tag:blogger.com,1999:blog-5799246.post7287619472314268722..comments2022-04-11T21:08:12.414+00:00Comments on a very occasional diary: The Hunt for Addi(c)tive Monsternikitahttp://www.blogger.com/profile/09403336533089968821noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-5799246.post-26417566045969390022012-10-29T18:28:23.502+00:002012-10-29T18:28:23.502+00:00Treeowl,
interesting question!
Yes, any additiv...Treeowl, <br /><br />interesting question!<br /><br />Yes, any additive, but not linear mapping f : V -> U, between 2 vector spaces (not necessary infinite-dimensional) is necessary a monster... at least from some angle!<br /><br />Consider an arbitrary linear non-0 p : R -> V and similarly q : U -> R. The composition t(r) = q(f(p(r))) is an additive mapping t : R -> R.<br /><br />It is possible that some t is linear, while f isn't, e.g., f(x, y) = x + monster(y), p(x) = (x, 0), q(x) = x.<br /><br />But if all t's are linear, than f is: select (v_i) --- a basis in V and (u_j) --- a basis in U. Define p_i : R -> V, such that p_i(1) = v_i and q_j : U -> R such that q_j(\Sigma_j l_j * u_j) = l_j, where \Sigma_j l_j * u_j is an arbitrary vector in basis (u_j). That is, q_j projects to j-th basis element. The rest of the argument easily follows by linearity of p_i, q_j and additivity of f and assumed linearity of q_j(f(p_i(r))). nikitahttps://www.blogger.com/profile/09403336533089968821noreply@blogger.comtag:blogger.com,1999:blog-5799246.post-56732142746763029022012-10-27T23:14:40.069+00:002012-10-27T23:14:40.069+00:00Sorry, I meant INTO such spaces, but I suppose the...Sorry, I meant INTO such spaces, but I suppose the other question might also make sense.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5799246.post-10530538377926029792012-10-27T23:11:46.534+00:002012-10-27T23:11:46.534+00:00Are non-linear additive functions on infinite-dime...Are non-linear additive functions on infinite-dimensional real vector spaces necessarily monsters?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-5799246.post-9199643602111691862012-01-13T15:47:11.712+00:002012-01-13T15:47:11.712+00:00Nice, yes, that seems to do it. And I made a silly...Nice, yes, that seems to do it. And I made a silly mistake above, when I suggested linearity on yQ contradicted periodicity - of course adding a multiple of s to y will typically carry you off yQ. There will be plenty of periodic solutions, since we can always specify the value 0 on some basis element.<br /><br />Cheers,<br />ChrisAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-5799246.post-30114547222546035242012-01-11T21:52:11.914+00:002012-01-11T21:52:11.914+00:00Dear Chris,
thank you for your interest in the po...Dear Chris,<br /><br />thank you for your interest in the post---it's a pleasure that it helped you. You are absolutely correct that the proof of Statement 1 contains a gap and Engel's device of building a periodic additive function is very ingenious.<br /><br />However, I think that the proof can be very easily amended: indeed, if an additive function is continuous at 0, it is continuous at arbitrary x, because for any y, such that |x - y| < \delta, |f(x) - f(y)| = |f(x - y)| < \epsilon, where \epsilon and \delta are the same as for continuity at 0.nikitahttps://www.blogger.com/profile/09403336533089968821noreply@blogger.comtag:blogger.com,1999:blog-5799246.post-61956816056471258052012-01-11T12:09:16.019+00:002012-01-11T12:09:16.019+00:00Dear Nikita,
thanks for your post - I enjoyed rea...Dear Nikita,<br /><br />thanks for your post - I enjoyed reading it, and together with some other sources, it helped me prepare a talk on this subject.<br /><br />However, I believe there's a gap in your proof of Statement 2. Your proof of Statement 1 only works for x nonzero, because you divide by x/y at the last step. So it's not enough to just prove continuity at zero in order to deduce Statement 2.<br /><br />There's a nice argument in Engel's "Problem Solving Strategies" (page 273) that begins by using boundedness on (p,p+s) to prove boundedness on (0,s). You then look at g(x)=f(x)-f(s)x/s. This is additive and satisfies g(s)=0, and together these imply g is periodic with period s. Since it's also bounded on (0,s), this means it's bounded on the entire real line, and finally this implies that g is identically zero, which is what we want: if g(y) is nonzero then linearity on yQ leads to a contradiction with boundedness (or periodicity too, I suppose, which is perhaps a bit quicker).<br /><br />All the best,<br /><br />ChrisAnonymousnoreply@blogger.com