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visits | member for | 2 years, 5 months |
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stats | profile views | 2,337 |
Aug 5 |
awarded | Popular Question |
Jul 11 |
comment |
Measures, orthogonal to holomorphic functions
Assume $d=1$. Formally we have $ \mathcal H(G)=\ker \overline \partial, $ so that $$ \bigl(\mathcal H(G)\bigr)^\perp=\bigl(\ker \overline \partial\bigr)^\perp=\overline{\text{ran }\partial }. $$ As a result the orthogonal of $\mathcal H(G)$ is the set of Radon measures $\mu$ of the form $$ \mu=\frac{\partial \nu}{\partial z}, $$ where $\nu$ is an hyperfunction in $U$ such that $\frac{\partial \nu}{\partial z}$ is a Radon measure. |
Jul 2 |
awarded | Curious |
May 29 |
awarded | Revival |
May 23 |
comment |
is 1/max(i,j) a bounded matrix on hilbert spaces?
I do not see the connection with D. Serre's result, which is comparing the numerical radius to the norm. The (discrete) Hardy operator is hard stuff to handle, in particular not trace class. To get its $\ell^2$ boundedness, I used the Hilbert transform, and to get the boundedness of the latter, Fourier transform, which is bounded as a sign function. |
May 23 |
comment |
is 1/max(i,j) a bounded matrix on hilbert spaces?
Are you saying that you can prove the $L^2$ boundedness of the discrete Hilbert transform (matrix $(i-j)^-1)$) or of the Hardy operator (matrix $(i+j)^-1)$) that way ? |
May 21 |
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is 1/max(i,j) a bounded matrix on hilbert spaces?
@fedja I should have said that I did not understand your reference to a Schur test with weight, while the raised question was without weight. |
May 21 |
comment |
is 1/max(i,j) a bounded matrix on hilbert spaces?
@fedja The Schur test does not work: you would have to check $$\sup_{i\ge 1}\sum_{j\ge 1}\frac{1}{i+j}$$ which are all infinite. I hope that the explanations below could clarify the situation and qualify the question for MO. |
May 21 |
answered | is 1/max(i,j) a bounded matrix on hilbert spaces? |
May 21 |
comment |
Improper integral $\int^\infty_0 e^{-a x^2} \cosh (b\sqrt{1+x^2})$
@user113103 The indicatrix function of $\mathbb R_+$ is the standard Heaviside function, which is 1 on the positive half-line and 0 on the negative half-line. $J_k(a)$ appears essentially as the $k$-th derivative of a product: just apply Leibniz formula. |
May 20 |
answered | Improper integral $\int^\infty_0 e^{-a x^2} \cosh (b\sqrt{1+x^2})$ |
Apr 1 |
comment |
Square root of a complex matrix commuting with a given one
Yes, but just a remark here: this holomorphic method provides an explicit formula that can be numerically calculated and approximated. The algebraic proofs above require the knowledge of unknown quantities, such as the minimal polynomial, or the Jordan form. Although perfect theoretically, an algebraic method for this problem will require a very large time to provide a simple approximation of a commuting square-root: the numerical cost of the determination of the minimal polynomial or of the Jordan form is huge, compared to the simple algorithm to approximate a simple integral. |
Apr 1 |
answered | Square root of a complex matrix commuting with a given one |
Mar 18 |
asked | Embeddings of Sobolev spaces |
Mar 17 |
asked | Inequality for Laguerre polynomials |
Mar 13 |
answered | Multivariate Hermite Polynomials |
Mar 5 |
answered | Smooth but non-analytic kernel functions |
Mar 5 |
answered | Lower bounds for norms of commutators |
Mar 5 |
awarded | Yearling |
Mar 1 |
answered | Weak convergence in the space of Lipschitz Functions |