Ytukyg

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 $alphain (0,1]$ and $C,D>0$ such that $C(d_1(x,y))^alphaleq d_2(x,y)leq D (d_1(x,y))^alpha$ for all $x,yin X$.

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?

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.