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awarded  Revival
Mar
8
comment Is the Hölder random constant of the Brownian Motion Integrable?
@MichaelHardy you can think of $K$ and $B_t$ as measurable functions on some measurable space $(\Omega,\mathcal{F})$, the Wiener space, for example. I think the point is if a r.v. $K$ fits the inequality $|B_t-B_s|\leq K|t-s|^{\gamma} \ \text{a.s.}$ for all $t,s\in [0,1]$, then the r.v. $K\cdot P$, where $P\geq 1$ is any positive r.v. also fits the above inequality. But a more interesting question is whether is possible to find a positive r.v. $0<P<1$ such that $K\cdot P$ fits the inequality and $\mathbb{E}[K\cdot P]<\infty$. Since $K>K\cdot P$ then $KP$ is a better random constant than $K$.
Mar
4
accepted Is the Hölder random constant of the Brownian Motion Integrable?
Mar
4
comment Is the Hölder random constant of the Brownian Motion Integrable?
Thanks a lot John!
Mar
4
comment Is the Hölder random constant of the Brownian Motion Integrable?
@MichaelHardy I would like to fix $\gamma$ and ask if the "best constant" $K$ we can chose for this fixed choice of $\gamma$ has finite first moment.
Mar
4
asked Is the Hölder random constant of the Brownian Motion Integrable?
Jan
13
answered Example of an uncountable dense set with measure zero
Jan
8
comment Relation between weak convergence of probability measures and weak-* convergence
You can find a nice discussion about this problem on the following PDF: math.ucsd.edu/~bdriver/240-01-02/Lecture_Notes/current_versions/… What you are looking for starts at p. 17. After understand how to decompose the Linear functional and use Riesz-Markov theorem give a look at Appendix in Functional Analysis Conway's book.
Jan
7
answered Confusion in not extending results in convergence in probability to convergence in distribution
Jan
4
comment $TS=0$ if and only if $T(v)=0$
I guess there is some mistake in the statement since $T\equiv 0$ satisfies $TS=0$.
Jan
2
answered Quotient map $\pi : X \rightarrow X / \mathrm{ker}(A)$ is open for a bounded linear operator $A$
Jan
1
comment Show that the square root of a non-negative operator is unique
Beautiful answer! In the last paragraph I guess that have a typo $p_n(T)$ should be $p_n(A)$.
Dec
31
comment On a step in Egorov's measure theorem
you are welcome.
Dec
31
comment Maximum value of an expression
In the numerator by replacing all $\lambda$'s by $\lambda_1$ we get that the maximum is less than or equal to $\lambda_1$. By taking $x=e_1$ we can see that it is attained.
Dec
31
comment On a step in Egorov's measure theorem
For a fixed $m$ the sequence $E^m\setminus E^m_k \downarrow \emptyset$, when $k\to\infty$.
Dec
31
asked Isolated Eigenvalues on the Extensions.
Nov
25
comment Show a differentiable function has value of zero
Is this not just simple application of Gronwall's inequality ?
Nov
6
answered Ising Model on 2k-regular graphs
Aug
4
awarded  Yearling
Apr
1
comment Finding functions from a series
I disagree, for instance, if $z=-1$ then this series converges to $1$.