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Nov
14 |
answered | Convergence of an oscillatory integral |
Nov
13 |
comment |
A special function solution of a fourth-order ODE
Thanks to Johannes Trost and Carlo Beenakker. |
Nov
12 |
asked | A special function solution of a fourth-order ODE |
Oct
22 |
asked | Weak continuity of the Hilbert transform |
Oct
14 |
comment |
Fundamental solution of Discrete Laplace in the plane
Try the discrete dbar and its adjoint: both should have homogeneity $-1$ and their convolution should provide the fundamental solution of the Laplace operator. |
Oct
14 |
revised |
Composition algebra of Gevrey function for $s<1$
added 17 characters in body |
Oct
14 |
revised |
Composition algebra of Gevrey function for $s<1$
added 1436 characters in body |
Oct
14 |
comment |
Composition algebra of Gevrey function for $s<1$
@CPJ Thanks for your comment. In fact with $M_n=n^{ns}$, we get $(M_n/n^n)^{1/n}=n^{s-1}$ which is increasing for $s\ge 1$. This suggests that the composition algebra property holds for $s\ge 1$ but not for $s<1$. |
Oct
13 |
comment |
Composition algebra of Gevrey function for $s<1$
@Piero D'Ancona You mean $s'<s$ since the Gevrey space $G^{(s)}$ with the notation above is increasing with $s$, e.g. analytic is $G^{(1)}$ is included in $G^{(2)}$ which contains compactly supported functions. On the other hand, I believe that the answer to the question is positive and is a matter of writing a precise Faà de Bruno formula. |
Oct
11 |
revised |
Composition algebra of Gevrey function for $s<1$
deleted 98 characters in body |
Oct
9 |
revised |
Composition algebra of Gevrey function for $s<1$
added 98 characters in body |
Oct
9 |
asked | Composition algebra of Gevrey function for $s<1$ |
Sep
21 |
comment |
Schwartz kernel
There was a typo (now erased) with an unwanted ' in the last sentence. |
Sep
21 |
revised |
Schwartz kernel
deleted 1 character in body |
Sep
20 |
answered | Schwartz kernel |
Sep
11 |
asked | The Schwartz space is not normable |
Aug
28 |
comment |
Trivial zeroes of the Riemann Zeta function are simple
@Igor Rivin It is indeed possible to define the Riemann Zeta function on the whole complex plane as a meromorphic function with a single simple pole at 1, using a variation of Euler-MacLaurin formula, much simpler to handle than the functional equation. |
Aug
28 |
asked | Trivial zeroes of the Riemann Zeta function are simple |
Jun
26 |
revised |
Logarithmic integral, $π(x)$ and $x/(\ln x)$
added 129 characters in body; edited title |
Jun
26 |
comment |
Logarithmic integral, $π(x)$ and $x/(\ln x)$
Thanks. I have added a related question. |