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No. Consider the one-dimensional situation: $X = \mathbb{R}, Y = \{0\}$ (the notion of conormal distribution is local, so the fact that $X$ is not closed doesn't really matter). Then any distribution …

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 …

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) / …

The answer is no, and a simple counterexample can be obtained by taking, say, $E = \{1,\dots,n\}$, where $n \ge 3$, and $K(x,y) = C (\delta_{xy} - \frac{1}{n})$, where $C > n$. In fact, what we will s …

For an infinite-dimensional Gaussian random vector $X$ consider the Ornstein-Uhlenbeck maximal operator: $M f(X) := \sup_{\rho \in [0,1]} \mathsf{E} [f(\rho X + (1-\rho^2)^{1/2} X^\prime) \mid X]$ ( …

It's not sequential because its closed subspace $M[0,1] = (C[0,1])^\ast$ is not sequential. Here is an example of a set $A \subset M[0,1]$ that is sequentially $\tau_v$-closed but not $\tau_v$-cl …

In the simpler dyadic setting the answer to your question is no. Consider independent Bernoulli ($\pm 1$) random variables, aka Rademacher functions, $X_1,X_2,\dots$. Take any function that is quadrat …

Let $\gamma$ be the standard product Gaussian measure in $\mathbb{R}^\infty$, and let $\mu$ be a finite variation measure, not necessarily positive, such that $\mu \ll \gamma$. Is the following true? …

Here is a counterexample: Take any vector $v \in \mathbb{R}^\infty \setminus \ell^1$. Then on the set $C := [0,1]^\infty + \{tv, 0 \le t \le 1\}$ there are infinitely many "Lebesgue measures", namely …

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 …

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 …

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 …

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 …

Here is a nice exposition due to Maurey: http://www.numdam.org/numdam-bin/fitem?id=SAF_1972-1973____A10_0 http://www.numdam.org/numdam-bin/fitem?id=SAF_1972-1973____A11_0

You haven't specified what role is played by the topology of $I$, so my default assumption will be that the kernel is continuous, in which case the answer to your question is negative. For a countere …