Let $\,f \colon [0,1] \to [0,\infty)\,$ be continuous . Suppose .. I am stuck on the following problem that says:

Let $\,f \colon [0,1] \to [0,\infty)\,$ be continuous . Suppose $$\int_o^xf(t)\mathrm{d}t \ge f(x) \,\,\quad\forall x \in [0,1] \tag{1}$$   Then

*

*No such function exists


*There are infinitely many such functions


*There is only one such function


*There are exactly two such functions

My Attempts: After taking derivatives both sides of (1) with respect to $x$, we get  $$f(x) \ge f'(x) \implies {d{f(x)}\over f(x)} \le 1$$ Also,I noticed there is at least one such function namely $f(x)=0$ which satisfies (1). But I am not sure if it is the only one such function.
Now,I am not sure which way to go. Can someone explain? Thanks in advance for your time.
 A: Let $g(x) = \int_0^x f(t)dt$. Then $g(x) \ge f(x)$, and thus
$$\int_0^xg(t)dt \ge \int_0^x f(t)dt = g(x)$$
so $g$ satisfies the same conditions. 
Further, $g(0)= 0$ and $g$ is continuously differentiable. Repeating this, we can assume $g$ is as many times continuously differentiable as we want, so no issues in taking derivatives. In particular, $g(x) \ge g'(x)$.
Now, consider $h(x) = g(x)e^{-x}$. Then $h'(x) = e^{-x}(g'(x)-g(x))  \le 0$. But, $h(x) \ge 0$ and $h(0) = 0$. The only way this is possible is if $h = 0$. So $g = 0$, and hence $f = 0$.
Thus, there is only one such function.
This is the major part in a particular proof of uniqueness of solutions of first order differential equations on $\mathbb{R}$.
A: I think there are infinitely many functions.
If you differentiate the given relationship, we have $f(x) \ge f'(x)$. We may also express this as, using Leibniz notation, as $f(x) \ge {d f(x)\over dx}$, which may be interpreted as claiming that the arbitrary function we consider is greater than it's derivative for all x on the unit interval. This is possible.
A: Hint: Integrating the inequality $\frac{f'(x)}{f(x)}\leq1$ gives
$$\int_0^xdx{{f'(x)}\over f(x)} = \log{\frac{f(x)}{f(0)}} \le \int_0^xdx\cdot1=x .$$
Exponentiating,
$$\frac{f(x)}{f(0)}\leq e^{x}$$
or
$$f(x)\leq f_0e^x.$$
