Hausdorff dimension of graph of function This question came up on an exam
Decide the Hausdorff dimension of the graph of the following function for $x>0$
$$y = \log(1+x)\sin\frac{1}{x}$$
In the course, we only touched upon the subject and the example given in the course was of the Cantor set. How does one go about solving this problem or more specifically, where do you start when you want to decide the Hausdorff dimension of some arbitrary graph $y = f(x)$?
 A: Seems like the dimension is $1$. Note that the word "hence" below often means you have some details to work out:
If $Q$ is a square in the plane let $\ell(Q)$ be the side length of $Q$. If $I$ is an interval let $|I|$ be the length of $I$. Say $G$ is the graph.
If $G\subset\bigcup Q_j$ then $(0,\infty)\subset \bigcup I_j$, where $|I_j|=\ell(Q_j)$. Hence $\sum\ell(Q_j)=\infty$. So $h_1(G)=\infty$.
For $n\in\Bbb Z$ let $I_n=(2^{n-1},2^n]$. Let $G_n=\{(x,f(x))\,:\,x\in I_n\}$.
Suppose $n<0$. Then $\log(1+2^n)\sim 2^n$. So $G_n$ is contained in the union of boundedly many squares of side length $2^n$. Hence $$h_1\left(\bigcup_{n=-\infty}^0G_n\right)<\infty,$$so $$h_\alpha\left(\bigcup_{n=-\infty}^0G_n\right)=0\quad(\alpha>1).$$
On the other hand, $f'(x)$ is bounded for $x>1$. Hence $$h_\alpha\left(\bigcup_{n=0}^\infty G_n\right)\le ch_\alpha((1,\infty))=0\quad(\alpha>1).$$
A: The graph of $y=f(x)$ where $x\in \mathbb{R}^m$ and $f(x)\in \mathbb{R}^n$ is a set of $\mathbb{R}^{m+n}$. If $f$ is smooth then Hausdorff dimension coincide with the euclidian dimension. Otherwise, I think you need do for each case, considerating coverings and using the definitions and auxiliar results.
A good book is Falconer, Fractal Geometry.
