# 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 $V\to V$ ...

17

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

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. 12 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 (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 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 pth 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 ... 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}$$Add. We get f(x)+f(-x)=2E(x), and therefore E(x)=\frac{f(x)+f(-x)}{2}. 9 I'm not sure if this is 100% standard, but I've always understood this notation to mean: f is a function from \Omega to \mathbb{R} which has compact support (the subscript 0), and is smooth/infinitely differentiable (the superscript \infty). I just checked, for example, that Stein and Shakarchi use this notation in their Functional Analysis. I ... 9 Show that T and S are closed. Then apply the closed graph theorem to conclude that T and S are continuous. To show that T is closed, suppose \{ x_{n} \} converges to x and suppose \{ Tx_{n} \} converges to y, and show that Tx=y; to do this observe that$$ (Tx,z) = (x,Sz) = \lim_{n}(x_{n},Sz) = \lim_{n}(Tx_{n},z)=(y,z),\;\;\; z \in ...

9

By the Cauchy Schwarz Inequality, for any integrable function $f(x)$: $\displaystyle\left(\int_a^b f(x) \cdot f(x)^2\,dx\right)^2 \le \left(\int_a^b f(x)^2\,dx\right) \left(\int_a^b (f(x)^2)^2\,dx\right)$ $\displaystyle\left(\int_a^b f(x)^3\,dx\right)^2 \le \left(\int_a^b f(x)^2\,dx\right) \left(\int_a^b f(x)^4\,dx\right)$ But by the given conditions, we ...

9

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 ...

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 ...

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 ...

8

While your question could have multiple answers, perhaps the closest to what you are looking for is the notion of a non-metrizable vector space. In the general setting of topological vector spaces, we consider (as one might guess from the name) vector spaces endowed with a topology so that we can discuss ideas like the continuity of linear operators. Normed ...

8

The answer is yes, any such space must be a Banach space. This result was proved by Victor Klee in 1952 and answered a question first asked by Banach in 1932. V. L. Klee, Invariant metrics in groups (Solution of a problem of Banach), Proc. Amer. Math. Soc., 3 (1952), 484–487.

8

It is true, which can be readily shown in ${\mathbb{R}^2}$. Let $U = \left\{ {\left( {x,0} \right):x \in \mathbb{R}} \right\}$ and $V = \left\{ {\left( {x,y} \right):x - y = 0} \right\}$. Let ${P_U}\left( {x,y} \right) = \left( {x + y,0} \right)$ and ${P_V}\left( {x,y} \right) = \left( {x,x} \right)$. Note that both are projections, since $P_U^2 = {P_U}$ ...

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

We want to mimic $f(x) = x^{-1/2}$ on $[0,1]$, but then cut $[0,1]$ into lots of intervals, and then on each interval remake the function so that its integral over that interval remains the same, but the support of the function is much smaller. So let $x_n \to 0$ be a decreasing sequence with $x_0 = 1$. Write $x_{n} = (1+\epsilon_n) x_{n+1}$, and suppose ...

8

$x = (1,1,0,0, \ldots )$ $y = (1,-1,0,0, \ldots)$ By the way, note that $\ell_1$ is complete and separable but has non-separable dual. On the other hand, Hilbert spaces are isomorhic to their dual spaces. This shows that $\ell_1$ is not linearly isomorphic to an inner-product space.

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