Can we extend a continuous function with keeping Hausdorff dimension? - MathOverflow 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 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 Answer by Tapio Rajala for Can we extend a continuous function with keeping Hausdorff dimension? 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>