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Banach spaces, function spaces, real functions, integral transforms, theory of distributions, measure theory.
9
votes
2
answers
307
views
Local-to-global inequalities for measures: Brunn-Minkowski, Ahlswede-Daykin, what else?
This question is motivated by an obvious formal analogy between two well-known inequalities:
Log-concavity and Brunn-Minkowski inequality
Let $\mu(dx) := m(x) dx$ be an absolutely continuous …
8
votes
2
answers
582
views
Does every operator from a Hilbert space to $L^0$ factor through a canonical one?
Let $A:H\to L^0(S, \mu)$ be a continuous operator from a Hilbert space to the space of (equivalence classes of) measurable functions on a probability measure space $S$ with convergence in measure. Let …
7
votes
Accepted
Is the ideal of functions vanishing at a set complementable in $C(X)$?
Not in general.
It's well-known in Banach space theory that the ideal $c_0$ in $\ell^\infty$ is not complemented (see e.g. Albiac & Kalton).
By the Gelfand representation, $\ell^\infty \simeq C(\bet …
7
votes
When $L^\infty$ is 1-injective
For a compact Hausdorff space $K$ the algebra $C(K)$ is $1$-injective iff $K$ is extremally disconnected iff its Boolean algebra of idempotents is complete. So to construct a counterexample we need an …
7
votes
1
answer
242
views
Is there a nice "minimum" of two symmetric operators?
Let $A$ and $B$ be two bounded symmetric positive operators in Hilbert space, such that $A-B$ is trace class. If needed, $A$ and $B$ may be assumed reasonably "small", let's say, Hilbert-Schmidt.
Doe …
7
votes
1
answer
652
views
Compactness of Sobolev embedding for domains of finite measure
Let $\Omega \subset \mathbb{R}^d$ be a domain of finite Lebesgue measure, not assumed to be smooth or bounded. Is it true that the embedding of, say, $W^{1,p}_0(\Omega)$ (Sobolev functions with zero b …
7
votes
1
answer
439
views
Is an infinite-dimensional "Lebesgue measure" uniquely determined by a set of positive finit...
Let $\mu$ be a probability measure on a subset $C \subset \mathbb{R}^\infty$ of the space of sequences, and assume, for simplicity, that $C$ is closed and convex.
We say that $\mu$ admits shifts if f …
6
votes
0
answers
242
views
Operator arithmetic-harmonic mean inequality with operator-valued weights
Let $\Lambda_1,\dots,\Lambda_n$ be strictly positive definite operators in the Euclidean space $\mathbb{R}^d$. By an operator arithmetic-harmonic mean inequality with weights $\Lambda_i$ I mean the fo …
6
votes
1
answer
290
views
Does the topological Varopoulos algebra consist of functions that are continuous and Varopou...
Let $X_1,\dots,X_n$ be compact Hausdorff spaces. Let's define the Varopoulos algebra as the projective tensor product: $$V(X_1,\dots,X_n) := C(X_1) \hat{\otimes} \dots \hat{\otimes} C(X_n),$$ i.e. the …
5
votes
0
answers
119
views
L^1 maximal inequalities for the Ornstein-Uhlenbeck semigroup in infinite dimension
For an infinite-dimensional Gaussian random vector $X$ consider the Ornstein-Uhlenbeck maximal operator:
$M f(X) := \sup_{\rho \in [0,1]} \mathsf{E} [f(\rho X + (1-\rho^2)^{1/2} X^\prime) \mid X]$
( …
5
votes
0
answers
161
views
$L^p$ estimates for Ornstein-Uhlenbeck: what is known beyond hypercontractivity?
Consider an infinite-dimensional Gaussian random vector $X$, and a positive random variable $f(X) \in L^p, p > 1$. Let $f(X) \sim \sum_n f_n(X)$ be its (formal) chaos expansion. Let $(U_\rho, \rho \in …
5
votes
Accepted
If Gaussian measures on a Hilbert space converge weakly to 0, how do their covariance operat...
Denote the Gaussian random vectors by $X(n)$.
Clearly, $\mathrm{tr} \, S(n) = \mathsf{E} \, \Vert X(n) \Vert^2$, so $\mathrm{tr} \, S(n) \to 0$ is certainly sufficient for weak convergence to $0$. An …
4
votes
Accepted
Convergence a.e and $L^1$ boundedness implies convergence in which sense?
There is convergence in some non-locally convex spaces, e.g. $L^p, 0 < p < 1$.
More generally, for any concave function $\Psi : \mathbb{R}_+ \to \mathbb{R}_+$, such that $\Psi(0) = 0$ and $\Psi(x) / …
4
votes
1
answer
155
views
For Hilbert spaces, does weak analyticity with respect to a dense subspace of functionals im...
Let $i : X \hookrightarrow Y$ be a dense embedding of complex Hilbert spaces.
Let $f : \mathbb{D} \to X$ be a function, such that $i \circ f$ is holomorphic ($\mathbb{D}$ is the open unit disk). I …
4
votes
Do semi-continuous functions generate bounded Borel measurable functions as a $C^*$-algebra?
Any upper or lower semicontinuous function is continuous almost everywhere in the sense of Baire category (since it is a pointwise limit of a sequence of continuous functions, at least when $\Omega$ i …