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Anthony Quas
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The only values of $\alpha$ for which $F(\alpha)$ is non-empty are 0$-\infty$ and $\log 2$. This is because the map is conjugate (by the conjugacy $H\colon x\mapsto \sin^2(\pi x/2)$ to the full tent map $S$ (that is, $T\circ H=H\circ S$).

In particular, $T^n\circ H=H\circ S^n$, so that for any $x$, $|(T^n)'(H(x))|H'(x)=H'(S^nx)\cdot 2^n$, so that for any $x$ away from the endpoints, $$ |(T^n)'(x)|=2^n\frac {H'(S^n(H^{-1}x))}{H'(H^{-1}x)}. $$ If $x$ is any element of the co-countable set of points whose orbit does not hit the point 0, then the limit superior of $|(T^n)'(x)|^{1/n}$ is $2$.

The only values of $\alpha$ for which $F(\alpha)$ is non-empty are 0 and $\log 2$. This is because the map is conjugate (by the conjugacy $H\colon x\mapsto \sin^2(\pi x/2)$ to the full tent map $S$ (that is, $T\circ H=H\circ S$).

In particular, $T^n\circ H=H\circ S^n$, so that for any $x$, $|(T^n)'(H(x))|H'(x)=H'(S^nx)\cdot 2^n$, so that for any $x$ away from the endpoints, $$ |(T^n)'(x)|=2^n\frac {H'(S^n(H^{-1}x))}{H'(H^{-1}x)}. $$ If $x$ is any element of the co-countable set of points whose orbit does not hit the point 0, then the limit superior of $|(T^n)'(x)|^{1/n}$ is $2$.

The only values of $\alpha$ for which $F(\alpha)$ is non-empty are $-\infty$ and $\log 2$. This is because the map is conjugate (by the conjugacy $H\colon x\mapsto \sin^2(\pi x/2)$ to the full tent map $S$ (that is, $T\circ H=H\circ S$).

In particular, $T^n\circ H=H\circ S^n$, so that for any $x$, $|(T^n)'(H(x))|H'(x)=H'(S^nx)\cdot 2^n$, so that for any $x$ away from the endpoints, $$ |(T^n)'(x)|=2^n\frac {H'(S^n(H^{-1}x))}{H'(H^{-1}x)}. $$ If $x$ is any element of the co-countable set of points whose orbit does not hit the point 0, then the limit superior of $|(T^n)'(x)|^{1/n}$ is $2$.

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Anthony Quas
  • 23.2k
  • 5
  • 63
  • 98

The only values of $\alpha$ for which $F(\alpha)$ is non-empty are 0 and $\log 2$. This is because the map is conjugate (by the conjugacy $H\colon x\mapsto \sin^2(\pi x/2)$ to the full tent map $S$ (that is, $T\circ H=H\circ S$).

In particular, $T^n\circ H=H\circ S^n$, so that for any $x$, $|(T^n)'(H(x))|H'(x)=H'(S^nx)\cdot 2^n$, so that for any $x$ away from the endpoints, $$ |(T^n)'(x)|=2^n\frac {H'(S^n(H^{-1}x))}{H'(H^{-1}x)}. $$ If $x$ is any element of the co-countable set of points whose orbit does not hit the point 0, then the limit superior of $|(T^n)'(x)|^{1/n}$ is $2$.