I am trying to calculate Hausdorff dimension for the trajectory of Brownian motion over $[0,1]$. I read the book of Morters and Peres and know that the dimension will be $\frac{3}{2}$. I tried to use different approach from the book and I have some concerns as follow:
Is it true that: If $f$ is a Lipschitz function or has a continuous derivative or has bounded variation, then graph $f $ has dimension of $1$. If this is true, can someone please explain to me or point me to some relevant material.
I calculate the total variation of Brownian motion (BM) and got the result as follow: $$\sum_{i=1}^{n} |B(t_i)-B(t_{i-1})| \longrightarrow \sqrt{\frac{2}{\pi h}}$$ this shows that Brownian motion has unbounded variation as $h \to 0$ $(h=t_i-t_{i-1})$. If 1. is true then this implies 1-dimensional Brownian motion has dimension of greater than $1$. If I look at the total variation of BM, the total variation will become larger as $h$ become smaller and the order of magnitude of the sum is $\frac{1}{2}$ (the exponent of $h$). I wonder if this is the Hausdorff dimension of BM because it is not $\frac{3}{2}$ as mentioned in a lot materials.
I really appreciate any helps or comments
Every continuous function of bounded variation (BV) has a graph of Hausdorff dimension $1$. Indeed, being BV means precisely that the graph is rectifiable, i.e., has finite $1$-dimensional Hausdorff measure. Being Lipschitz continuous implies BV. Having continuous derivative implies Lipschitz continuity, hence implies BV.
It's true that sample paths of Brownian motion have infinite variation. By itself, infinite variation does not imply Hdim $>1$. One has to look more carefully at the sample path (quadratic variation, modulus of continuity...) to conclude that Hdim $=3/2$ for almost every path. This is not an easy result; it was proved in 1953 by S.J. Taylor. A proof can be found in An Invitation to Sample Paths of Brownian Motion by Yuval Peres; see Theorem 6.1.