Show that $f(x)=0$ for all $x \in [a,b]$. I have the following problem:

Suppose that $f$ is continuous on $[a,b]$ and suppose that for all $x \in [a,b]$, $f(x) \geq 0$ and $f(x)\leq \int_a^x f(t)dt$. Show that $f(x)=0$ for all $x \in [a,b]$.

By integrating repeatedly, we obtain: $$f(x) \leq \int_a^x f(t_1)dt_1 \leq \cdots \leq \int_a^x \int_a^{t_1} \cdots \int_a^{t_n} f(s)dsdt_n \cdots dt_1$$ Thus, if $M_x$ denotes the maximum of $f$ on the interval $[a,x]$, we have that $f(x) \leq M_x(x-a)^n$. Taking $n \to \infty$, we find that $f(x)=0$ on the interval $\left[a,a+1\right)$.
I am now having trouble proving that $f(x)=0$ on $[a,b]$, although I believe induction is involved. Any help would be appreciated!
 A: To continue your work we have $f(x)=0$ for all $x\in[a,a+1]$ but this implies that:
$$f(x)\leq\int_{a}^xf(t)dt= \int_{a+1}^xf(t)dt$$
and here you apply the same reasoning and you get $f(x)=0$ on $[a+1,a+2]$ again, and repeating the process you get the result.

Another method :
we have $f(a)=0$ now let $m=\max\{x\in[a,b]\mid\forall y\leq x f(y)=0\}$,
Assume that $m<b$ there exist $\epsilon$ such that $f(m+\epsilon)> 0$ for some $0<\epsilon< 1$ and $\epsilon \leq b-m $, and let $f(t)=\max_{x\in[m,m+\epsilon]}(f(x))$ we have:
$$0 \leq f(t)\leq \int_{a}^tf(t)dt=\int_{m}^tf(t)dt\leq \int_{m}^{m+\epsilon}f(t)dt\leq \epsilon f(t)$$
thus $f(t)=0$, this is absurd because $f(t)=0$ implies $f(m+\epsilon)=0$
Finally $m=b$ and $f=0$.
A: Let $F(x)=\int_a^x f(t) dt$.
The inequality $0\leq f\leq F$ implies both $F\geq 0$ and $$\forall x\in [a,b], F(x)=|F(x)|\leq \int_a^x |f(t)|dt\leq \int_a^x F(t) dt$$
Note that this inequality is the same as $f(x)\leq \int_a^x f(t) dt$, with a huge difference: $F$ is continuously differentiable, so we're allowed to use some calculus.
Let $g(x)=\int_a^x F(t) dt$ . We have proven that $g'-g\leq 0$.
Let $h(x)=e^{-x}g(x)$. Note that $\forall x\in [a,b], h'(x)=e^{-x}(g'(x)-g(x))$.
Hence $h'\leq 0$
Integrating from $a$ to $x$ yields $h(x)-h(a)\leq 0$, and since $h(a)=0$, $h(x)\leq 0$
This, in turn implies that $g(x)\leq 0$.
$x$ being arbitrary, we have $g\leq 0$.
Now, recall that $F\geq 0$. Hence $g\geq 0$ (by definition of $g$).
Hence $g=0$, thus $g''=f=0$ and we're done.
