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.



Can we extend a continuous function with keeping Hausdorff dimension?
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 $Q\_i$.


