Zero function with integral inequality [duplicate]

If $0\leq f(x)\leq c \int_0^x f(t) dt$, then does that imply that $f'(x)\leq cf(x)$ on $(0,b)$? Assuming f is continuous and differentiable.

(note c>0) I cant think of a counter example..anybody know more about this?

maybe since integral f(t)dt has series expansion (at x=0) = f(0)t + f'(0)t^2/2+ ... that has something to do with it?

Here is another take. This is essentially the Bellman-Gronwall inequality except it is computed iteratively.

Define the operator $L:C[0,b] \to C[0,b]$ by $(Lf)(x) = c\int_0^x f(t) dt$. Note that $L$ preserves order in the sense that if $f(x) \le g(x)$ for all $x$, then $(Lf)(x) \le (Lg)(x)$ for all $x$.

We are given $f(x) \le (Lf)(x)$ for all $x$. Applying $L$ to both sides and using the original inequality again gives $f(x) \le (L^2f)(x)$ for all $x$. Repeating gives $f(x) \le (L^n f)(x)$ for all $x$ and $n$.

Expanding gives $(L^n f)(x) = c^n\int_{t_n=0}^x \cdots \int_{t_1=0}^{t_2} f(t_1) dt_1 \cdots dt_n$. Let $K = \sup_x |f(x)|$, then we have $(L^n f)(x) \le c^n K \int_{t_n=0}^x \cdots \int_{t_1=0}^{t_2} dt_1 \cdots dt_n$, which evaluates to $(L^n f)(x) \le K { (cx)^n \over n!} \le K { (cb)^n \over n!}$.

Taking limits shows that $f(x) \le 0$ for all $x$, from which we get $f = 0$.