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Consider $d=1$, $s=1$, $\Omega=(-1,1) \subset \mathbb{R}^1$. We can find a function $g \in L^2(-1,1)$ (or even continuous) for which $g(0)$ is arbitrarily large but $\|g\|_{L^2(-1,1)}$ is arbitrarily …

It certainly is measurable. In fact, you may find an explicit formula for it. If we take $I = (0,1)$, then $g_s$ is simply given by $g_s(t) = \operatorname{min}(s,t) - st$. How did I find this? We …

Take $d=1$. In this case all functions in $H^1(\mathbb{R})$ are (absolutely) continuous, so evaluation at a point is well defined . Let $X = \{f \in H^1 : f(0) = 0\}$ which is a well-defined closed …

By a cutoff function argument, it suffices to assume $f$ is compactly supported, so we can integrate by parts without picking up boundary terms. Thus $$\int (xf')^2 = \int (x^2f') f' = -\int 2xf'f - …

Let $\sigma(f)$ be the smallest $\sigma$-algebra on $[0,1]$ which makes $f$ measurable. I claim that $\overline{P_f} = L^2(I, \sigma(f), m)$, the space of all square-integrable $\sigma(f)$-measurable …

Yes, you've got it right. Given an unbounded self-adjoint operator $A$ with domain $D(A) \subset H$, using Zorn's lemma you can produce an everywhere defined operator $A'$ on $H$ which extends $A$. …

The error is in this line: The standard argument shows that $\widetilde{\mathcal{M}}$ is an RKHS of functions on $X$. In fact, this is not generally true. The completion $\widetilde{\mathcal{M} …

It seems, from comments, that the question is based on a misreading. The text does not assert that the inclusion map is always injective; it only gives a necessary and sufficient condition for the ma …

They are not equivalent. For an explicit counterexample, let $\{e_1, e_2, \dots\}$ be an orthonormal basis for $H$, and let $C$ be the diagonal operator $C e_n = \frac{1}{n^2} e_n$. Let $D =2C$. Th …

No. For a very simple example, take $\Omega = \{a,b,c\}$ consisting of three points, with $\mathcal{F} = 2^\Omega$ and $P(A) = |A|/3$ the uniform measure. Let $\mathcal{G} = \{\{a\}, \{b,c\}, \Omega …

Here's one solution. There may be cleaner ones. Let $E$ be as supposed. The natural inclusion $T : L^\infty([0,1]) \hookrightarrow L^2([0,1])$ is bounded, so $E = T^{-1}(E)$ is therefore also close …

(Rewritten to give an answer more useful to future visitors.) First of all, as noted in comments, there is no (countably additive) Gaussian measure on $H$ with covariance operator the identity. Ho …

Not an answer really, but a collection of several comments. The "skew-symmetric" condition is not really natural for an operator on a complex Hilbert space, since it isn't preserved by unitary trans …