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 Yearling
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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.