11. (20 points) Given a function $f(x)$ defined on $[0,1]$, $f(0)=0, f(1)=1$, and satisfying the following conditions:
(a) For any $x \in[0,1]$, $f(x) \geqslant 0$;
(b) For any two real numbers $x_{1}, x_{2}$ satisfying $x_{1} \geqslant 0, x_{2} \geqslant 0, x_{1}+x_{2} \leqslant 1$, $f\left(x_{1}+x_{2}\right) \geqslant$ $f\left(x_{1}\right)+f\left(x_{2}\right)$.

Find the smallest positive number $c$, such that for any function $f(x)$ satisfying the above conditions and any real number $x \in[0,1]$, $f(x) \leqslant c x$.
(Feng Yuefeng)
11. For any $0 \leqslant x \leqslant y \leqslant 1$, we have $0 \leqslant y-x \leqslant 1$. By condition (a), we have $f(y-x) \geqslant 0$. Therefore, by condition (b), we have
$$
f(y)=f(y-x+x) \geqslant f(y-x)+f(x) \geqslant f(x) .
$$

Thus, for any $0 \leqslant x \leqslant y \leqslant 1$, we have
$$
f(x) \leqslant f(y) \leqslant f(1) \leqslant 1 .
$$

Moreover, by condition (b), for any $x \in\left[0, \frac{1}{2}\right]$, we have
$$
f(2 x) \geqslant f(x)+f(x)=2 f(x),
$$

Combining this with mathematical induction, we can see that for any positive integer $k$ and $x \in\left[0, \frac{1}{2^{k}}\right]$, we have
$$
f\left(2^{k} x\right) \geqslant 2^{k} f(x) \text {. }
$$

Next, we prove that for any $x \in[0,1]$, we have $f(x) \leqslant 2 x$.
First, when $x=0$, $f(0)=0 \leqslant 2 \times 0$. So when $x=0$, $f(x) \leqslant 2 x$ holds.
Assume there exists $x_{0} \in(0,1]$, such that $f\left(x_{0}\right)>2 x_{0}$. Since $x_{0} \neq 0$, there must exist a positive integer $k$ such that $2^{k-1} x_{0} \leqslant 1$, but $2^{k} x_{0}>1$. For such $k$, by equation (2), we have
$$
f\left(2^{k-1} x_{0}\right) \geqslant 2^{k-1} f\left(x_{0}\right)>2^{k-1}\left(2 x_{0}\right)=2^{k} x_{0}>1 .
$$

But $2^{k-1} x_{0} \leqslant 1$, by equation (1), we have
$$
f\left(2^{k-1} x_{0}\right) \leqslant f(1)=1,
$$

which is a contradiction.
Therefore, for any $0 \leqslant x \leqslant 1$, $f(x) \leqslant 2 x$, so $c=2$ is valid.
Next, we prove that $c \geqslant 2$.
Assume $c < 2$.
Notice that the range of $f(x)$ is $[0,1]$. To make $f(t)$ as large as possible, we can take $f(t)=1$, which by the definition of $f(x)$, must mean $\frac{1}{2} c t < 1$, i.e., $t < \frac{2}{c}$. Since $c < 2$, we have $t < 1$. Thus, $x=t$ does not satisfy $f(x) \leqslant c x$, which is a contradiction.
In conclusion, the minimum value of $c$ is 2.