Prove $f(x) < 0 \forall x$ Let $f$ is a function from $\mathbb{R}^+$ to $\mathbb{R}$ twice continuously differentiable and:


*

*$f''(x)\leq f(x)$    $\forall x$

*$f(0)=0$

*$f'(0)=0$
How can I prove $f(x)\leq 0 $    $\forall x \in\mathbb{R}^+$? (it intuitively seems obvious) I just have basic knowledge of analysis so please keep it simple, thanks!
 A: Let us work on the interval $[0,1]$. Now assume that there is $a \in (0,1)$ such that 
$f(a) >0$
 then by mean value there is 
$$f(a)=f^{\prime}(b)a$$
 and 
$$f^{\prime}(b)=f^{\prime\prime}(c)b$$
with $c<b<a$
This implies that 
$$f(a) \leq f(c)ab$$
and this gives that 
$$\frac{f(a)}{a^2}\leq\frac{f(a)}{ab}\leq  f(c)$$
Thus we can define a sequence $a_n$ decreasing so that 
$$\frac{f(a)}{a^{2n}}\leq f(a_n)$$ and since $a<1$ this is impossible.
A: You have $f^{\prime\prime}(x)\leq f(x)$. Hence $f^{\prime\prime}(x)+f^{\prime}(x)\leq f(x)+f^{\prime}(x)$. Multiplying by $\exp(x)$, you get $(\exp(x) f^{\prime}(x))^{\prime}\leq (\exp(x) f(x))^{\prime}$
and so the function $\exp(x)(f(x)-f^{\prime}(x))$ is increasing. Now show  that $f(x)-f^{\prime}(x)\geq 0$ and use a similar trick to finish the proof.
A: Note that, $f''(x) \le f(x) \implies f''(x)-f'(x) \le f(x) - f'(x) $ . 
Define a function,  $$ G(x) = f(x)-f'(x) $$
Note that, $G(x)$ is also differentiable function.  So we have, 
$$ -G'(x) \le G(x) \implies e^x(G(x)+G'(x)) \ge 0 \implies (G(x)e^x)' \ge 0 $$
And, so using monotonic behavior of $ G(x)e^x$, we have $$G(x)e^x \ge G(0)e^0 = 0 \forall x \in \mathbb{R^{+}} $$
This gives $ f(x)-f'(x) \ge 0 $. 
Again using the same idea, we have , $$ e^{-x}f(x) - e^{-x}f'(x) \ge 0 \implies (e^{-x}f(x))' \le 0 $$
And so, $$e^{-x}f(x) \le e^{0}f(0) = 0 \forall x \in \mathbb{R^{+}}$$
So we have $f(x) \le 0$ and we are done! $\Box$
