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7h
comment Is Wiener amalgam spaces $W^{2,1}(\mathbb R)\subset C_0(\mathbb R)$?
I'm still confused. For the Wikipedia definition to make sense, we should have $f : \mathbb{R} \to \mathcal{F} L^1$. But you have $f : \mathbb{R} \to \mathbb{R}$ it seems. (And if you mean that $f(x)$ is the constant function 1 when $x$ is rational, note that $1$ is not in $\mathcal{F} L^1$ since its Fourier transform is not a function.) Anyway, it seems pretty clear from the definition that it's not affected by null sets, so any appropriate function that vanishes off the rationals would trivially be in $W(X, L^p)$.
8h
comment Is Wiener amalgam spaces $W^{2,1}(\mathbb R)\subset C_0(\mathbb R)$?
Um, if you're using Lebesgue measure, $f$ is clearly not in $L^2(\mathbb{R})$. Also, if I'm reading the Wikipedia page correctly, the space $W(X, L^2(\mathbb{R}))$ should be a space of $X$-valued functions on $\mathbb{R}$, not $\mathbb{R}$-valued.
May
21
awarded  Nice Answer
May
18
comment When Banach indicatrix is measurable?
"Measure class preserving" means "maps null sets to null sets"?
May
18
comment Contraction semigroup
Did you mean to write $e^{-tA}$?
May
18
comment When Banach indicatrix is measurable?
You may find it helpful to know that if $X$ is a Borel set and $f$ is Borel measurable, then for any Borel set $A \subset X$, the set $f(A)$ is analytic and hence universally measurable, so the hypothesis of the theorem from Federer is satisfied. (Presumably part of the conclusion of that theorem is the statement that $N(y,f)$ is measurable, otherwise the integral makes no sense.)
May
18
comment When Banach indicatrix is measurable?
In light of Gerald Edgar's counterexample, and especially for non-readers of Russian, could you elaborate on the hypotheses which are needed for Rokhlin's theorem?
May
18
comment $L_{\infty}$-norm of a $\delta(t)$-“function”?
Both these notions are defined for functions (or equivalence classes of functions) on the measure space which is the domain. The Dirac delta is not a function in this sense, so the question does not make sense. This is not a research level question and should be asked on math.stackexchange.com instead of here.
May
12
answered Integral kernels of self-adjoint operators
May
10
comment Is $B(t-1)$ an Ito process?
If it were an Ito process, wouldn't we have to have $b_s = \sigma_s = 0$ for $0 \le s < 1$? Then it ought to be possible to show that $Y$ is $\sigma(B_t - B_1 : t \ge 1)$-measurable, hence independent of $\sigma(B_t : t \le 1\}$. That is absurd since $Y_{3/2} = B_{1/2}$.
May
8
comment Asymptotics of “heat” semigroup
If $f \in L^2$ then $f$ is not well defined pointwise, so asking for pointwise convergence doesn't make sense.
May
8
comment Asymptotics of “heat” semigroup
What are you assuming about $f$?
May
8
revised Principal bundles and Subriemannian Geometry
fix formatting
May
7
answered Principal bundles and Subriemannian Geometry
May
7
answered probability in galton watson processes
May
5
comment Analogue of Cayley Hamilton theorem for operators on Hilbert space
This does of course show that if $T$ has finite spectrum $\sigma(T) = \{\lambda_1, \dots, \lambda_n\}$ then taking $f(\lambda) = \prod_{i=1}^n (\lambda - \lambda_i)$ will give $f(T) = 0$. So this is sort of a trivial analogue of Cayley-Hamilton.
May
4
comment Heisenberg group: function without vertical derivative
@TaQ: Won't any diffeomorphism $g$ preserve the Lie bracket? So $g$ can't pull back $\partial_x, \partial_y$ to $X,Y$ since the former commute and the latter do not. I think this is true even if $g$ is only $C^1$.
May
1
comment Poincare inequality for the measure of Brownian path
In this context I think it's more usual to consider the Malliavin derivative (and its Cameron-Martin norm) instead of Frechét. The Poincaré inequality does hold in that case (of course you have to restrict to $f$ with $\int f\,d\mu = 0$). You can find this, and an involved discussion of the many notions of differentiability (which I have not taken the time to unwind) in Chapter 5 of V. Bogachev's book Gaussian Measures.
Apr
30
comment convergence of integral for each bounded function in probability
Is it even true for real-valued random variables $X_n$ that if $X_n \to X$ i.p. then you can find an a.s. convergent subsequences which consists of almost all members in your sense? Is it true for the standard "typewriter sequence" counterexample?
Apr
30
comment convergence of integral for each bounded function in probability
I am not really sure what you mean by "consisting of almost all members", can you be more explicit? You can certainly say that there is a subsequence $(\mu_{n_k})$ which converges weakly almost surely; this is a standard fact for real-valued random variables and the proof works for random variables taking values in any metrizable topological space, such as the weak topology on a bounded set of measures on a Polish space.