most recent 30 from http://mathoverflow.net 2013-06-18T21:04:23Z http://mathoverflow.net/feeds/question/100693 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/100693/can-we-extend-a-continuous-function-with-keeping-hausdorff-dimension luka 2012-06-26T15:37:00Z 2012-06-26T21:02:51Z

<p>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.</p>

http://mathoverflow.net/questions/100693/can-we-extend-a-continuous-function-with-keeping-hausdorff-dimension/100718#100718 Tapio Rajala 2012-06-26T21:02:51Z 2012-06-26T21:02:51Z

<p>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 $X \setminus K$.</p> <h3>Overly complicated way to construct the extension:</h3> <p>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 $$K \cup \bigcup_{j &lt; i} Q_j$$ which is closest to $x$ and then extend $F$ piecewise linearly to $Q_i$.</p>