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Show that $F: (\mathbb{R}^n, \varepsilon) \rightarrow (\mathbb{R}^n, \rho), \quad \mathbf{x} \mapsto \mathbf{x}$ is continuous, where $\varepsilon$ is the Euclidean metric and $\rho(\mathbf{x},\mathbf{y}) = \sqrt{\sum_{i,j=1}^n a_{ij} (x_i-y_i)(x_j-y_j)}$, and $\mathbf{A}$ is a symmetric positive-definite real $n \times n$ matrix.

This feels like it should be rather easier than I'm currently finding it, considering how close in form the metrics are:
$\sqrt{\sum_{i=1}^n (x_i-y_i)^2} < \delta \Rightarrow \sqrt{\sum_{i,j=1}^n a_{ij} (x_i-y_i)(x_j-y_j)} < \mu$

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up vote 0 down vote accepted

Hint: there are positive constants $c$, $d$ such that for all $\bf x$, $ c\, {\bf x}^T {\bf x} \le {\bf x}^T {\bf A x} \le d\, {\bf x}^T {\bf x}$.

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Thanks, that really worked out well, but do you think I'd be required to prove that it's true? If so, how would one do that? I mean, I'm pretty sure I understand it - the positive definite quadratic forms are paraboloids with their apices at the origin, so they have the same order of growth, meaning that ${\bf x}^T{\bf Ax} \in \Theta({\bf x}^T{\bf x})$? – user9008 Apr 2 '11 at 9:33

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