1)\leqslant f(x_0)$, similarly, we can find an $x_2>\max\{2, x_1\}$ such that $0\leqslant f(x_2)\leqslant f(x_1)$. Continue this procedure, we can find an $x_{n}>\max\{n, x_{n-1}\}$ such that $0 \leqslant f(x_{n+1}) \leqslant f(x_{n})$ for all $n\geqslant 0$. Arccording to mean value theorem, there has a $\xi_{n}\in (x_{n}, x_{n+1})$ such that $f(x_{n+1})-f(x_n) = f'(\xi_n)(x_{n+1}-x_n)$. It implies $f'(\xi_n) \leqslant 0$ for all $n$ and $\xi_n \to \infty$.
Next, consider function $g(x)=e^xf(x)$ on $[0, \infty)$. We have $g'(x)=e^x(f(x)+f'(x))$ and we will show that $g'(x)\leqslant 0$ on $[0, \infty)$. Define $h(x)=e^{-x}(f(x)+f'(x))$ we get from the assumption that
$$h'(x)=e^{-x}(-f(x)+f''(x))\geqslant 0.$$
Thus, $h(x)$ is increasing on $[0, \infty)$ and then $\lim_{x\to \infty}h(x)$ exists (may be infinity). Now,
$$\lim_{x\to \infty} h(x) = \lim_{x\to \infty}[e^{-x}f(x)+e^{-x}f'(x)]=\liminf_{x\to \infty}[e^{-x}f(x)+e^{-x}f'(x)] $$$$= \liminf_{x\to \infty}[e^{-x}f(x)]+\liminf_{x\to \infty}[e^{-x}f'(x)]=0 + \liminf_{x\to \infty}[e^{-x}f'(x)] \leqslant \liminf_{n\to \infty}[e^{-\xi_n}f'(\xi_n)]\leqslant 0.$$
Here we used the property $\liminf(u+v)=\liminf u + \liminf v$ whenever $\lim u$  exists.
Since $h(x)$ is increasing and $\lim_{x\to \infty} h(x) \leqslant 0$, we derive that $h(x)\leqslant 0$ for all $x$. Hence, $f(x)+f'(x)\leqslant 0$ for all $x$ and this implies that $g'(x)\leqslant 0$ on $[0, \infty)$. Up to now, we proved that $g(x)$ is decreasing on $[0, \infty)$ and hence for all $x\geqslant 0$,
$$e^{x}f(x)=g(x)\leqslant g(0)=f(0).$$
