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6 misunderstood

As to the border case. An example that you might like to consider is given by the blanc-mange curve, $f_{\lambda}:\mathbb{R}\to\mathbb{R}$, that for any value of $0\leq\lambda\leq 1,$ is defined as the unique bounded solution of the fixed point functional equation

$f(x)=\mathrm{dist }(x,\mathbb{Z})+\lambda f(2x)$

(by the contraction principle there is a unique such function; it is continuous and 1-periodic, with an immediate series expansion coming from the iteration).

Consider $f_{\lambda}$ on the unit interval. If you take $\lambda=1/4$ you find a parabola; with $\lambda < 1/2$ it's Lipschitz (hence the graph has a finite length) with constant $(1-2\lambda)^{-1}$; if $\lambda > 1/2$ it's Hoelder continuous with an exponent depending from $\lambda$. The parameter 1/2 is critical: you find a curve that is not Lipschitz, but it's Hoelder of all exponents $\alpha>0$, precisely, it has a modulus of continuity ct|log(t)|, and looking at it a bit more closely, it is not of bounded variation on any nontrivial interval (so the graph is not rectifiable even locally), nor is BV for any $\lambda \geq 1/2.$

As to the geodesic in the Sierpinski carpet. The SC is the set of (x,y) in the unit square such that either x or y are in the Cantor set, that is, admit a base 3 expansion with no 1's. To find a shortest path from (x,y) to (x',y') my guess is: working on the base 3 expansions, transform (x,y) into (x',y') always leaving fixed the Cantor set coordinate. This implies making a continuous broken line with horizontal or vertical segments. The resulting length has an expression in terms of (the base 3 expansion of) x,y,x'y' and I guess is the shortest possible.

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As to the border case. An example that you might like to consider is given by the blanc-mange curve, $f_{\lambda}:\mathbb{R}\to\mathbb{R}$, that for any value of $0\leq\lambda\leq 1,$ is defined as the unique bounded solution of the fixed point functional equation

$f(x)=\mathrm{dist }(x,\mathbb{Z})+\lambda f(2x)$

(by the contraction principle there is a unique such function; it is continuous and 1-periodic, with an immediate series expansion coming from the iteration).

Consider $f_{\lambda}$ on the unit interval. If you take $\lambda=1/4$ you find a parabola; with $\lambda < 1/2$ it's Lipschitz (hence the graph has a finite length) with constant $(1-2\lambda)^{-1}$; if $\lambda > 1/2$ it's Hoelder continuous with an exponent depending from $\lambda$. The parameter 1/2 is critical: you find a curve that is not Lipschitz, but it's Hoelder of all exponents $\alpha>0$, precisely, it has a modulus of continuity ct|log(t)|, and looking at it a bit more closely, it is not of bounded variation on any nontrivial interval (so the graph is not rectifiable even locally), nor is BV for any $\lambda \geq 1/2.$

As to the geodesic in the Sierpinski carpet. The SC is the set of (x,y) in the unit square such that either x or y are in the Cantor set, that is, admit a base 3 expansion with no 1's. To find a shortest path from (x,y) to (x',y') my guess is: working on the base 3 expansions, transform (x,y) into (x',y') always leaving fixed the Cantor set coordinate. This implies making a continuous broken line with horizontal or vertical segments. The resulting length has an expression in terms of (the base 3 expansion of) x,y,x'y' and I guess is the shortest possible.

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An example that you might like to consider is given by the blanc-mange curve, $f_{\lambda}:\mathbb{R}\to\mathbb{R}$, that for any value of $0\leq\lambda\leq 1,$ is defined as the unique bounded solution of the fixed point functional equation

$f(x)=\mathrm{dist }(x,\mathbb{Z})+\lambda f(2x)$

(by the contraction principle there is a unique such function; it is continuous and 1-periodic, with an immediate series expansion coming from the iteration).

Consider $f_{\lambda}$ on the unit interval. If you take $\lambda=1/4$ you find a parabola; with $\lambda < 1/2$ it's Lipschitz (hence the graph has a finite length) with constant $(1-2\lambda)^{-1}$; if $\lambda > 1/2$ it's Hoelder continuous with an exponent depending from $\lambda$. The parameter 1/2 is critical: you find a curve that is not Lipschitz, but it's Hoelder of all exponents $\alpha>0$, precisely, it has a modulus of continuity ct|log(t)|, and looking at it a bit more closely, it is not of bounded variation on any nontrivial interval (so the graph is not rectifiable even locally), nor is BV for any $\lambda \geq 1/2.$

As to the geodesic in the Sierpinski carpet. The SC is the set of (x,y) in the unit square such that either x or y are in the Cantor set, that is, admit a base 3 expansion with no 1's. To find a shortest path from (x,y) to (x',y') my guess is: working on the base 3 expansions, transform (x,y) into (x',y') always leaving fixed the Cantor set coordinate. This implies making a continuous broken line with horizontal or vertical segments. The resulting length has an expression in terms of (the base 3 expansion of) x,y,x'y' and I guess is the shortest possible.

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