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How can one prove the following result on the Hausdorff dimension of the graph of the sum of two continuous functions:

Let $f,g:[0,1] \to \mathbb R$ be two continuous functions. Suppose that $$\operatorname{dim}_H(\operatorname{graph} f) < \operatorname{dim}_H(\operatorname{graph} g). $$

Then $$\operatorname{dim}_H(\operatorname{graph} (g+f)) = \operatorname{dim}_H(\operatorname{graph} g)$$

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  • $\begingroup$ There is no way. It is merely false. Think of your favorite construction of a function $g$ with the Hausdorff dimension of the graph greater than $1$ as a convergent series $\sum_k g_k$ of Lipschitz functions. Now split the sum into blocks choosing every next block much longer than the previous one, Then the sum over odd blocks will have the graph of Hausdorff dimension $1$ and so will the sum of even blocks. The problem with the Hausdorff measure is that it is controlled just by some sequence of scales and that sequence may be different for $g+f$ and $f$. $\endgroup$ – fedja May 16 at 17:50
  • $\begingroup$ @fedja Thanks for your remark. I've two follow up questions: (1) Could you add a more detailed answer based on your comment? (2) Is the statement true for the box-counting dimension? $\endgroup$ – Dal May 16 at 17:57
  • $\begingroup$ 2) For the box counting dimension it, obviously, holds. The equivalent reformulation is that $dim(graph(f+g))\le\max(dim(graph(f)),dim(graph(g)))$ and the box dimension of $graph(f)$ is just the infimum of $p$ such that you can split the interval $[0,1]$ into $C\delta^{-p}$ subintervals $I_j$ such that $|I_j|\le\delta$ and $osc_{I_j}f\le\delta$ for all $\delta\in(0,1)$. Making the common splitting for 2 functions merely doubles the number of intervals. As to 1), what is unclear that you want me to comment upon it in detail? $\endgroup$ – fedja May 16 at 18:14
  • $\begingroup$ P.S. That is about the upper box dimension. With the lower box dimension you have the same problem as with Hausdorff. $\endgroup$ – fedja May 16 at 18:20
  • $\begingroup$ @fedja What properties of the sequence $(g_k)$ do you intend to use, and how do you choose the blocks? For example, the $g_k$ appearing in the odd blocks could be each identically zero, and then the sum of the even blocks would be the constructed function with dimension greater than $1$ ? $\endgroup$ – Lutz Mattner May 16 at 18:47
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OK, it looks like I, indeed, need to spell a few things out.

First, continuous functions with Hausdorff dimension of the graph greater than $1$ exist. I'll skip this part.

Let $g$ be any such function. It can be written as the sum of a uniformly convergent series $\sum_k g_k$ of Lipschitz functions such that $\|g_k\|_\infty\le 2^{-k}$. If it is not already constructed like that, then just consider any sequence of piecewise linear approximations $G_k$ such that $\|G_k-g\|\le 2^{-k-2}$ and put $g_1=G_1$, $g_k=G_k-G_{k-1}$ for $k\ge 2$. Of course, the Lipschitz constants $L_k$ of $g_k$ will grow pretty fast.

Now let the first block be just $\{1\}$. Suppose that we have already constructed the blocks $J_1,\dots,J_n$ and $N$ is the last index of $J_n$. Let $L=\sum_{k=1}^N L_k$. Then the sum of $g_k$ over any subset of the union of the first $n$ blocks (which is just $[1,N]$) is $L$-Lipschitz, so for every $\delta>0$, we can cover the graph of that sum by about $L\delta^{-1}$ disks of radius $\delta$. Choose $\delta_n\in(0,2^{-n})$ so small that $L\delta_n^{-1}\le \delta_n^{-p_n}$ where $p_n\in(1,2)$ is some fixed sequence tending to $1$. Now let the $n+1$-st block to be $J_{n+1}=[N,M]$ where $M$ satisfies $2^{-M}\le\delta_n$. Then the graph of the sum of $g_k$ over any subset of indices disjoint with $J_{n+1}$ can be covered by $\delta_n^{-p_n}$ balls of radius $3\delta_n$, say (the tail beyond $M$ is just too small to really matter on that scale).

Now, by construction, the sum of $g_k$ over odd blocks $J_n$ has a graph that can be covered by $\delta_n^{-p_n}$ balls of radius $3\delta_n$ for all odd $n$. Since $\delta_n\to 0$ and $p_n\to 1$, we conclude that its Hausdorff (or even lower box) dimension is $1$. The same is true for the sum over even blocks.

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  • $\begingroup$ Just take the centers to be $(x,f(x))$ where $x$ runs over some $L^{-1}\delta/2$ net in $[0,1]$ and use the Lipschitz condition. $\endgroup$ – fedja May 16 at 21:22
  • $\begingroup$ (1) We do not need it but we can do it. (2) To make sure that $\sum_{k>M}\|g_k\|_\infty\le 2\delta_n$ (3) We split the sum into that over the indices to the left of $J_{n+1}$ (whose graph can be covered by disks of radius $\delta_n$ and that over the indices to the right of $J_{n+1}$, which is less than $2\delta_n$ at every point. $\endgroup$ – fedja May 16 at 21:27
  • $\begingroup$ Their sum is $g$. As to why the sums have graphs ofdimension $1$, ask a more specific question. I provided the proof of that fact in the last paragraph. What exactly is unclear about it? $\endgroup$ – fedja May 17 at 11:56
  • $\begingroup$ By the construction we have that if the set of indices does not intersect $J_{n+1}$, then we have a good cover at the scale $\delta_n$. Now, if infinitely many blocks are missing from the set of indices, good covers occur at arbitrarily small scales. The easiest way to ensure that infinitely many blocks are missing in each sum is to take every other block, i.e., to split into odd and even. However, this is not the only option, of course: you can just as well make one sum consisting of blocks indexed by perfect squares. $\endgroup$ – fedja May 17 at 12:16
  • $\begingroup$ $dim(graph(f_1))=1=dim(graph(g-f_1))<dim(graph(g))$ (so $f_1$ is $f_1$ but $g$ is what you called $f_2$). $\endgroup$ – fedja May 17 at 20:50

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