Skip to main content
MathOverflow
epsilon's user avatar
epsilon's user avatar
epsilon's user avatar
epsilon
  • Member for 5 years, 3 months
  • Last seen more than 2 years ago
Profile Activity

15 Actions

All Accepts Posts Badges Comments Revisions Reviews Suggestions
comment
What is lost after RKHS embedding of the L1 space?
any thoughts or comments?
asked
What is lost after RKHS embedding of the L1 space?
awarded
 Supporter
awarded
 Scholar
accepted
How does convergence in $\ell^2$ norm imply convergence in $\ell^\infty$-norm with Lipschitz conditions?
comment
How does convergence in $\ell^2$ norm imply convergence in $\ell^\infty$-norm with Lipschitz conditions?
@losif Pinelis: hh you are right. But at last we still need an order of $r_c$ in $h_{r_c}$ right?
comment
How does convergence in $\ell^2$ norm imply convergence in $\ell^\infty$-norm with Lipschitz conditions?
Thanks so much for the clear answer. I am thinking does it still require a lower bound like $g_r(x)>c'\cdot r^d$ for some constant $c'>0$? Otherwise we can not make sure its minimum over $S$ is greater than zero.
revised
How does convergence in $\ell^2$ norm imply convergence in $\ell^\infty$-norm with Lipschitz conditions?
added 21 characters in body
revised
How does convergence in $\ell^2$ norm imply convergence in $\ell^\infty$-norm with Lipschitz conditions?
added 21 characters in body
comment
How does convergence in $\ell^2$ norm imply convergence in $\ell^\infty$-norm with Lipschitz conditions?
@JochenGlueck Yes you are right. Let me edit the question to make it clear.
awarded
 Student
awarded
 Editor
awarded
 Informed
revised
How does convergence in $\ell^2$ norm imply convergence in $\ell^\infty$-norm with Lipschitz conditions?
added 5 characters in body
asked
How does convergence in $\ell^2$ norm imply convergence in $\ell^\infty$-norm with Lipschitz conditions?