Nate Eldredge
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 9h comment A computation using the Ito integral Your third equality, in which you interchange a Riemann and an Ito integral, should be carefully justified. 2d comment Proving that if $f\in\mathcal{F}C^{1}_{b}(X)$ then $f\in W^{1,p}(X,\gamma$) for $p>1$ If $A$ is any unbounded operator, then by very definition, the domain of the closure of $A$ contains the domain of $A$. Therefore $W^{1,p}(X, \gamma)$ most assuredly contains $\mathcal{F} C^1_b(X)$. I don't think I understand what your question is about. 2d comment Proving that if $f\in\mathcal{F}C^{1}_{b}(X)$ then $f\in W^{1,p}(X,\gamma$) for $p>1$ What's the precise definition of the space $W^{1,p}(X,\gamma)$ for you? For some authors, it's defined as the closure of an appropriate space of cylinder functions under an appropriate norm, so your question might be tautological. Also, does $C^1_b$ mean that the first partial derivatives of $\varphi$ are bounded, or only $\varphi$ itself? 2d comment Congruent measurable sets Can you give some background on what you already know about Lebesgue measure and integration? If you have to start from first principles, this will take quite a lot of work. 2d answered Weak convergence in $L^p$ equivalent to pointwise almost everywhere convergence 2d comment Is it possible to find a perfect cube like 111…11? I still don't see how you conclude $k^2+k+1\equiv0\pmod{10}\text{ or } k=10m+1$. It appears you are taking your equation $9(k-1)(k^2+k+1)=10(10^{n-1}-1)$ mod 10 to read $-(k-1)(k^2+k+1) \equiv 0 \pmod{10}$ and concluding that therefore one factor must be congruent to 0, mod 10. But that conclusion isn't valid in general: 10 is composite so the integers mod 10 are not a field; there are zero divisors. For instance, $5 \cdot 2 \equiv 0 \pmod{10}$ yet neither factor is congruent to 0. One would need another argument and I don't see what it is. Feb 4 comment Explanation of “weight function” of inner product in Hilbert space You don't need the weight function either in finite or infinite dimensions, but you can include it in either case. An inner product creates a geometry on a vector space, and in finite dimensions, you can see adding the weight function as "distorting" that geometry in a linear way (e.g. stretching in some directions more than others). Maybe the same analogy is helpful for infinite dimensions. Jan 31 comment Functions to represent set operations?! How would $f(a_i + a_j)$, which is a number, "represent" $A_i \cap A_j$, which is a set? What exactly do you mean by "represent"? Jan 30 awarded Revival Jan 30 answered What does “modulo equivalence relationship” mean? Jan 30 comment Question about weak convergence of random variables @Student: I added a counterexample for that. A natural question is whether one could find an example with continuous (or even smooth) densities. I thought about modifying my example but couldn't quite make it work. Jan 30 revised Question about weak convergence of random variables another example Jan 30 answered Question about weak convergence of random variables Jan 29 answered Random function of random variable Jan 29 comment Understanding the concept of measurability of random variables I don't understand what you are asking here. It doesn't seem to correspond to a precise mathematical statement that can be answered yes or no. Jan 29 comment Why is the drift of an Itō process considered to be a Riemann integral even when it's not even Riemann integrable? For your second question, sure, replace $X_t$ by $0$ on $N$, or anything else you like. Nobody cares what happens on a null set. All the theorems you want to prove say "a.s." everywhere anyway. Jan 29 comment Why is the drift of an Itō process considered to be a Riemann integral even when it's not even Riemann integrable? "But unless $b$ has (surely) continuous paths, it is not Riemann integrable": Why do you say that? There are lots of functions that are Riemann integrable which are not continuous. For instance, every cadlag function has at most countably many discontinuities and hence is Riemann integrable. Jan 29 revised Application of Doob's inequality edited tags Jan 26 comment Are the polynomial functions on $S^1$ dense in $C(S^1,ℂ)$? @AreaMan: Then Stone-Weierstrass says you do have a dense set. Jan 25 comment How to picture a first countable space? At a higher level, "picturing" a first countable space should in some sense be easy, because pretty much every space you can picture is first countable. Spaces that are not first countable tend to be weird, pathological spaces that are hard to visualize (at least from an analyst's perspective).