I'm interested to compute the $L^q$ spectrum for absolutely continuous measures $\nu$ with respect to Lebesgue measure denoted by $\lambda$ with support in $[0,1]$, that is for $q\geq 0$ \begin{align*}\tau(q)= \limsup_{n\rightarrow \infty}\dfrac{\log\left( \sum_{k=0,\nu((k/2^n,(k+1)/2^n)>0}^{2^n-1}\nu((k/2^n,(k+1)/2^n])^q \right)}{\log(2^n)}. \end{align*} I think it is sufficient to prove this for $\nu=\textbf{1}_Ad \lambda $ where $A\subset [0,1]$ is a Borel set with $\lambda(A)>0$.
In the case that there exists $(c,d)\subset A$ a.e. it follows \begin{align*} \sum_{k=0,\nu((k/2^n,(k+1)/2^n)>0}^{2^n-1}\nu((k/2^n,(k+1)/2^n])^q \geq 2^{n(1-q)}\left(\frac{\left\lfloor d2^{n}-1\right\rfloor -\left\lceil c2^{n}\right\rceil +1}{2^{n}}\right) \end{align*} This implies $\tau(q)\geq 1-q$ in tandem with the convexity of $\tau$ and $\tau(0)\leq 1,\tau(1)=0$, we obtain $\tau(q)=1-q$.
I'm conjecture that we always have $\tau(q)=1-q$ for absolutely continuous measures and all $q\in [0,1]$.
Unfortunately, there are Borel sets with positive measure which contains no open interval. Is there a way to fix this problem?
One way to fix this is to use the usual Holder's inequality. More precisely, suppose that $\frac{d\nu}{d\lambda}=f$ then $$2^{nq}\nu\left( k/2^n, (k+1)/2^n) \right]^q =\left( 2^n\int_{k/2^n}^{(k+1)/2^n}f(x)d\lambda(x) \right)^{q}\ge 2^{n}\int_{k/2^n}^{(k+1)/2^n}f^q(x)d\lambda(x)$$ for all $q \in (0,1)$ . Hence $$\tau(q) \ge \lim \frac{\log \left( 2^{n(1-q)}\int_{0}^1 f^q d\lambda \right)}{\log(2^n)}= 1-q$$ $\square$