Let $(f_n)$ be Cauchy in $L^\infty(E,\mathcal A,\mu)$. Is it true that $$\lim_{n\to\infty}\sup_{k>n}\|f_n-f_k\|_\infty=0?$$This equality appears in a proof I'm reading. It looks like a limsup of some kind, but it's a bit mysterious with the $f_n$ in there.
I think it's true. By definition of Cauchy, we have for $n,m\geq N$ that $$\|f_n-f_m\|_\infty<\epsilon.$$Now if we fix some $n\geq N$ and then $$\|f_n-f_m\|_\infty\leq\sup_{m>n}\|f_n-f_m\|_\infty\leq\epsilon.$$But this is the definition of the nonnegative sequence $h_n=\sup_{m>n}\|f_n-f_m\|_\infty$ tending to $0$. Bingo.
I apologize for asking a question though there are pretty much questions on math.stackexchange with the same title, but the answers on them are still not clear for me.
I have this linear operator:
$$
Ax = (2x_1-x_2-x_3, x_1-2x_2+x_3, x_1+x_2-2x_3);
$$
And I need to find the basis of the kernel...
I'm very confused at the following question:
Find the basis for the image and a basis of the kernel for the following matrix:
$\begin{bmatrix} 7 & 0 & 7 \\ 2 & 3 & 8 \\ 9 & 0 & 9 \\ 5 & 6 & 17 \end{bmatrix}$
I just don't know how to do any of this. We find the image by doing the following...
@psie you might think about if the infinity norm is replaced by arbitrary metric $d$, is the same result true i.e. $\lim_n \sup_{k > n} d(f_n, f_k) = 0$
you will find out that this indeed has nothing to do with this particular setting, and holds for any Cauchy sequence in a metric space
Is it common practice to write $\tau=\langle \mathcal{B} \rangle$ for a given topological basis $\mathcal B$, to denote the topology $\tau$ being generated by the basis $\mathcal B$?
well maybe when you say that, I do usually encounter something like $\tau(\mathcal{A})$
similarly to how $\sigma(\mathcal{A})$ denotes the $\sigma$-algebra generated by $\mathcal{A}$
but nonetheless, I think brackets are still popular choice by many, maybe not in elementary textbooks, but its definitely something anyone could denote topology generated by a given family of sets and often seem to do, from my experience with this site
@BenSteffan you shouldn't take my word for it, I seem to sometimes hallucinate that things are true when they aren't necessarily so. Lately, this happened to me, where apparently "$x$ is smaller than $y$" always means $x < y$ in English, and cannot mean, as I thought is true, that $x\leq y$
in high school, I used to think the parametrization for a hyperbola that gives its coordinates the form of the hyperbolic trig functions is actually the angle of the line joining (0,0) and the point on the hyperbola. I found out a while later that it is actually the area of the (triangle - the hyperbola). It is one of the things that I never checked for using pen and paper, and just hallucinated out of thin air
Can someone help me understand if convergence in $L^\infty$ norm is equivalent to uniform convergence almost everywhere? I think I understand how convergence in $L^\infty$ implies uniform convergence almost everywhere, but the other direction confuses me.
Attempt Suppose $f_n\to g$ uniformly on $A^c$ with $\mu(A)=0$. Then for $t\in A^c$ and $n> N$, $|f_n(t)-g(t)|< \epsilon$. Now fix an $n>N$. Then $$|g(t)|=|g(t)-f_n(t)+f_n(t)|\leq |g(t)-f_n(t)|+|f_n(t)|< \epsilon + \| f_n(t)\|_\infty$$for almost every $t$. So $g$ is bounded and hence $g\in L^\infty$. Next, how do I establish that $\|f_n-g\|_\infty\to0$ as $n\to\infty$? The definition is $\|f\|_\infty=\inf\{C\in[0,\infty]:|f|\leq C,\mu\text{ a.e.}\}$.
Hmm, maybe it's not so hard. $\epsilon$ is an upper bound to $|f_n(t)-g(t)|$ and $\|f_n-g\|_\infty$ seems to be the smallest upper bound to $|f_n(t)-g(t)|$ (it's weird though how it's defined in terms of an infimum...), so $\|f_n-g\|_\infty\leq\epsilon$, i.e. $\|f_n-g\|_\infty\to0$.
We have to take a mandatory course on mechanics (, tho I am majoring in Mathematics). Can someone suggest some good books that I may follow to have a clear understanding of it? Side note: I am not at all good in physics. I feel horrible while studying physics coz I think the way physics books delves into concepts is highly unrigorous. (I know that I might be wrong, but that's just my opinion)
For a better glimpse into the matter, our syllabus consists of:
Co-planar forces. Astatic equilibrium. Friction. Equilibrium of a particle on a rough curve. Virtual work. Forces in three dimensions. General conditions of equilibrium. Centre of gravity for different bodies. Stable and unstable equilibrium.
Simple Harmonic Motion. Velocities and accelerations in Cartesian, polar, and intrinsic coordinates. Equations of motion referred to a set of rotating axes. Central forces. Stability of nearly circular orbits. Motion of a projectile in a resisting medium. Stability of nearly circular orbits. Motion under the inverse square law. Slightly disturbed orbits. Motion of artificial satellites. Motion of a particle in three dimensions. Motion on a smooth sphere, cone, and on any surface of revolution.
Degrees of freedom. Moments and products of inertia. Momental Ellipsoid. Principal axes. D'Alembert's Principle. Motion about a fixed axis. Compound pendulum. Motion of a rigid body in two dimensions under finite and impulsive forces. Conservation of momentum and energy.
@ThomasFinley I suggest Kleppner and Kolenkov's mechanics, perhaps till the chapter that deals with completely general rigid body motion. Another source I'd suggest is David Morin's book on mechanics- again, till the most general rigid body motion chapter. the D'alembert's stuff is not that important, but its there in Goldstein. For the most general rigid rotator (in CM), any of Landau or Goldstein have it. As long as youre not going into advanced classical mech, this should be fine.
all of the above are as mathematically rigorous as a physics text can get
If $G$ is a virtually simple group, does $G$ have only finitely many normal subgroups? Intuitively, it seems true, but I can't seem to write down a proof...
@leslietownes Yes, that is what virtually simple means. And in this case case, I think we can even assume that the simple subgroup of finite index is also normal by looking at the normal core of the subgroup.
I say that it seems intuitively true because virtually simple means that the group is very close to being simple, so I suspect it should have very few normal subgroups (maybe even finitely many)...But maybe I am wrong...maybe it is possible for a virtually simple group to have infinitely many normal subgroups...Maybe that's why I can't write down a proof.
Hmm...maybe this is false...But I can't think of a counterexample...I don't think direct products will work...Like, if $G$ is simple and $F$ is any finite group, then $G$ is a finite-index subgroup of $G \times F$...But I don't think $G \times F$ will have infinitely many normal subgroups...Maybe there is some other group construction that will give us a counterexample...Like, semi-direct products, (amalgamated) free groups, HNN-extensions...or maybe something more strange...hmm...
I have a basic topology question. Let X be the set of natural numbers {0, 1, 2, ...} and let Y be the set {0} ∪ {1, 1/2, 1/3, ...}, both with the subspace topology from the real line. Are they homotopy equivalent spaces?
I'm very confused at the following question:
Find the basis for the image and a basis of the kernel for the following matrix:
$\begin{bmatrix} 7 & 0 & 7 \\ 2 & 3 & 8 \\ 9 & 0 & 9 \\ 5 & 6 & 17 \end{bmatrix}$
I just don't know how to do any of this. We find the image by doing the following...
