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May
25
revised Compactness of an operator involving the resolvent of laplacian
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May
17
revised Problem about Ricci Flow
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Apr
25
answered Radii problem in a power series
Mar
22
comment show $\sum_{i=1}^Nx_i\bar{y_i}$ is defined
Square summability would be enough, which is weaker than absolute convergence.
Mar
21
awarded  Necromancer
Mar
17
awarded  Enlightened
Mar
17
awarded  Nice Answer
Feb
11
comment oblique derivative smoothness of harmonic functions
@Andrew: I overlooked the condition $f\in C(S)$, but I am pretty sure your statement is true. I take back the smoothness statement though. I am not sure what you mean by "which assume that solutions are from ..." but the Schauder estimates you mention imply additional regularity, given that the solution has some minimal regularity to start with.
Feb
11
answered oblique derivative smoothness of harmonic functions
Feb
1
revised Convergence of a crazy power series
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Feb
1
answered Convergence of a crazy power series
Feb
1
comment A power series problem, find ROC
$|\sin x|\leq|x|$ is a good bound when $x$ is small, but for large $x$ one should use $|\sin x|\leq1$.
Jan
31
answered Radius of convergence powers series s.t seriex $ \sum |a_n| $diverges
Jan
21
awarded  Nice Answer
Dec
16
awarded  Caucus
Dec
15
revised Method of characteristics for systems of PDE (vs. Lewy's example)
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Dec
15
answered Method of characteristics for systems of PDE (vs. Lewy's example)
Dec
14
comment Sobolev space is an algebra
@MathematicalPhysicist: Write $\|\hat u\|_{L^1}=\int<\xi>^{-s}<\xi>^{s}|\hat u(\xi)|\mathrm{d}\xi$, and apply Cauchy-Schwarz. The condition $2s>n$ gives integrability of $<\xi>^{-2s}$.
Dec
14
comment For which $s\in\mathbb R$, is $H^s(\mathbb T)$ a Banach algebra?
possible duplicate of Sobolev space is an algebra
Nov
28
comment Is this bootstrap argument correct?
@BeniBogosel: For $\phi$ an arbitrary smooth function on the boundary, we have $\int_{\partial\Omega}v_n\phi=0$. By weak convergence, $\int_{\partial\Omega}w\phi=0$. Since $\phi$ is arbitrary, $w=0$ on $\partial\Omega$.