Let $f$ be a function on $[0,1]$. Say that $f$ is reverse Hölder continuous of exponent $\beta > 0$ if there is a $C >0$ such that for any $s<t\in [0,1]$, there exists $s',t'\in [s,t]$ such that $|f(s') - f(t')| \geq (t-s)^\beta$. I am trying to show that if this is the case, then the Hausdorff dimension of the graph of $f$ is at least $2-\beta$. I then want to show that standard Brownian motion satisfies this condition with probability 1 for $\beta > 1/2$, so that the Hausdorf dimension of its graph is at least $3/2$.
This is based on an exercise from "Brownian Motion" by Morters and Peres.
For the first part, I considered a cover of the graph $G$ of $f$ by open balls and tried to show that the projections of various rescalings of these balls still cover $[0,1]$. But, this did not get me anywhere and I still fail to see where the $2-\beta$ comes in. For the second part, involving Brownian motion, I tried to break up $[s,t]$ into small independent increments and show that at least one of these increments is large if there are sufficiently many, but this did not seem to work either (the probability that an increment is large decays too quickly). Does anyone have any suggestions?
With the given conditions, the graph of the function need not have Hausdorff dimension at least $2-\beta$. Consider the characteristic function on the rationals, $1_\mathbb{Q}$.
For any $s<t<1$, pick $s',t' \in [s,t]$ so that $s' \in \mathbb{Q}, t' \notin \mathbb{Q}$. Then $|f(s')-f(t')|=1\geq (t-s)^{1/2}$. Hence this is reverse Holder continuous with exponent 1/2.
However, the graph of $f$ is essentially $[0,1]\times\{0\}$ (with countably many points removed). You can cover this by $(-1,2)\times\{0\}$. Hence its 1-dimensional Hausdorff content is finite, and non-zero. Its Hausdorff dimension is definitely not > 1.5.
You might need to impose a continuity condition on $f$.