Let X be a compact subset of R^d, and K be a compact subset of X, such that Dim_H(X)=Dim_H(K). Let F be a continuous function on K, Can we extend F from K to X, with keeping the continuous and the Hausdorff dimension of the gragh.
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$\begingroup$ What is the range of the function $F$? Is it also Euclidean space? Without specifying the range, the "Hausdorff dimension of the graph" makes no sense. $\endgroup$– Lee MosherCommented Jun 26, 2012 at 16:30
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$\begingroup$ Yes,I am sorrry, I mean taht F is a real-valued continuous function. $\endgroup$– lukaCommented Jun 27, 2012 at 3:36
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Yes, because you can extend any continuous mapping defined on $K$ to the whole space $\mathbb R^d$ so that it is locally Lipschitz outside $K$. Now the graph of $F$ on $X \setminus K$ has the same dimension as $X \setminus K$.
Overly complicated way to construct the extension:
First take a Whitney decomposition of $Q\setminus K$, where $Q \subset \mathbb R^d$ is some dyadic cube containing $K$. Then enumerate the decomposition cubes $Q_i$ so that the diameter of $Q_i$ is decreasing. Next iteratively define $F$ on $Q_i$ as follows: For each corner $x$ of the cube $Q_i$ define $F(x)$ to be the value of $F$ at one of the points on $$K \cup \bigcup_{j < i} Q_j$$ which is closest to $x$ and then extend $F$ piecewise linearly to $Q_i$.
answered Jun 26, 2012 at 21:02
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$\begingroup$ (Please note that one has to be careful with the above construction: for example one should not select cubes to the decomposition which agree with the boundary of $K$ at other points besides the corners, one has to make sure that the extension agree on faces etc. All these things should be doable, but I left such details out.) $\endgroup$ Commented Jun 27, 2012 at 6:35