In Falconer's book Fractal Geometry, he proves that the dimension of a Brownian trail is $2$. I have all of it understood except this one part.

Let $X(t)$ be a Brownian trail in $\mathbb{R}^{n}$, for $n\geq2$. Then, with probability $1$, $\dim_{H}(X(t))=\dim_{B}(X(t))=2$.

Proof. For every $\lambda<\frac{1}{2}$ the Brownian trail $X:[0,1]\to\mathbb{R}^{n}$ satisfies $|X(t+h)-X(t)|\leq b|h|^{\lambda}$. So $$\dim_{H}(X([0,1]))\leq\frac{1}{\lambda}\dim_{H}([0,1])\leq\frac{1}{\lambda}.$$

Here he says "with a similar inequality for the box dimension."

Could someone help me see what this inequality is please? I just can't seem to find one.

  • 1
    $\begingroup$ The second inequality should be $=$ since the dimension of $[0,1]$ is $1$. $\endgroup$
    – user357151
    Aug 26, 2017 at 16:02

1 Answer 1


Generally, when $f$ is a $\lambda$-Hölder map, meaning $|f(a)-f(b)|\le C|a-b|^\lambda$ for some $\lambda\in (0,1]$, the various fractal dimensions: Hausdorff, box/Minkowski (upper and lower), packing (upper and lower) satisfy $$ \dim f(E) \le \frac{1}{\lambda}\dim E $$ The proof is the same:

  1. Cover $E$ by suitable sets $E_k$ that nearly achieve the bound required by $\dim E$
  2. Observe that $\operatorname{diam} f(E_k) \le C(\operatorname{diam} E_k)^\lambda$
  3. Make the conclusion about $\dim f(E)$ based on the cover $\{f(E_k)\}$.
  • $\begingroup$ That is really helpful, thank you. I knew I was missing something. $\endgroup$
    – JSharpee
    Aug 27, 2017 at 17:28

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