Function dominated by its integral Let $g:[a,b] \rightarrow \mathbb{R}$ be a continuous function. If there exist a positive constant $K$ such that 
\begin{align}
|g(x)| \leq K \int_a^x |g(t)| \, dt
\end{align}
for every $x\in [a,b]$, then show that $g(x)=0$ for every $x\in [a,b]$.
My partial answer:
Since $|g(a)|\leq K \int_a^a |g(t)| \, dt=0$, then we have $g(a)=0$. 
 A: If $g$ is continuous so is its absolute value. write it as $$|g(x)| \le \alpha(x) + \int_a^x \beta(t)|g(t)|\ dt$$
where, of course, $\alpha(x) \equiv 0$ and $\beta(t) \equiv K$.
Now the thesis easily follows from gronwall's integral inequality, see for example http://en.wikipedia.org/wiki/Gronwall's_inequality
A: Let $M$ be an upper bound of $|g(x)|$ on $[a,b]$, we have:
$$\begin{align}
&|g(x)| \le K \int_{a}^{x} M dt = MK (x - a)\\
\implies &|g(x)| \le K \int_{a}^{x} MK(x-a) = M \frac{(K(x-a))^2}{2!}\\
&\vdots\\
\implies &|g(x)| \le K \int_{a}^{x} M\frac{(K(x-a))^{n-1}}{(n-1)!} = M \frac{(K(x-a))^n}{n!}\\
&\vdots
\end{align}$$
This means for all $n > 0$, we have:
$$|g(x)| \le M \frac{(K(x-a))^n}{n!} \le M \frac{(K(b-a))^n}{n!}\tag{*}$$
Since R.H.S of $(*)$ converges to zero as $n \to \infty$, we get $g(x) = 0$.
A: What I'm going to write is probably very similar to what you would find in a proof of Gronwall's Inequality, and that proof very well might be in the link in the previous answer.  
We only care about $|g|$, not $g$, so let $h = |g| \geq 0$, and define $H(x) = \int_a^x h(t)\,dt$. Then you have
$$ H'(x) \leq KH(x),\quad H'(x) - KH(x) \leq 0.$$
Multiply both sides of the inequality by the positive "integrating factor" $e^{-Kx}$ and rewrite the left-hand side as $\frac{d}{dx}(e^{-Kx}H(x))$.  I'll leave the rest to you.
A: As the other answers point out, this is sort of a watered-down Gronwall's inequality. I think Stefan Smith's answer is a pretty good hint for how you can prove this special case. I just want to point out that the phrase "watered down" is actually going a bit far. This inequality is already strong enough to prove quite nice ODE uniqueness theorems, for example:

Theorem. Consider a "time-dependent vector field" $F=F(X,t)$ mapping $\mathbb{R}^n \times [a,b] \to \mathbb{R}^n$ that is "uniformly Lipschitz" in the sense that there is a constant $K$ such that 
  $$\| F(X,t) - F(Y,t)\| \leq K \cdot \|X - Y\|$$
   holds for all $X,Y \in \mathbb{R}^n$ and $t \in [a,b]$. Then, at most one curve $X : [a,b] \to \mathbb{R}^n$ solves the IVP 
  \begin{align*}
\frac{dX}{dt} = F(X(t),t) && X(a) = X_0.
\end{align*}
Proof. Suppose that $X,Y$ are two solutions. For any $t \in [a,b]$, applying the fundamental theorem of calculus (component-wise) yields
  \begin{align*}
 X(t) - Y(t) &= X_0 - X_0 + \int_a^t \frac{dX}{ds} \ ds -  \int_a^t \frac{dY}{ds} \ ds  
= \int_a^t \bigg[  F(X(s),s) - F(Y(s),s) \bigg] \ ds
\end{align*}
  whence
  $$ \|X(t) - Y(t) \| \leq \int_a^t \|F(X(s),s)- F(Y(s),s)\| \ ds \leq K \cdot \int_a^t \|X(t) - Y(t)\|.$$
  But then, your integral inequality implies that $t \mapsto \|X(t) - Y(t)\| : [a,b] \to \mathbb{R}$ is identically zero, so $X = Y$. 

...in answer to a question you absolutely did not ask :)
