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!

• $|u(x)-u(y)|\geq|x-y|^{1/2}$ implies $u$ has infinite derivative at any point. – Gabriel Romon Sep 12 '17 at 6:53
• The claim that a continuous function $v$ exists so that $v(u(x))=x$ is completely unfounded. $u(x) = 1$ on $[0,1]$ is bounded, but there is no inverse map. – adfriedman Sep 12 '17 at 7:09
• @adfriedman If $u$ satisfies $\lvert u(x)-u(y)\rvert\geq \lvert x-y\rvert^{1/2}$, then $u$ clearly must be injective, so $u$ has an inverse. – Michael Lee Sep 12 '17 at 7:40

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.

$$|u(k/n) - u(l/n)| \leq \frac{1}{\sqrt{n}} \leq |k/n-l/n|^{1/2}$$

• Very nice argument. – uniquesolution Sep 12 '17 at 7:24
• @Long At the end it should be $< \frac{1}{\sqrt{n}}\leq \sqrt{|k/n - l/n|}$. Am I right? – Robert Z Sep 12 '17 at 7:36
• @RobertZ, yes. Also $n > R^2$ does not suffice entirely, should be something like $n > (R+\frac{1}{\sqrt{n}})^2$... didn't want to add another paragraph just for that detail though. – Long Sep 12 '17 at 7:40
• Thank you very much, it is really a delicate proof! – user88544 Sep 12 '17 at 8:45
• @Robert Z There is no essential problem. (@Long At the very beginning, it might be better to assume $0\leq u(x)\leq R-1.$) We can replace $\frac{1}{2\sqrt n}$ by $\frac{1}{3\sqrt n}$, and take $n\in\mathbf Z_+$ with $\frac{2\sqrt n}{3}>R,$ then $$\left|u\left(\frac{k}{n}\right)-u\left(\frac{l}{n}\right)\right|\leq\frac{2}{3\sqrt n}<\left|\frac{k}{n}-\frac{l}{n}\right|^{1/2}.$$ – user88544 Sep 12 '17 at 11:25

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).

• A good proof, thank you very much! – user88544 Sep 12 '17 at 8:57