bio | website | |
---|---|---|
location | Rehovot, Israel | |
age | 24 | |
visits | member for | 2 years, 4 months |
seen | 17 hours ago | |
stats | profile views | 644 |
I'm a PhD student in Weizmann Institute, specializing in Probability.
Aug 12 |
comment |
Is the ideal of functions vanishing at a set complementable in $C(X)$?
@SergeiAkbarov: Actually, it is. $\mathbb{N} \subset \beta \mathbb{N}$ consists of isolated points, so it's open. And in any case, your ideal $I$ only depends on the closure of the set. |
Aug 12 |
answered | Is the ideal of functions vanishing at a set complementable in $C(X)$? |
Aug 7 |
awarded | Popular Question |
Aug 4 |
revised |
When does a stochastic process have its sample paths a.s. in the reproducing kernel hilbert space (RKHS) induced by its covariance function?
added 104 characters in body |
Aug 4 |
revised |
When does a stochastic process have its sample paths a.s. in the reproducing kernel hilbert space (RKHS) induced by its covariance function?
added 185 characters in body |
Aug 4 |
answered | When does a stochastic process have its sample paths a.s. in the reproducing kernel hilbert space (RKHS) induced by its covariance function? |
Jul 12 |
comment |
Equivalent Gaussian measures
Shouldn't it be "Hilbert-Schmidt" instead of "trace class"? Anyway, I believe that for most $v$ your $I - G^{-1/2} G^\prime G^{-1/2}$ is not even bounded. |
Jul 2 |
awarded | Curious |
Jun 18 |
awarded | Nice Answer |
Jun 17 |
comment |
Why differential forms are important?
Another thing you can integrate is a measure. So I wouldn't call differential forms "the" natural way to express "the" idea of integration. :) |
Jun 17 |
comment |
Why differential forms are important?
@LeeMosher: Right, but that requires a full-blown tensor calculus, not just differential forms. |
Jun 17 |
answered | Why differential forms are important? |
May 30 |
answered | Does every operator from a Hilbert space to $L^0$ factor through a canonical one? |
May 22 |
answered | The borel $\sigma-$algebra of the set of probability measures |
May 6 |
comment |
smooth Luzin theorem
Any such function must have an "$L^0$ derivative", in the sense that $\frac{1}{t}(f(\cdot + t) - f(\cdot))$ has a limit in measure as $t \to 0$. |
Apr 23 |
comment |
How to define the distributional Hessian for a convex function on a $C^0$ Riemannian manifold?
My comment would only apply if the manifold and the metric were smooth, in which case $f \mapsto \nabla^2 f$ would be a smooth differential operator. In your nonsmooth case I don't know how to make sense of this. |
Apr 23 |
comment |
How to define the distributional Hessian for a convex function on a $C^0$ Riemannian manifold?
Any differential operator with smooth coefficients acts on distributions, |
Apr 11 |
comment |
When does topological homogeneity imply algebraic homogeneity? Pseudo-arc and Hilbert cube
If I remember correctly, the pseudoarc has the property that it admits no interesting binary operations at all: namely, every continuous map $X \times X \to X$ factors through the projection onto one of the coordinates. |
Apr 8 |
awarded | Yearling |
Apr 3 |
comment |
Finiteness of “novel variance” from a kernel on a compact space
Then I don't have a guess what you are computing in the case of more than 2 vectors. |