Question 17: Let the function $f(x)$ be defined on $[0,1]$, satisfying: $f(0)=f(1)$, and for any $x, y \in[0,1]$ there is $|f(x)-f(y)|<|x-y|$. Try to find the smallest real number $m$, such that for any $f(x)$ satisfying the above conditions and any $x, y \in[0,1]$, we have $|f(x)-f(y)|<m$.
Question 17, Solution: First, we prove that for any function $f(x)$ satisfying the conditions and any $x, y \in [0,1]$, we have $|f(x)-f(y)| < \frac{1}{2}$. If $|x-y| > \frac{1}{2}$, then $|f(x)-f(y)| = |f(x)-f(0)+f(1)-f(y)| \leq |f(x)-f(0)| + |f(1)-f(y)| < (x-0) + (1-y) = 1 - (y-x) < \frac{1}{2}$.

On the other hand, we take $f(x) = \begin{cases} \lambda x, & \text{when } x \in \left[0, \frac{1}{2}\right] \\ \lambda(1-x), & \text{when } x \in \left(\frac{1}{2}, 1\right) \end{cases}$, where $0 < \lambda < 1$. It is easy to verify that this function satisfies the given conditions, and $\left|f\left(\frac{1}{2}\right) - f(0)\right| = \frac{\lambda}{2} < m$. We take $\lambda \rightarrow 1^{-}$, then $m \geq \frac{1}{2}$. In summary, the minimum value of $m$ is $\frac{1}{2}$.