Let $T:[0, 1]\rightarrow [0, 1]$ be map such that $T(x)=4x(1-x)$. For any $\alpha \in \mathbb{R}$, we define the level set as follows $$F(\alpha)=\{x\in [0,1]: \lim_{n\rightarrow \infty}\frac{1}{n}\log |(T^{n})^{'}(x)|=\alpha\}.$$
Is not $\alpha \mapsto h_{top}(F(\alpha))$ concave? $h_{top}(.)$ defines in this paper
I do not know how to prove it.
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 $2$.
