# Tag Info

98

This is how I used to imagine projections: If a mouse: gets run over by a steamroller: It will look like this: Now if it gets run over by a steamroller another time, it will still look like this:

23

A compact set must be bounded. Otherwise we can take $\{ x_n \}_{n=1}^\infty$ such that $\| x_n \| \geq n$. This will have no convergent subsequence, which we can prove by showing that it has no Cauchy subsequence. A compact set must be closed. Otherwise we can pick a sequence which converges to a point in the closure which is not in the set. This will ...

21

It certainly cannot be "only because the unit ball becomes non-compact", because there are important algebraic differences between finite and infinite dimension that don't even depend on having a norm available (so "unit ball" would be meaningless). One of the simplest differences is that if $V$ is finite-dimensional, then a linear transformation $V\to V$ ...

18

Preduals of $\ell_1$ are very interesting creatures. The question is slightly ill-posed but let me comment on that anyway. We have quite a lot of isometric preduals of $\ell_1$. Indeed, for any countably infinite ordinal number $\alpha$ (endowed with the order topology) we have $C_0(\alpha)^* \cong \ell_1$. This essentially follows from the ...

16

Let's call $f$ an $n$-SOP if we can write $$f(x,y) = \sum_{k = 1}^n g_k(x)\cdot h_k(y)$$ with continuous functions $g_k, h_k \colon [0,1] \to \mathbb{R}$. If $f$ is an $n$-SOP, for every family $x_1 < x_2 < \dotsc < x_r$ of $r > n$ points in $[0,1]$, the set $$\left\{ \begin{pmatrix} f(x_1,y) \\ f(x_2,y) \\ \vdots \\ f(x_r,y)\end{pmatrix} : y ... 16 Here's the way I like to think about it. I'll start with the finite dimensional space \Bbb{R}^n because it looks like that's where you are, but I'll give an analogy for infinite dimensional spaces as well. The quantity z^Tx represents a linear functional on \Bbb{R}^n, that is a linear function which eats a vector and spits out a real number:$$ ...

16

$S^3$ is compact, while $\mathbb{R}^3$ is not. Since any continuous function $f:S^3\rightarrow \mathbb{R}^3$ maps compact subsets of $S^3$ to compact subsets of $\mathbb{R}^3$, it can't be surjective (or else $f(S^3)=\mathbb{R}^3$ is also compact).

15

Let $V$ be an $n$-dimensional vector space over the field $\mathbb{F}$. Given a linear map $T : V \to V$, there is an induced linear map $\bigwedge^nT : \bigwedge^n V \to \bigwedge^n V$ given by $\left(\bigwedge^nT\right)(v_1\wedge\dots\wedge v_n) = (Tv_1)\wedge\dots\wedge(Tv_n)$. As $\bigwedge^nV$ is one-dimensional, $\bigwedge^nT = ... 13 Vector spaces are, by default, unnormed. A norm is extra structure we add to a vector space, to define a normed vector space. 13 A vector space is just a set in which you can add and multiply by elements of the base field. You can add polynomials together and multiply them by real numbers (in a way satisfying the axioms,) so polynomials form a vector space. A vector is nothing more or less than an element of a vector space, so polynomials can be seen as vectors. 12 Note that$C_c \subset C_0$, but$C_c \neq C_0$. For example,$f(x) = \dfrac{1}{x^2+1}$belongs to$C_0$but not$C_c$. What you seem to be assuming is that$\lim_{|x|\to\infty}f(x) = 0$implies that there is some$N > 0$with$f(x) = 0$for all$|x| > N$. This is not true, as the above example demonstrates. That is, a function can limit to zero at ... 12 Take your open cover of a ball of radius 5. Shrink it by a factor of 5. It's an open cover of the ball of radius 1. Therefore it has a finite subcover. Now reflate the subcover by a factor of 5. Et Voila! It covers the ball of radius 5. 12 Let's take$L^p$space for example: how can it be in both real and functional analysis? If one uses "$L^p$" as notation for the set of functions that are Lebesgue integrable to the$p$th power, and proceeds to study the properties of individual functions in this set, this is Real Analysis. The mode of thinking here is not much different from studying ... 12 (Update. Added a self-contained proof that under ZF+DC+BP there is no norm.) Martín-Blas Pérez Pinilla is on the right track. You can't even put a norm on$C(\mathbb{R})$without using the axiom of choice in an essential way. Dependent choice is not enough. Claim. It is consistent with ZF+DC that there does not exist any norm on$C(\mathbb{R})$. ... 11 Let's go ahead and prove Eric Wofsey's claim (see his comment under the question), that if$H$is an infinite-dimensional Hilbert space with an orthonormal basis$U$of cardinality$\kappa$, then any Hamel basis$B$of$H$has cardinality$|B| = \kappa^{\aleph_0}$. Notice this implies that some (real or complex) vector spaces do not admit a Hilbert space ... 11 Take any constant function. Then$y_n$is trivially convergent for any$x_n$. 11 Literally, the claim is wrong, since the space has dimension$2^{\aleph_0}$, and there are Banach spaces with the same dimension (e.g. the$\ell^p(\mathbb{N})$spaces). Since the rationals are dense in$[a,b]$, there is a linear injection of$C^\infty([a,b])$into$\mathbb{R}^{\mathbb{Q}\cap [a,b]}$, and the latter space has cardinality ... 10 Suppose that $$f(x)=E(x)+O(x),$$ for all$x$, where$E(x)$is even and$O(x)$is odd. Then for all$x$, $$f(-x)=E(-x)+O(-x)=E(x)-O(x).\tag{2}$$ Add. We get$f(x)+f(-x)=2E(x)$, and therefore$E(x)=\frac{f(x)+f(-x)}{2}$. 10 Generally, without completeness, you can't deduce that a weak$^\ast$convergent sequence is bounded. Let$X = c_{00}$be the space of sequences with only finitely many nonzero terms, endowed with the$\lVert\,\cdot\,\rVert_\infty$-norm. Its completion is$c_0$, the space of sequences converging to$0$, and its dual therefore isometrically isomorphic to ... 9 I'm not sure if this is 100% standard, but I've always understood this notation to mean:$f$is a function from$\Omega$to$\mathbb{R}$which has compact support (the subscript$0$), and is smooth/infinitely differentiable (the superscript$\infty$). I just checked, for example, that Stein and Shakarchi use this notation in their Functional Analysis. I ... 9 Pick an enumeration of the rationals$r_1,r_2,r_3,...$. Let$d(f,g) = 0$if$f = g$. Else let$r_n$be the first rational in the enumeration for which$f(r_n) \neq g(r_n)$. Define$d(f,g) = f(r_n) - g(r_n)$. This is well defined since if$f \neq g$, there must be some rational number for which$f(r) \neq g(r)$, since$f$and$g$are continuous, and the ... 9 In any metric space, there are an infinite number of ways to write down balls with a given center. But some of the balls might actually be the same. For instance, in the "discrete metric"$d(x,y)=0$if$x=y$and$1$otherwise, all balls$B_r(x)$for$r \leq 1$are the same (they are just$\{ x \}$) while all balls$B_r(x)$for$r>1$are also the same ... 9 Sorry to resurrect such an old post, but I would like to supply a proof of this without invoking Jordan normal form or Schur decomposition. So we want to show the following statement. Theorem. Let$T$be a linear transformation on a finite-dimensional complex vector space$V$, say on$\mathbb{C}^n$for simplicity and without loss of generality; given ... 9 Show that$T$and$S$are closed. Then apply the closed graph theorem to conclude that$T$and$S$are continuous. To show that$T$is closed, suppose$\{ x_{n} \}$converges to$x$and suppose$\{ Tx_{n} \}$converges to$y$, and show that$Tx=y$; to do this observe that $$(Tx,z) = (x,Sz) = \lim_{n}(x_{n},Sz) = \lim_{n}(Tx_{n},z)=(y,z),\;\;\; z \in ... 8 This is not true. Simply take f = \chi_{[-1,1]} (the indicator function of the interval [-1,1]). Then f \in L^p for all p \in (0,\infty], but if \widehat{f} \in L^1 was true, then Fourier inversion would imply that$$ f = \mathcal{F}^{-1} \widehat{f} \in C_0 $$would be (almost everywhere equal to) a continuous function. This is clearly not the ... 8 Here is an outline for proving your statement. \ell^p is a space of (some) functions \mathbb N\to\mathbb R, and so is L^p (with the choices indicated by OP). (I chose to work with the reals, but this choice is irrelevant.) The counting measure gives nonzero measure to all nonempty sets, so each equivalence class in L^p only consists of one ... 8 The setup For completeness I want to state the parts from the book that are needed: We are given a (in general nonlinear) operator$$A: W^{1,p}(\Omega) \rightarrow W^{1,p}(\Omega)^*,$$which is coercive (see Lemma 2.35 in the given literature). At that, coercivity means the existence of a mapping \xi: \mathbb{R}^+ \rightarrow \mathbb{R}^+ such that \xi ... 8 It is certainly not dense. The linear functional$$ T(x_n) = \sum_{n=0}^{\infty} \frac{x_n}{n}$$Is continuous on$l^2$. Thus, the set$T^{-1}(0)$must be a closed subspace of$l^2$. If it were dense, we'd have$l^2 = T^{-1}(0)$. But the sequence${\frac{1}{n}}$obviously isn't in this space. 8 We know that,$L_\infty$is not separable, so neither does its dual$L_\infty^*$. It is remains to recall that$L_1\$ is separable.

8

I think most of the differences can ultimately be traced back to the following properties of infinite sets (where the relevant sets would be bases of the vector space or functions of the basis vectors): There exist bijections between an infinite set and proper subsets of that infinite set. For example, there's a bijection between the integers and the even ...

Only top voted, non community-wiki answers of a minimum length are eligible