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
$$F(\alpha)=\{x\in [0,1]: \lim\_{n\rightarrow \infty}\frac{1}{n}\log |(T^{n})^{'}(x)|=\alpha\}.$$



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> Is not $\alpha \mapsto h\_{top}(F(\alpha))$ concave? $h\_{top}(.)$ defines in [this paper](https://www.ams.org/journals/tran/1973-184-00/S0002-9947-1973-0338317-X/S0002-9947-1973-0338317-X.pdf)
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I do not know how to prove it.



Entropy spectrum is not concave
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 $2$.


