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This is the definition given for the Hilbert Cube: enter image description here

The metric defined here is basically just the L2 metric. I was wondering if we can use any of the Lp metric instead as long as p>1. It can be shown using the comparison test that they all converge, so any of those metric should also be a well-defined metric correct?

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No doubt that those are also well-defined metrics, yes. For any $x, y \in I^\infty$ we have: $$ d_{\ell^p}(x, y) = \sqrt[p]{\sum_{i = 1}^\infty \lvert x_i - y_i \rvert^p}\leq \sqrt[p]{\sum_{i = 1}^\infty 2^{p-1}\lvert x_i \rvert^p + 2^{p-1}\lvert y_i \rvert^p} \leq \sqrt[p]{\sum_{i = 1}^\infty \frac{2^p}{i^p}} $$ This is finite iff $p>1$.

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  • $\begingroup$ what identity did you use from the first to the second inequality? $\endgroup$
    – Bill
    Commented Apr 4, 2021 at 10:54
  • $\begingroup$ I used $\lvert x_i - y_i \rvert^p \leq (\lvert x_i \rvert + \lvert y_i \rvert)^p and then that (a+b)^p\leq 2^{p-1}(a^p+b^p)$ for all $a, b \in \mathbb{R}$. The latter you prove by exploiting convexity of $g(t) := t^p$, i.e. $g\left(\frac{1}{2}(a+b) \right) \leq \frac{1}{2}g(a) + \frac{1}{2}g(b)$. $\endgroup$ Commented Apr 4, 2021 at 11:29

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