Recent content by Undecided Guy

For completeness, I thought I'd go ahead and update this with the solution. First I'll need the following fact: If \mu is some positive measure, g \in L^1(\mu) , and for every measurable set E, 0 \le \int_E g(x) d\mu(x) , then g \ge 0 almost everywhere (wrt \mu ). Proof...

Oh dear, sorry, what I meant is that I have a \psi \in L^1(|\lambda|) so that d\lambda = \psi d|\lambda|. Sorry for the mistake. Anyway, what I've ended up doing so far is showing one of the two ineqaulities. Fix a partition \{X_j\}_{j=1}^{\infty} of \mathbb{R}^d. Then...

Hey, I know this is commonly a homework question, but it came up in my own studies; so this isn't a homework question for me. I hope it's alright that I put it here. I'm trying to show that if f dx = d\lambda for some f \in L^1(\mathbb{R}^d) and complex Borel measure \lambda then |f| dx...

Suppose a point x in P is isolated. Then there's an \epsilon > 0 so that B(x; \epsilon) contains no other point of P. Since x is in P, this ball contains uncountably many points of E. Note that we may write B(x; \epsilon) = \bigcup_{j \in J} B(x_j; r_j) for each r_j < \epsilon and x_j...

The way you've worded the statement, it's not possible. Suppose that v \in L^1[0, T] satisfies 0 < (g*v)(0) < v(0). Let v'(t) = v(t) for all t other than 0 and v'(0) = .5(g*v)(0). Then v = v' in the sense of L1, but (g*v)(0) > v'(0).

Specifically, I was thinking something like this: the statement is obvious for simple functions. Let f be some nonnegative measurable function on \mathbb{R}. Construct a sequence of simple functions \{s_n\}_{n=1}^{\infty} so that s_n(x) \uparrow f(x) as n \to \infty . Then we also have...

There are two answers here. I'll give the more pertinent one to academics first. The thing that bothers me in math is the lack of practicality. When I took calculus, I was incredibly bothered by the lack of justifications. My professors would assure me that things were true, but I had no...

Stat: Using my notation above, we can reword 1 and 2 as follows. 1. Let f be a function from X to Y. Then F^-1(Y -B) = X - F^-1(B). 2. Let f be a function from X to y. Then F(X - A) = Y -F(A) if and only if f is one to one and onto. If f is 1-1 and onto, there is a well-defined function...

I found it easier. Calculus professors have a tendency to wave their hands at justifications that are at the heart of most conceptual misunderstandings. If you do not understand a single line in a proof, you should not move on until you understand it completely. To do otherwise is cheating.

f^-1 is not defined on a single element. It's not a function on Y at all, actually. This is not immaterial. It is at the heart of the matter.

The statement can be easily shown via the MCT.

That does make things seem a little difficult to impossible. I can't say why any graduate school should take me. My experience is entirely in pure mathematics. I have complete confidence in my ability to do the work, but I think a lot of people can boast that. And a graduate program wouldn't...

I feel like you're confusing yourself with an abuse of notation. Let f:X \to Y be any function. Define F:P(X) \to P(Y) by F(A) = f(A) . Note that while f(A) is merely a slight notational abuse to describe a set, F is actually a function defined on the subsets of X. F is associated to f...

Is it more theoretical? I'll admit, the mathematical sloppiness in Griffith's book has caused me quite a bit of stress (much to the humor of my physics friends). Also, am I correct then in thinking that the sort of math I've taken is not directly applicable to physics, possibly outside of...

Hi, I'm new. I just graduated with a BS in mathematics (pure bent) from a state university in the US. I ended up taking 7 graduate courses - two semesters of measure theory, a semester of functional analysis, two semesters of complex analysis, a semester of general topology and a semester of...