Properties of Function Satisfying an Integral Equation

To find the properties of the function $f$, we begin with the given integral equation:
$\int_0^x \sqrt{1-(f'(t))^2} \, dt = \int_0^x f(t) \, dt, \quad 0 \le x \le 1$

Differentiating both sides with respect to $x$ using the Fundamental Theorem of Calculus (Leibniz Rule):
$\sqrt{1 - (f'(x))^2} = f(x)$

Since $f(x) \ge 0$ for all $x \in [0, 1]$, we can square both sides:
$1 - (f'(x))^2 = (f(x))^2$
$(f'(x))^2 + (f(x))^2 = 1$

This gives:
$f'(x) = \pm \sqrt{1 - (f(x))^2}$

Given that $f(0) = 0$ and $f(x)$ is non-negative on $[0, 1]$, $f(x)$ must be non-decreasing near $x = 0$, so we take the positive square root:
$f'(x) = \sqrt{1 - (f(x))^2}$

Separating the variables:
$\frac{df}{\sqrt{1 - f^2}} = dx$

Integrating both sides:
$\arcsin(f(x)) = x + C$

Using the initial condition $f(0) = 0$:
$\arcsin(0) = 0 + C \implies C = 0$

Thus, the function is:
$f(x) = \sin x, \quad \text{for } x \in [0, 1]$

We know that for all $x > 0$:
$\sin x < x$

Applying this inequality for $x = \frac{1}{2}$ and $x = \frac{1}{3}$:
$f\left(\frac{1}{2}\right) = \sin\left(\frac{1}{2}\right) < \frac{1}{2}$
$f\left(\frac{1}{3}\right) = \sin\left(\frac{1}{3}\right) < \frac{1}{3}$

Therefore, both $f\left(\frac{1}{2}\right) < \frac{1}{2}$ and $f\left(\frac{1}{3}\right) < \frac{1}{3}$ hold true.

The correct option is C.