Timeline for Theorems that are 'obvious' but hard to prove
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 4, 2019 at 13:27 | comment | added | Tom Leinster | @Brahadeesh The citation is Jussi Väisälä, A proof of the Mazur-Ulam theorem, American Mathematical Monthly 110 (2003), 633-635, doi.org/10.1080/00029890.2003.11920004. I couldn't quickly find a free copy online, but maybe you'll have better luck. | |
| May 20, 2019 at 15:00 | comment | added | user82537 | @TomLeinster The link to the pdf seems to be broken. Could you take a look at it, please? | |
| Mar 5, 2011 at 15:39 | comment | added | Todd Trimble | Since the inner product is linear in each argument, it quickly follows that if e_i is orthonormal, then the inner product of f(av + bw) against all the f(e_i) matches the inner product of af(v) + bf(w) against all the f(e_i). Since f(e_i) is an orthonormal basis, this shows f preserves linear combinations av + bw. | |
| Mar 5, 2011 at 15:34 | comment | added | Todd Trimble | Sorry Tom, didn't see your comment until now. Thanks for your comment. I had assumed "isometry" meant with respect to the usual Euclidean distance (2-norm). Here is a sketch of the proof I had in mind. First, by translation, assume WLOG that the origin is taken to itself. Then |f(v)|^2 = |v|^2 for all v. Since d(fv, fw) = d(v, w) for all v, w, it follows that |fv - fw|^2 = |v - w|^2. Since the inner product of v and w can be defined in terms of |v|^2, |w|^2, and |v-w|^2, it follows that f preserves inner products. So it takes an orthonormal basis to an orthonormal basis. Finally, [cont'd] | |
| Jan 13, 2011 at 18:22 | comment | added | Tom Leinster | But maybe you know all that, Todd... | |
| Jan 13, 2011 at 18:21 | comment | added | Tom Leinster | Todd, I think something closely resembling David's statement might count as hard: the Mazur-Ulam Theorem, that every bijective isometry between real normed spaces is affine. An indication that it's not obvious is that it's false without the word "bijective": e.g. you can cook up a non-affine isometry from R to (R^2 with the infinity-norm). Reference: helsinki.fi/~jvaisala/mazurulam.pdf . I don't know whether it's appreciably easier if you stick to two dimensions. | |
| Jan 10, 2011 at 15:20 | comment | added | Todd Trimble | Is this considered hard? | |
| Jan 10, 2011 at 7:39 | history | answered | David Sprehn | CC BY-SA 2.5 |