Does it follow that f =0 on [0,1]? Let $f$ be a continuous, real-valued function on $[0,1]$ such that there is $K>0$ for which
$$\vert{f(x)}\vert \le K\int_{0}^{x}{\vert{f}\vert}\qquad \text{for all }x\in[0,1].$$
Does it follow that $f=0$ on $[0,1]$?
 A: There's a much more straightforward way to do this, though I believe it's in the same spirit as the Gronwall's inequality.
Since $|f(x_1)|\le K\int_0^{x_1}|f(x_0)|dx_0$ we have for all $x_2\in[0,1]$
$$\int_0^{x_2}|f(x_1)|dx_1\le \int_0^{x_2}K\int_0^{x_1}|f(x_0)|dx_0dx_1\le KM\int_0^{x_2}x_1dx_1=\frac12KMx_2^2.$$
Where $M=\max_{t\in[0,1]}|f(t)|$.
Note that $|f(x_2)|\le K\int_0^{x_2}|f(x_1)|dx_1$, so we have $|f(x_2)|\le \frac12K^2Mx_2^2$. Iterate again, we have for all $x_3\in[0,1]$
$$|f(x_3)|\le K\int_0^{x_3}|f(x_2)|dx_2\le K\int_0^{x_3}\frac12K^2Mx_2^2dx_2=\frac16K^3Mx_3^3.$$
Inductively, get $|f(x_n)|\le\frac1{n!}MK^nx_n^n\le\frac1{n!}MK^n$ for all $x_n\in [0,1]$. Or $M\le\frac1{n!}MK^n$. Since $\frac1{n!}K^n\to 0$ as $n\to\infty$, the only possible choice of $M$ is $0$.
A: Yes, just apply Gronwall's inequality which says that if $g:[a,b]\to\mathbb{R}$ is differentiable and
$$g'(x)\le \beta(x)g(x),$$
for some continuous function $\beta$, then $$g(x)\le g(a)\exp\left(\int_a^x\beta (t)\,dt\right).$$
Edit
If you don't want to use Gronwall's inequality, you can adapt the proof. Define $g(x)=\varepsilon+\int_0^x |f(t)|\,dt,$ so that $g'(x)\le K g(x)$ and since $g>0$ you have 
$$\frac{g'(t)}{g(t)}\le K.$$
Integrating you get
$$\log g(x)-\log g(0)\le Kx$$ and so
$$g(x)\le g(0)\exp(Kx),$$ which gives 
$$\varepsilon+\int_0^x |f(t)|\,dt\le  \varepsilon \exp(Kx).$$
Send $\varepsilon$ to zero. 
(see the link below)
Gronwall
