11. Given a function $f(x)$ defined on $[0,1]$, $f(0)=0, f(1)=1$, and satisfies the following conditions:
(a) For any $x \in[0,1], f(x) \geq 0$;
(b) For any two numbers $x_{1} \geq 0, x_{2} \geq 0, x_{1}+x_{2} \leq 1$, it holds that $f\left(x_{1}+x_{2}\right) \geq f\left(x_{1}\right)+f\left(x_{2}\right)$.
Find the smallest positive number $c$, such that for any function $f(x)$ satisfying the above conditions and for any $x \in[0,1]$, we have $f(x) \leq c x$.
11. Solution: The smallest positive number $c$ is 2.

We first prove that for any $0 \leq x \leq y \leq 1, f(x) \leq f(y)$. Since $0 \leq x \leq y \leq 1$, it follows that $0 \leq y-x \leq 1$, thus $f(y) \geq f(x) + f(y-x) \geq f(x)$.

Next, we prove that for any $x \in [0,1]$, we have $f(x) \leq 2x$. From the given condition, we know: $f(2x) \geq 2f(x)$. Suppose there exists $x_{0} \in [0,1]$ such that $f(x_{0}) > 2x_{0}$, then there exists a positive integer $k$ such that $x_{0}$ must be in some interval $x_{0} \in \left[\frac{1}{2^{k}}, \frac{1}{2^{k-1}}\right]$, then $f(2x_{0}) > 4x_{0}, f(4x_{0}) > 8x_{0}, \ldots, f(2^{k-1}x_{0}) > 2^{k}x_{0}$. By $x_{0} \in \left[\frac{1}{2^{k}}, \frac{1}{2^{k-1}}\right]$, we know $\frac{1}{2}1$, hence $f(2^{k-1}x_{0}) \leq f(1) = 1$, and $f(2^{k-1}x_{0}) > 2^{k}x_{0} > 1$, which is a contradiction! Therefore, there does not exist $x_{0} \in [0,1]$ such that $f(x_{0}) > 2x_{0}$, i.e., for any $x \in [0,1]$, we have $f(x) \leq 2x$.
Finally, we prove that for any $0 < c < 2$, there exists $x \in [0,1]$ such that $f(x) > cx$.
Thus, the smallest positive number $c$ that satisfies the condition is 2.