Show that there exist $x,y\in[0,1]$ such that $|u(x)-u(y)|<|x-y|^{1/2}$. Suppose $u:[0,1]\to\mathbf R$ is a bounded function, show that there exist $x,y\in[0,1]$ such that $|u(x)-u(y)|<|x-y|^{1/2}.$
I have some stupid trials but failed. It seems that we could argue via contradiction. 
If for each $x,y\in[0,1]$ with $x\neq y,$ there holds $$|u(x)-u(y)|\geq|x-y|^{1/2}.$$ We denote by $R=\{u(x);\ x\in[0,1]\},$ then we can define a continuous function $v:R\to[0,1]$ such that $$v(u(x))=x\ \hbox{for all $x\in[0,1].$}$$ According to our assumption, we have $$|v(s)-v(t)|\leq|s-t|^2\ \hbox{for all $s,t\in R$}.$$ Let $\bar R$ be the closure of $R$ in $\mathbf R,$ then we can continuously extend the function $v:R\to[0,1]$ to $\bar v:\bar R\to[0,1].$ It is easy to check $\bar v:\bar R\to[0,1]$ is a continuous and closed map, and also a surjection. In additional, if we could prove that $\bar v:\bar R\to[0,1]$ is an injection, then $\bar v:\bar R\to[0,1]$ is a homeomorphism, a contradiction. But it seems difficult to verify this argument. 
Another trial. If $U:[0,1]\to\mathbf R$ satisfies $$|U(x)-U(y)|\leq|x-y|^2\ \hbox{for all $x,y\in[0,1]$,}$$ then actually $U$ is constant. According to this result, if there is some interval $[a,b]\subset\bar R,$ then $\bar v|_{[a,b]}$ is constant. However, it seems impossible to show that $\bar R$ contains some interval, though the cardinal number of the set $R$ is $\aleph.$
Hope someone could give me some hint, and any comments will be welcome. Thank you very much! 
 A: Suppose without loss that $u : [0,1] \to [0,R]$ and consider the graph of 
$u$ in $[0,1]\times[0,R]$. Note that the area of the big rectangle $[0,1]\times[0,R]$ is $R$.
Let $n$ be a positive integer.
For each integer $0 \leq k \leq n$ cover the point $(k/n,u(k/n))$ with the rectangle $[0,1]\times (u(k/n) + (-\frac{1}{2\sqrt{n}},\frac{1}{2\sqrt{n}}))$.
If the rectangles were disjoint, their union would cover an area of more than $n\cdot \frac{1}{\sqrt{n}} = \sqrt{n}$, which is not possible for $n > R^2$. So for $n > R^2$ we find two distinct points $(k/n,u(k/n))$ and $(l/n,u(l/n))$ whose rectangles overlap.
$$ |u(k/n) - u(l/n)| \leq \frac{1}{\sqrt{n}} \leq |k/n-l/n|^{1/2} $$
A: Let's prove the problem statement by contradiction. Suppose that $\lvert u(x)-u(y)\rvert\geq \lvert x-y\rvert^{1/2}$ for all $x, y\in [0, 1]$. Then, let $x_n = \frac{6}{\pi^2}\sum_{k=1}^n \frac{1}{k^2}$ so that $\lim_{n\to \infty} x_n = 1$ (recall that $\zeta(2) := \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6}$). Then, for all $n\in \mathbb{N}$, $u(x_n)$ must be a minimum distance of $r_n = \frac{\sqrt{6}}{\pi n}$ away from any $u(x_m)$, $m\neq n$ (i.e. $B_{r_n}(u(x_n))\cap \{u(x_m)\}_{m=1}^{\infty} = \{u(x_n)\}$). From this, it is easy to show that $\{u(x_n)\}$ cannot be bounded. First, we note a lemma:

Lemma 1: There is no $(n_1, n_2, n_3)\in \mathbb{N}^3$ such that $B_{r_{n_1}}(u(x_{n_1}))\cap B_{r_{n_2}}(u(x_{n_2}))\cap B_{r_{n_3}}(u(x_{n_3}))\neq \emptyset$.

Proof: Assume that there is a $y\in B_{r_{n_1}}(u(x_{n_1}))\cap B_{r_{n_2}}(u(x_{n_2}))\cap B_{r_{n_3}}(u(x_{n_3}))$. Then, we must have that $y$ is less than $u(x_{n_1})$, between $u(x_{n_1})$ and $u(x_{n_2})$, between $u(x_{n_2})$ and $u(x_{n_3})$, or greater than $u(x_{n_3})$. It is easy to see that in each of these cases, we must have $u(x_{n_i})\in B_{r_{n_j}}(u(x_{n_j}))$ for some $i, j\in \{1, 2, 3\}$, $i\neq j$.
Therefore, $\{B_{r_n}(u(x_n))\}_{n=1}^{\infty}$ can at most doubly-cover any point of $\mathbb{R}$. This implies that for any $N\in \mathbb{N}$, $$m\left(\bigcup_{n=1}^N B_{r_n}(u(x_n))\right)\geq \frac{1}{2}\sum_{n=1}^N \frac{2\sqrt{6}}{\pi n} = \frac{\sqrt{6}}{\pi}\sum_{n=1}^N \frac{1}{n}$$ so therefore, $$m\left(\bigcup_{n=1}^{\infty} B_{r_n}(u(x_n))\right) = \lim_{N\to \infty} m\left(\bigcup_{n=1}^N B_{r_n}(u(x_n))\right) = \infty$$ which contradicts that $\{u(x_n)\}$ can be contained in a bounded interval $I$ of $\mathbb{R}$ (if it could, then $\bigcup_{n=1}^{\infty} B_{r_n}(u(x_n))$ would be contained in a set of measure at most $m(I)+\frac{2\sqrt{6}}{\pi}$, i.e. $I$ with "buffer zones" of radius $\frac{\sqrt{6}}{\pi}$ at either end).
