# show that $f(x)>0~(0\leq x\leq1)$

I would appreciate if somebody could help me with the following problem:

Q: If $f''(x)$ is exist, $f'(0)>0, f(0)>0, f''(x)>f(x)~(0\leq x\leq 1)$ then $f(x)>0(0\leq x\leq 1)$

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Assume there exists $x\in[0,1]$ such that $f(x)\leq 0$. By the intermediate value theorem, the compact set $$\{x\in[0,1]\;;\;f(x)=0\}$$ is nonempty. Let $x_0$ be its minimum. Note that $x_0\in(0,1]$ and remember that $f(x_0)=0$. Then $f(x)>0$ for all $x\in[0,x_0)$. So $f''(x)>0$ for all $x\in[0,x_0)$. Hence $f'$ is increasing on $[0,x_0]$. Since $f'(0)>0$, we get $f'(x)>0$ on $[0,x_0]$. So $f$ is increasing on $[0,x_0]$. Since $f(0)>0$, we get $f(x_0)>0$. Contradiction.