Lipschitz maps with Hölder inverse preserve the doubling property Let $K$ be a compact doubling metric space, $X$ be a metric space and $f:K\rightarrow X$ be Lipschitz with $\alpha$-Hölder inverse, where $0<\alpha<1$.  Does $f(K)$ need to be doubling?
 A: If I have understood the definitions correctly, $f(K)$ need not be doubling.
For example consider a map $f$ from $[0,1]$ to the Hilbert space $\mathbb{R}\times l^2$ defined in the following way. Let $(e_k)_{k=1}^\infty$ be the usual basis for $l^2$. For each $n\geq0$ we will define the map in the interval $[2^{-(n+1)},2^{-n}]$.
Let $f(2^{-n})=(2^{-n},0)$, and $f\left(2^{-n}\left(1-\frac{k}{2n}\right)\right)=\left(2^{-n},2^{-n}\frac{e_k}{n+1}\right)$ for $k=1,\dots,n$. Then interpolate linearly, using that $f(2^{-(n+1)})=(2^{-(n+1)},0)$.
To give an intuition using the picture of $[0,1]$ below, in the green intervals $[2^{-(n+1)},3\cdot2^{-(n+2)}]$ $f$ changes the first coordinate from $2^{-(n+1)}$ to $2^{-n}$, while in the red intervals $[3\cdot2^{-(n+2)},2^{-n}]$ $f$ traces some piecewise linear path through an $n$-sphere in the second coordinate.

The resulting map is Lipschitz, and its image is not doubling because the open ball $B((2^{-n},0),2\cdot\frac{2^{-n}}{n+1})$ in $f([0,1])$ cannot be covered with $<n$ open balls of radius $\frac{2^{-n}}{n+1}$.
To see that the inverse is $\frac{1}{2}$-Hölder, it is enough to prove $\forall n\geq4$ that if $y>x+2^{-n}$ then $d(f(x),f(y))\geq2^{-2n}$. This is true if $x\leq2^{-n}$, because then $f(x)$ and $f(y)$ differ by at least $2^{-(n+2)}\geq2^{-2n}$ in the first coordinate. If $x\geq2^{-n}$, then either $f(x)$ and $f(y)$ differ at least $2^{-n-1}$ in the first coordinate (if they contain a green segment of length $\geq2^{-n-1}$) or they are at distance $\geq\frac{2^{-n}}{4n}\geq2^{-2n}$ in some other coordinate.
