distance between two functions by considering invariance

Question

For two functions $f(x): \mathbb R^n \mapsto \mathbb R^m$, $g(x): \mathbb R^n \mapsto \mathbb R^m$. An usual way to define their distance may be $$ \|f-g\| = \sqrt{\int_x \|f(x)-g(x)\|^2 dx}. $$ I think we can prove this distance is a metric. Now suppose I allow linear invariance, i.e.:

If there exists a matrix $W \in \mathbb R^{m \times m}$ such that for any $x$, $f(x) = W g(x)$, then the distance between $f$ and $g$ is zero.

How could I modify the definition of distance above to take care of this invariance? Can we still prove that the modified distance is also a metric?

You must require some conditions on $f$ and $g$, otherwise your integral might not make sense. — Falcon, May 06 '22 at 19:20
I assume they have to be $L^2(\mathbb R^n, \mathbb R^m)$, so that your distance is just the one associate to the usual $L^2$-norm. — Falcon, May 06 '22 at 19:21
What you want by dividing by that equivalence class is called a projective space. Natural metrics on projective spaces are angular in nature, so your current definition would not be a metric on the new space - this answer explains why — Ninad Munshi, May 06 '22 at 19:23
I see @NinadMunshi. This is quite helpful. Anyway, do you think $\min_W | f - Wg|$ a reasonable definition? — Jiaji Huang, May 06 '22 at 20:25
I have given an appropriate metric below as an answer that has the invariance you want. This describes an angular distance between "rays" in your space. — Ninad Munshi, May 06 '22 at 20:36

score 1 · Answer 1 · answered May 06 '22 at 20:36

1

Following on from the intuition in projective spaces, the following would make a good metric for your new space

$$d(f,g) = \cos^{-1}\left[\int_{\Bbb{R}^n}\hat{f}\cdot \hat{g}\:dx\right] = \cos^{-1}\left[\frac{\int_{\Bbb{R}^n}f\cdot g\:dx}{\sqrt{\int_{\Bbb{R}^n}||f||^2\:dx}\sqrt{\int_{\Bbb{R}^n}||g||^2dx}}\right]$$

where $\hat{f}$ and $\hat{g}$ are the chosen representatives of their equivalence classes with unit norm in the original space.

answered May 06 '22 at 20:36

Ninad Munshi

34,407

I could see that this definition is scale invariant. Although not fully address the linear invariant case, it's still very helpful. – Jiaji Huang May 06 '22 at 21:03

distance between two functions by considering invariance

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