$\int_{0}^{x} f(t) dt \ge {f(x)}$ holds or not on $[0,1]$ Let $$f\colon [0,1]\rightarrow [0, \infty )$$  be  continuous. Suppose that
$$\int_{0}^{x} f(t)\,\mathrm dt \ge {f(x)} \quad\text{for  all }x\in[0,1].$$
Then 
$A.$ No  such  function  exists.
$B.$ There  are  infinitely  many  such  functions.
$C.$ There is  only  one  such  function
$D.$ There  are  exactly  two  functions.
Now  I  was  thinking,  since $\int_{0}^{x} f(t) dt$  is  the  area  under  the  graph  of  $f(t)$  from $0$  to  $x$ so it  can  be  easily  made  greater  than  the  value  of  $f(t)$  at  one  point, namely  $x$.  So  there   will  be  infinitely  many  such  functions  satisfying  this  condition. 
Is  my  reasoning  correct or  not?
Thanks.
 A: Let us modify the problem slightly.
Assume that the function
$$
f:[0,a]\to[0,\infty)
$$
is continuous and that
$$
\int_{0}^{x}f(t)\, dt \ge f(x) \quad \text{for all } x\in[0,a].
$$
Then $f(x) = 0$ for all $x\in[0,a]$.
Proof. We can rewrite the inequality as
$$
\dfrac{d}{dx}\left(e^{-x}\cdot\int_{0}^{x}f(t)\, dt\right) \le 0 \quad \text{for all } x\in[0,a].
$$
Consequently the function
$$
x\mapsto e^{-x}\cdot\int_{0}^{x}f(t)\, dt
$$
is decreasing and it takes its maximum at $x=0$. 
Thus
$$
e^{-x}\cdot\int_{0}^{x}f(t)\, dt \le 0 \Leftrightarrow \int_{0}^{x}f(t)\, dt \le 0.
$$
However $f$ is non-negative and continuous. We conclude that
 $f(x) = 0$ for all $x\in[0,a]$.
A: Let $f$ be such a function.
If $0<x< 1$ then by the MWT, there exists $\xi\in (0,x)$ with
$$\int_0^xf(t)\,\mathrm dt=xf(\xi)\stackrel{(1)}\le x\int_0^\xi f(t)\,\mathrm dt\stackrel{(2)}\le \int_0^\xi f(t)\,\mathrm dt\stackrel{(3)}\le \int_0^xf(t)\,\mathrm dt.$$
We conclude that $(1)$, $(2)$,  and $(3)$ are in fact equalities.
The second means that $\int_0^\xi f(t)\,\mathrm dt=0$, the third that $\int_\xi^x f(t)\,\mathrm dt=0$, so that in fact $\int_0^x f(t)\,\mathrm dt=0$. 
As this holds for all $x\in(0,1)$ and $f$ is continuous, we conclude $f(x)=0$ for all $x\in[0,1]$.
So, $C$ - final answer.
A: Your reasoning has a big hole - the condition is supposed to hold for all $x$, and you talk about just one $x$.
Say $f\le c$. Then the inequality shows that $f(x) \le cx$. Now integrate $cx$ and you see that $f(x)\le c x^2/2$. And then $f(x)\le cx^3/6$. By induction $f(x)\le cx^n/n!$. Let $n\to\infty$ and you see $f(x)=0$. So there's exactly one such function.
