Uniqueness of a function Can there exist a continuous function $f$ from $[0,1]$ to $[0,\infty)$  such that 
$ \int^{x}_{0} f(t) \, dt \geq f(x) $ $ \forall x\in [0,1]$
Can such a function be constructed uniquely?
 A: If such a function $f$ exists, define $\displaystyle F(x) = \int_0^x f(t) \, dt$. Then $F(0) = 0$, and since $F'(x) = f(x) \ge 0$, $F$ is nondecreasing. Thus $F(x) \ge 0$ for all $x \in [0,1]$.
You are given the condition that $F'(x) \le F(x)$. This leads to $-e^{-x} F(x) + e^{-x} F'(x) \le 0$, i.e., $[e^{-x}F(x)]' \le 0$. Thus $e^{-x} F(x)$ is nonincreasing.  This means that $e^{-x} F(x) \le e^{-0} F(0) = 0$ so that $F(x) \le 0$ for all $x \in [0,1]$.
You conclude that $F(x) = 0$ and thus $f(x) = 0$.
A: Suppose that such an $f$ exists. Note that $f(0) \leq \int_0^0 f(t)\, dt = 0$, implying that $f(0)=0$.
Since $f$ is continuous on $[0,1]$, which is compact, $f$ attains maximum at some point (not necessarily unique) $a \in [0,1]$. Then, $\int_o^a f(t) \, dt \leq af(a) \leq f(a)$. So, the best we can hope for, in the case of $a$, is equality, i.e. $a=1$ or $f(a)=0$. Also, the maximum can occur at $a=0$, but then, that implies that $f(0)=0$.
The $f(a)=0$ case and $f(0)=0$ both imply that $f \equiv 0$ on $[0,1]$.
What happens in the $a=1$ case, i.e. $f$ attains its maximum at $1$?
