991 reputation
212
bio website mathtm.blogspot.com
location University of California Los Angeles, CA
age
visits member for 1 year, 11 months
seen Apr 11 at 19:38

I am a recent graduate in applied mathematics at UCLA and currently trying to break into the quantitative investment/trading industry while also continuing to pursue advanced graduate-level mathematics as both a hobby and career necessity.


Sep
6
asked Counter-Example (or Proof) to $\int_{0}^{1}f_{n}\;dx\to0$ Implies $f_{n}\to0$ a.e. $x$ Whenever $f_{n}\geq0$.
Sep
6
comment Limit of Integral of Difference Quotients of Measurable/Bounded $f$ Being $0$ Implies $f$ is Constant
@Jonathan I incorporated your comments in the most recent edit.
Sep
6
revised Limit of Integral of Difference Quotients of Measurable/Bounded $f$ Being $0$ Implies $f$ is Constant
added 1133 characters in body
Sep
5
comment Limit of Integral of Difference Quotients of Measurable/Bounded $f$ Being $0$ Implies $f$ is Constant
Thank you for making me realize my naivety. Approximations to the identity much... (:
Sep
5
comment Limit of Integral of Difference Quotients of Measurable/Bounded $f$ Being $0$ Implies $f$ is Constant
Isn't this true though when $g\geq0$? How can you have $\int_{0}^{1}g(x)\;dx<M$ with $g(x)>M$ on a set of positive measure?
Sep
5
comment Limit of Integral of Difference Quotients of Measurable/Bounded $f$ Being $0$ Implies $f$ is Constant
math.ucla.edu/grad/handbook/hbquals.shtml -- See problem #1 from Analysis Spring 2013. I don't understand your point though; there are no singularities or otherwise obstructions to speak of that would prevent us from integrating along $[0,1]$...
Sep
5
asked Limit of Integral of Difference Quotients of Measurable/Bounded $f$ Being $0$ Implies $f$ is Constant
Jul
19
answered $f^{-1} $ is continuously differentiable.
Jul
13
comment (Qual Question) Example of a non-measurable function $a_{ij}:\mathbb{Z}\times\mathbb{Z}\to\mathbb{R}$
Well, my apologies then. By the way, I also mentioned this fact in paragraph #4. :P I do understand your point though, but I think there is still a fine line between saying the title says it all and having no body in the question, between what I have (and have in most of my questions, where I have said such things in the past).
Jul
12
comment (Qual Question) Example of a non-measurable function $a_{ij}:\mathbb{Z}\times\mathbb{Z}\to\mathbb{R}$
When somebody says that, they typically don't literally mean just read the title and proceed to answering the question. It's just a summing up of the current progress/assessment of the question which is described in the body of the post, to which I mentioned the appropriate measure on $\mathbb{Z}\times\mathbb{Z}$. In other words, my feelings at the time of posting the question (in terms of direction in obtaining a solution) were characterized by the question's tittle.
Jul
12
accepted (Qual Question) Example of a non-measurable function $a_{ij}:\mathbb{Z}\times\mathbb{Z}\to\mathbb{R}$
Jul
12
comment (Qual Question) Example of a non-measurable function $a_{ij}:\mathbb{Z}\times\mathbb{Z}\to\mathbb{R}$
But it's because I misunderstood what was meant by absolute convergence of the inner and outer sums (see Daniel's answer). I upvoted your question, but I feel compelled to accept his as the answer. But much thanks!
Jul
12
comment (Qual Question) Example of a non-measurable function $a_{ij}:\mathbb{Z}\times\mathbb{Z}\to\mathbb{R}$
Does that double sum converge absolutely?
Jul
12
asked (Qual Question) Example of a non-measurable function $a_{ij}:\mathbb{Z}\times\mathbb{Z}\to\mathbb{R}$
Jul
12
comment What is an example of a bounded, discontinuous linear operator between topological vector spaces?
As the above answer shows, it is easier to think of continuity in terms of open sets when considering topological (vector) spaces (at least for questions of this sort) than the usual $\epsilon-\delta$ definition when a norm is available (e.g. in Banach spaces). When considering two topological spaces $X$ and $Y$, the stronger the topology is on $X$, the more continuous maps there are from $X$ to $Y$, and conversely, the stronger the topology on $Y$, the fewer continuous maps there are.
Jul
7
awarded  Nice Question
Jul
3
awarded  Critic
Jun
19
comment Closed rectangle contained in set that does not have measure zero.
The answer is no, $\mathbb{R}-\mathbb{Q}$ being an example (as indicated in the answers). But there are some important instances where the answer is true that are helpful to keep in mind. For instance, the set on which a continuous function is positive (or negative) has positive measure, being that it contains a $k$-cell (rectangle).
May
21
revised Rudin Theorem 1.17
[Edit removed during grace period]
May
21
answered Rudin Theorem 1.17