Example 13 Given that the domain of the function $f(x)$ is $[0,1]$, and it satisfies the following conditions:
( I ) For any $x \in[0,1]$, we always have $f(x) \geqslant 3$;
( II ) $f(1)=4$;
(III) If $x_{1} \geqslant 0, x_{2} \geqslant 0, x_{1}+x_{2} \leqslant 1$, then $f\left(x_{1}+x_{2}\right) \geqslant f\left(x_{1}\right)+f\left(x_{2}\right)-3$.
(1) Try to find the value of $f(0)$;
(2) Try to find the maximum value of the function $f(x)$;
(3) Try to prove: When $x \in\left(\frac{1}{3}, 1\right]$, $f(x)<3 x+3$; When $x \in\left(\frac{1}{3^{n+1}}, \frac{1}{3^{n}}\right]\left(n \in \mathbf{N}^{*}\right)$, $f(x)<3 x+3$.
(1) Let $x_{1}=x_{2}=0$, then we have $f(0) \geqslant 2 f(0)-3$, which means $f(0) \leqslant 3$.
For any $x \in[0,1]$, we always have $f(x) \geqslant 3$, so $f(0)=3$.
(2) Take any $x_{1}, x_{2} \in[0,1], x_{1}<x_{2}$, so $x_{2}-x_{1}>0$, thus $f\left(x_{2}-x_{1}\right) \geqslant 3$, which means $f\left(x_{2}-x_{1}\right)-3 \geqslant 0$, so $f\left(x_{2}\right) \geqslant f\left(x_{1}\right)$. Therefore, when $x \in[0,1]$, $f(x) \leqslant f(1)=4$, so the maximum value of the function $f(x)$ is 4.
(3) Proof: When $x \in\left(\frac{1}{3}, 1\right]$, $f(x) \leqslant f(1)=4=1+3$.
And when $x \in\left(\frac{1}{3}, 1\right]$, $3 x>3 \times \frac{1}{3}=1$, hence $f(x)3 \times \frac{1}{3^{n+1}}=\frac{1}{3^{n}}$,
so when $x \in\left(\frac{1}{3^{n+1}}, \frac{1}{3^{n}}\right]$, $f(x) \leqslant f\left(\frac{1}{3^{n}}\right) \leqslant \frac{1}{3^{n}}+3<3 x+3$.