Nik Weaver
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 3h comment Extremal Lipschitz convex functions Seems like every extreme function should be of the form $\tilde{f}_L$ for some $L$, though. Given any extreme function $g$, let $L$ be the set where it achieves its minimum value and compare $g$ to $\tilde{f}_L$. WLOG $0 \in L$ to avoid having to normalize. We automatically have $g \leq \tilde{f}_L$ and it seems like any region where the inequality is strict could be used to falsify $g$ being extreme. 3h comment Extremal Lipschitz convex functions On the other hand, the $d = 1$ case already shows that $\tilde{f}_L$ might not be extreme when $L$ is an interval that disconnects $B_1$. 3h comment Extremal Lipschitz convex functions @AdamSmith: Hmm, you are right. If $L$ is any closed convex subset of the interior of $B_d$ then $\tilde{f}_L$ should be extreme. If $\tilde{f}_L = .5(g + h)$ then $\nabla \tilde{f}_L = .5(\nabla g + \nabla h)$, so that $\nabla g = \nabla h = \nabla f$ outside $L$. I think this implies that $f$, $g$, and $h$ differ by an additive constant outside $L$, and then inside $L$ convexity forces equality. 6h comment An unconditional convergent series in $\ell_2$? Looks like a homework problem. 20h answered Extremal Lipschitz convex functions 20h answered Modified interlacing of eigenvalues 1d comment Extremal Lipschitz convex functions E.g. $|x| = .5(1.5|x| + .5|x|)$, so the absolute value function still is not extreme. Do you mean to require Lipschitz constant at most 1, as Robert suggests? 1d comment Extremal Lipschitz convex functions $F_d$ has no extremal points --- every Lipschitz convex function $f$ can be written as $f = .5((f + 1) + (f-1))$. Unless you mean something else by "extremal". Can you make the question more precise? 2d comment Modified interlacing of eigenvalues Okay, I misread the problem. I don't know the answer, but did you notice that you could put $B = 2v$ without changing nonzero eigenvalues? So the question is about interlacing between $\left[\matrix{A& v\cr v^T&0}\right]$ and $\left[\matrix{A& 2v\cr 2v^T& 0}\right]$. 2d comment Definitions of Hilbert Bundles I've never heard of this and I'm not completely sure what you have in mind, but it could be interesting to look into. 2d comment Adapting arguments and plagiarism " In many mathematical realms, the actual achievement in research is that certain issues and ideas become easy to understand." --- I really like this insight. 2d revised Definitions of Hilbert Bundles added 138 characters in body 2d answered Definitions of Hilbert Bundles Apr 26 comment Modified interlacing of eigenvalues $P$ is any order $n$ principal submatrix of $D$ $\ldots$ so if $n \leq 4$, could $P$ be the zero matrix? Apr 23 awarded Yearling Apr 21 comment A problem from Sakai's book on derivations on C(K) and differential structure on K What does Sakai mean by "one-dimensional differential structure"? One gets a very well-behaved construction by considering weak*-continuous derivations from ${\rm Lip}(M)$ into $L^\infty(M)$ when $M$ is a metric measure space, and this covers a lot of non-classical objects (sub-Riemannian manifolds, various fractals, etc.). That was done in this paper of mine. Apr 18 comment $l^1$ versus $l^2$ @DenisSerre: but do you need an extra step to get from $L^1(d\mu_\infty)$ to $l^1$? Sorry to be obtuse. Apr 18 comment $l^1$ versus $l^2$ @GeraldEdgar: yeah, I guess I should have said either "isomorphism" or "linear homeomorphism". In the context of Banach spaces I think "isomorphism" is understood to include homeomorphism. Apr 18 comment $l^1$ versus $l^2$ Thank you, Bill, and everyone else for setting me straight. Apr 18 comment $l^1$ versus $l^2$ Thank you, I guess I was confused. I'm accepting Fedor's answer because it was first.