5
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ $|u(x)-u(y)|\geq|x-y|^{1/2}$ implies $u$ has infinite derivative at any point. $\endgroup$ – Gabriel Romon Sep 12 '17 at 6:53
  • $\begingroup$ 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. $\endgroup$ – adfriedman Sep 12 '17 at 7:09
  • $\begingroup$ @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. $\endgroup$ – Michael Lee Sep 12 '17 at 7:40
4
$\begingroup$

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.

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

$\endgroup$
  • $\begingroup$ Very nice argument. $\endgroup$ – uniquesolution Sep 12 '17 at 7:24
  • $\begingroup$ @Long At the end it should be $< \frac{1}{\sqrt{n}}\leq \sqrt{|k/n - l/n|}$. Am I right? $\endgroup$ – Robert Z Sep 12 '17 at 7:36
  • $\begingroup$ @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. $\endgroup$ – Long Sep 12 '17 at 7:40
  • $\begingroup$ Thank you very much, it is really a delicate proof! $\endgroup$ – user88544 Sep 12 '17 at 8:45
  • $\begingroup$ @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 $$\left|u\left(\frac{k}{n}\right)-u\left(\frac{l}{n}\right)\right|\leq\frac{2}{3\sqrt n}<\left|\frac{k}{n}-\frac{l}{n}\right|^{1/2}.$$ $\endgroup$ – user88544 Sep 12 '17 at 11:25
1
$\begingroup$

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).

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.