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This is a follow-up to my question here.  Let $d_1$ and $d_2$ be two metrics on the same set $X$.  Then $d_1$ and $d_2$ are uniformly equivalent if the identity maps $i:(M,d_1)\rightarrow(M,d_2)$ and $i^{-1}:(M,d_2)\rightarrow(M,d_1)$ are uniformly continuous.  And $d_1$ and $d_2$ are Holder equivalent if there exist constants $\alpha\in (0,1]$ and $C,D>0$ such that $C(d_1(x,y))^\alpha\leq d_2(x,y)\leq D (d_1(x,y))^\alpha$ for all $x,y\in X$.

Now if $d_1$ and $d_2$ are Holder equivalent, then they are uniformly equivalent and they have the same bounded sets.  My question is, is the converse true?  That is, if $d_1$ and $d_2$ are uniformly equivalent and have the same bounded sets, then are they Holder equivalent?

If not, is there an example of metrics which are uniformly equivalent and have the same bounded sets but are not Holder equivalent?

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  • $\begingroup$ How is Holder equivalence even symmetric? $\endgroup$ – SmileyCraft Jan 1 at 20:34
  • $\begingroup$ @SmileyCraft Maybe I stated it incorrectly? I want to say “$i$ and $i^{-1}$ are Holder continuous”. $\endgroup$ – Keshav Srinivasan Jan 1 at 20:35
  • $\begingroup$ So you mean "there exist constants $\alpha\, \beta \in (0,1]$ and $C,D>0$ such that $d_1(x,y)) \leq C (d_2(x,y))^\alpha$ and $d_2(x,y)) \leq D (d_1(x,y))^\beta$ for all $x,y\in X$." $\endgroup$ – Paul Frost Jan 2 at 11:36
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Here is a counterexample. Let $X = \mathbb{R}$ and $d_1(x,y) = \lvert y - x \rvert$. Define $d_2(x,y) = \lvert y - x \rvert$ if $\lvert y - x \rvert \le 1$ and $d_2(x,y) = \sqrt{\lvert y - x \rvert}$ if $\lvert y - x \rvert \ge 1$.

An obvious property of $d_2$ is that $\lvert y - x \rvert \le \lvert y' - x' \rvert$ implies $d_2(x,y) \le d_2(x',y')$. We shall moreover need the following well-known fact:

$(*)$ If $1 \le a \le b$, then $\sqrt{b} - \sqrt{a} \le b - a$.

Let us verify that $d_2$ is a metric. The only thing which is not trivial is the triangle inequality. For $x,y,z \in \mathbb{R}$ we have to show $d_2(x,y) \le d_2(x,z) + d_2(z,y)$. This is trivially true if $y = x$. In case $y \ne x$ we may w.l.o.g. assume that $y > x$, i.e. $y = x + r$ with $r > 0$.

(a) $z \le x$. Then $\lvert y - x \rvert \le \lvert z - x \rvert$, hence $d_2(x,y) \le d_2(x,z)$. The triangle inequality follows.

(b) $z \ge y$. Similar!

(c) $x < z < y$.

(c1) $r \le 1$. This reduces to the triangle inequality for $d_1$.

(c2) $r > 1$. Write $z = x + s$ with $0 < s < r$.

(c2.1) $s \le 1$ and $r - s \le 1$. Then we have to show $\sqrt{r} \le s + (r - s) = r$ which is true since $r > 1$.

(c2.2) $s \le 1$ and $r - s > 1$. Then we have to show $\sqrt{r} \le s + \sqrt{r - s}$. This follows from $(*)$.

(c2.3) $s > 1$ and $r - s \le 1$. Then we have to show $\sqrt{r} \le \sqrt{s} + (r - s) $. This follows agaim from $(*)$.

(c2.4) $s > 1$ and $r - s > 1$. Then we have to show $\sqrt{r} \le \sqrt{s} + \sqrt{r - s}$. This is obvious (take the squares of both sides).

$d_1$ are $d_2$ are uniformly equivalent because $d_2(x,y) \le d_1(x,y)$ and $d_2(x,y) < \min(1,\epsilon)$ implies $d_1(x,y) < \epsilon$ (in fact, we have $d_2(x,y) < 1$ which is only possible when $\lvert y - x \rvert < 1$ so that $d_1(x,y) = d_2(x,y) < \epsilon $).

$d_1$ and $d_2$ have the same bounded sets. If $B$ is bounded with respect to $d_1$, then it is obviously respect to $d_2$. Let $B$ be unbounded with respect to $d_1$. Then there exist $x_n,y_n \in B$ such that $d_1(x_n,y_n) \ge n$. But then $d_2(x_n,y_n) \ge \sqrt{n}$, hence $B$ be unbounded with respect to $d_2$.

$d_1$ and $d_2$ are not Hölder equivalent. Assume there are $C > 0$ and $\alpha \in (0,1]$ such that $d_1(x,y) \le C (d_2(x,y))^\alpha$. For $\lvert y - x \rvert \ge 1$ this means $\lvert y - x \rvert \le C \lvert y - x \rvert^{\alpha/2}$, i.e. $\lvert y - x \rvert^{1 - \alpha/2} \le C$. But $1 - \alpha/2 \in [1/2,1)$, hence $\lvert y - x \rvert^{1 - \alpha/2}$ cannot bounded by any constant $C$. This is a contradiction.

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