Here is the statement I'm trying to prove:
If $\Lambda(f)\in C([0,1])$ (space of bounded continuous functions on
$[0,1]$) satisfies $\Lambda(f)\leq C\sup_{x\in[0,1]}|f(x)|$, and
$\mu=\mu^+-\mu^-$ is a signed measure with
$\mu^+([0,1])+\mu^-([0,1])\leq C$, prove that $\Lambda(f)$ has the
represen...
The first problem is just proving that a bounded linear functional on C[0,1] can be decomposed into the difference of two nonnegative linear functionals, which I can handle. The problem is from a problem solving session in real analysis.
Ah wait, I think martin argerami reads the problems statement as saying "for all linear functionals and measures, this works", which of course isn't true
@Mike that would have been a helpful lemma for me to know though
I'm not sure. Maybe you are right. I have show that Lambda^+ can be represented uniquely by mu^+, and Lambda^- can be represented uniquely by mu^-, then I can apply decomposition to get the desired the result. The problem is that I cannot represent the negative part with the negative variation of the measure
I'm thinking about zeros of products of analytic functions and their order and I am wondering if the following is true. If $f(z)$ and $g(z)$ are analytic, and $z_0$ is a zero of $f(z)$ but not $g(z)$, can I say $f(z)$ is the only factor which "contributes" to the order of the zero $z_0$ of $f(z)g(z)$? More specifically, if $z_0$ is a zero of order $n$ for $f(z)$, will it also be of order $n$ for $f(z)g(z)$?
I also had a prof whose wife died about a year before he did, and he spent that entire year writing poetry and music for his late wife until he himself passed
If $k$is a field, $X(k)=Hom(G_k, \Bbb Q/\Bbb Z)$, and $Br(k)$ is the Brauer group of $k$. What does "The application $X(k)\times k^* \to Br(k)$ is bilinear" mean ? Does it just mean it's a group morphism ?
**2.37 Theorem:** If $E$ is an infinite subset of a compact set $K$, then $E$ has a limit point in K.
**Proof:** If no point of $K$ were a limit point of $E$, then each $q\in K$ would have a neighborhood $V_q$ which contains at most one point of $E$. It is clear that no finite subcollection of ${V_q}$ can cover $E$; and the same is true of $K$, since $E\subset K$. This contradicts the compactness of $K$.
have some difficulty understanding topology
Why If no point of $K$ were a limit point of $E$, then each $q\in K$ would have a neighborhood $V_q$ which contains at most one point of $E$?
Why at most one point? There is no concrete proof or theorem stating this but it becomes intutive when you draw ball but not every set is ball.
Why is it clear that no finite subcollection of ${V_q}$ can cover $E$?
I mean you can have neighborhood become as big as you want and cover E.
I'm thinking about zeros of products of analytic functions and their order and I am wondering if the following is true. If $f(z)$ and $g(z)$ are analytic, and $z_0$ is a zero of $f(z)$ but not $g(z)$, can I say $f(z)$ is the only factor which "contributes" to the order of the zero $z_0$ of $f(z)g(z)$? More specifically, if $z_0$ is a zero of order $n$ for $f(z)$, will it also be of order $n$ for $f(z)g(z)$?
@user193319 yes, more generally $\operatorname{ord}(fg,a)=\operatorname{ord}(f,a)+\operatorname{ord}(g,a)$ for non-essential isolated singularities $a$
enriched over Diff, i.e. the action on Hom-sets $\operatorname{Hom}(V,W)\rightarrow\operatorname{Hom}(FV,FW)$ is smooth (where the Hom-sets carry their unique differentiable structure induced by the vector space structure)
Such a thing should exist, but I have a feeling it must be ugly. The action restricts to a group morphism on GL when V=W so I believe that has to be non-measurable to not be continuous or something?
oh wait, I asked for not smooth, not discontinuous
Alright potentially dumb question: given a complex rep $V$ of $G$, and $\varphi \in \operatorname{End}_G(V)$ (endomorphisms of V commuting with the G-action) such that $\varphi \notin \Bbb C$, we have $\Bbb C(\varphi)$ is a transcendental extension contained in $\operatorname{End}_G(V)$. Whhhhy is $\lbrace (\varphi - a)^{-1} : a \in \Bbb C\rbrace$ a linearly independent subset of $\Bbb C(\varphi)$?
Yup, that'd be excellent. Yesterday my friend (the aforementioned noise <<< throat friend lol) gave a talk explaining the Gaussian isoperimetric inequality
Is there a single connected topological space $X$ such that maps $X\to Y$ are enough to test whether $Y$ is totally disconnected for some large classes of $Y$?
@Alessandro Just to recall, the "Gaussian measure on $\Bbb R^d$ centered at $0$" is the measure $\gamma_d$ such that the Radon-Nikodym derivative is $d\gamma_d/d\lambda_d = e^{-\|x\|^2/2}/(2\pi)^{d/2}$ where $\lambda_d$ is the Lebesgue measure
The Gaussian isoperimetric inequality states that amongst all Borel sets of the same Gaussian measure, the one which minimizes boundary measure is a "half plane", i.e., one side of a hyperplane in $\Bbb R^d$
I can send you my cover letter if you give me an email address, I don't mind sharing it, it's mostly "standard fluff nobody cares about - actual matter of fact description of the math I like - some more fluff but more specific for the place I'm applying to"
The position I was applying to was somewhat particular though, my advisor was promoted from postdoc to junior professor, so there was a position attached to her position as junior professor guaranteed for someone to work with her
Also note that the application asked for a summary and an excerpt of the thesis separately from the cover letter, so I barely mention my thesis in the cover letter, but I talked about it a lot in the application in other places
Hi all, brief question, provided I have an analytical expression for some function $g(t)$, can I obtain an expression for $f(t)$ if I know that $g(t) = \int^T_0 f(t) dt$?
@CharlieShuffler If $F$ is an antiderivative of $f$, write $\int_{0}^{T}f(t)\text{ d}t = F(T) - F(0)$. Then take derivatives and use the fact that $F^{\prime} = f$.
Given a complex vector space $V$, does the "evaluation pairing" $V^* \times V \to \Bbb C$ literally mean.. like.. evaluate a linear form at an element of $V$
"We now give a quick, but essentially complete, account of some invariant mea- sures on a locally profinite group and associated homogeneous spaces. Locally profinite groups are such that their measure theory is effectively algebraic in nature, and can be treated as an episode in their representation theory."
Looooord I thank thee
I sometimes wonder how far I could actually get without ever doing any analysis
Can someone explain the last part of Wikipedia's argument that nimber addition of finite ordinals is equivalent to their XOR addition?
Anonymous
I don't understand what this means: "thus γ is included as (β ⊕ γ) ⊕ β or as α ⊕ (α ⊕ γ), and hence α ⊕ β is the minimum excluded ordinal"
I am trying to determine the zeros and their order of the function $\frac{\cos z - 1}{z^2}$. So far I determined that it has zeros $2 \pi k$ for $k \in \Bbb{Z} \setminus \{0\}$, and they are of order $2$. Does that sound right?
Now I am wondering whether it has a zero at infinity. Call the function $h(z)$. Then $h(z)$ has a zero at infinity provided $h(\frac{1}{w})$ has a zero at $w=0$. Well, $h(1/w) = 0$ if and only if $w = 0$ or $w = \frac{1}{2 \pi k}$. The second types of zeros don't count because they aren't zero themselves. But $w=0$ doesn't count either, because the whole function isn't defined at $w = 0$?
If $M$ is a manifold and $p\in M$ and we want to construct category where objects are $C_{p}^{\infty}(U)$ for $U\subseteq M$, what should the morphisms be?
hi chat! in the frechet metric in $(\mathbb{R}^\omega,d)$ if we change the $2^{-n}$ coefficient with any other convergent sequence will we get the induced product topology?
So with the Frechet metric a sequence $(x^{(n)})_{n\in\mathbb{N}}$ converges to $x^{(0)}$ iff $x_j^{(0)}=\lim_{n\to\infty}x_j^{(n)}$, hence the frechet metric induces the product topology
i encountered some expressions like Frechet's metric $\sum_{n=1}^\infty 2^{-n}\frac{|x_n-y_n|}{1+|x_n-y_n|}$ where the only difference is that there another term $\mu(n)$ instead of $2^{-n}$ so i'm wondering if with them we still get the induced product topology
Mine is chess , puzzle solving . Any one else watch movies as a hobby ? I did not used to but recently i watched few movies and i thought my minds grasping power reduced
Music, tennis, basketball, hanging out with friends, coffee, reading math articles, walking, sitting outside on a nice day drinking coffee listening to music on my computer, chatting in the math chat, talking
@Thorgott If p is not in the closure of f(M), pick a chart around p contained in the complement of f(M). On that chart (assume it's a 1-ball, with phi(p) = 0) construct the map to the sphere sending the 1/2 ball to the bottom hemisphere and the outer annulus to the upper OR LOWER hemisphere, complement of the chart getting sent to the north or south pole.
