# Tag Info

18

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 ...

16

Fredholm theory, originally developed for studying (systems of) differential equations, had been around for several decades when Gel'fand et al., working in the '50s, noticed that the Fredholm index was homotopy invariant, i.e., if $F_t$, $a \leq t \leq b$ is a continuous path of Fredholm operators, then $\operatorname{Index}(F_t)$ is constant in $t$, and so ...

14

First, a general remark. A common way to prove that some operator $T$ is not compact is to exhibit an infinite-dimensional subspace $M$ on which $T$ has a lower bound: that is, there exists $c>0$ such that $$\|Tx\|\ge c\|x\|,\quad \forall\ x\in M \tag{1}$$ If (1) holds, then the image of unit ball under $T$ contains a ball of radius $c$ in the ...

13

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: $$... 13 Preduals of \ell_1 are very interesting creatures. The question is sligthly ill-posed but let me comment on that anyway. We have quite a lot isometric preduals of \ell_1. Indeed, for any countably infinite ordinal number \alpha (endowed with the order topology) we have C_0(\alpha)^* \cong \ell_1. This essentially follows from the ... 12 (Update. Added a self-contained proof that under ZF+DC+BP there is no norm.) Martín-Blas Pérez Pinilla is on the right track. You can't even put a norm on C(\mathbb{R}) without using the axiom of choice in an essential way. Dependent choice is not enough. Claim. It is consistent with ZF+DC that there does not exist any norm on C(\mathbb{R}). ... 12 The problem is that an element of a Hamel basis might be an infinite linear combination of the other basis elements. Essentially, linear dependence changes definition. 12 Let V be an n-dimensional vector space over the field \mathbb{F}. Given a linear map T : V \to V, there is an induced linear map \bigwedge^nT : \bigwedge^n V \to \bigwedge^n V given by \left(\bigwedge^nT\right)(v_1\wedge\dots\wedge v_n) = (Tv_1)\wedge\dots\wedge(Tv_n). As \bigwedge^nV is one-dimensional, \bigwedge^nT = ... 11 The reason is that we would like to define a norm by the following formula:$$ \|f\|_p:=\left(\int_X |f|^p d\mu \right)^{1/p}. $$Therefore, we need to have triangle inequality (Minkowski's Inequality) which is available only for p\geq 1. See more details in Chapter 6 of Folland's book: Real Analysis. 10 Maybe a good point to start is this useful corollary of Baire Cathegory Theorem the cardinality of an Hamel base of a Banach Space can be finite or uncountable. It can't be countable The proof is a delightful application of Baire theorem. Now to give an explicit example, we can consider the space \ell^2  which has the standard base  M:= \lbrace ... 10 Let me provide a detailed proof: Assume that v_1,\ldots, v_n\in S are orthonormal functions, i.e., \int_0^1 v_iv_j\,dx=\delta_{ij}, and for a fixed a=(a_1,\ldots,a_n)\in\mathbb R^n define \varPhi_a :\mathbb R^n\to \mathbb R as$$ \varPhi_a(x)=\sum_{j=1}^n a_jv_j(x). $$Then$$ \|\varPhi_a\|_{L^2}^2=\sum_{j,k=1}^n\int_0^1a_ja_kv_jv_k ...

10

For any $f \in L^1$ we have $$\lim_{\omega \to \infty} \underbrace{\int_{\mathbb{R}} f(x) \cdot e^{-\imath \, x \omega} \, dx}_{=:\hat{f}(\omega)} = 0, \tag{1}$$ this result is known as Riemann-Lebesgue lemma. As $f \in L^1$ we can choose a sequence $(f_n)_{n \in \mathbb{N}}$ of simple functions such that $f_n \stackrel{L^1}{\to} f$. Since ...

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

Your question is equivalent to the following: does every pre-Hilbert space (= inner product space) $M$ admit an orthonormal basis? It turns out that the answer is "no". A counter-example can be found in N. Bourbaki's Topological Vector Spaces, Exercise V.2.2. I tried to solve the exercise, but I'm not quite sure that I understood it. Let ...

9

Let $\varepsilon>0$, we shall show that there exists an $n_0=n_0(\varepsilon)$, such that, $$n\ge n_0\quad\Longrightarrow\quad \lvert\, f_n(x)-f(x)\rvert<\varepsilon,$$ for all $x\in K$. As $\{f_n\}$ is equicontinuous, there exists a $\delta>0$, such that $$d(x,y)<\delta\quad\Longrightarrow\quad \lvert\, ... 9 By the Cauchy Schwarz Inequality, for any integrable function f(x): \displaystyle\left(\int_a^b f(x) \cdot f(x)^2\,dx\right)^2 \le \left(\int_a^b f(x)^2\,dx\right) \left(\int_a^b (f(x)^2)^2\,dx\right) \displaystyle\left(\int_a^b f(x)^3\,dx\right)^2 \le \left(\int_a^b f(x)^2\,dx\right) \left(\int_a^b f(x)^4\,dx\right) But by the given conditions, we ... 8 By the hypothesis the Neumann series \sum\limits_{n=0}^\infty T^n is absolutely convergent so convergent and we have$$(I-T)\sum\limits_{n=0}^\infty T^n=\left(\sum\limits_{n=0}^\infty T^n\right)(I-T)=I$$so I-T is invertible and its inverse is \sum\limits_{n=0}^\infty T^n, moreover we have$$\left\|\sum\limits_{n=0}^\infty ...

8

People keep mentioning the restriction on the size of a Schauder basis, but I think it's more important to emphasize that these bases are bases with respect to different spans. For an ordinary vector space, only finite linear combinations are defined, and you can't hope for anything more. (Let's call these Hamel combinations.) In this context, you can talk ...

8

Even worse. You can define a $L^p$-distance but the space is not locally convex. See "Examples of spaces lacking local convexity" in http://en.wikipedia.org/wiki/Locally_convex_topological_vector_space.

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.

8

You can apply a theorem of Grothendieck to the closure of $S$ in $L^2$ which is (as you show) contained in $C([0,1]) \subseteq L^\infty$. Grothendieck's theorem says that every closed subspace of $L^p(\mu)$ (where $\mu$ is a probability measure on some measurable space and $0<p<\infty$) which is contained in $L^\infty(\mu)$ is finite dimensional. A ...

8

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 ...

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

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

You do not need the axiom of choice to prove "There exists a normed space with a discontinuous linear functional". Consider the space $c_{00}$ of all sequences of real numbers with only finitely many nonzero terms, with the supremum norm. The functional $f(x) = \sum_{n=1}^\infty n x(n)$ is discontinuous. Everything in this argument is completely explicit ...

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 answer is: Not in general. For example, $\Omega=(0,2\pi)$, $u_n(x)=\sin nx$. Then $u_n\rightharpoonup 0$, since $$\int_0^{2\pi} f(x)\,\sin nx\,dx\to 0=u,$$ for all $f\in L^2[0,2\pi]$. Meanwhile $$v_n(x)=u_n^2(x)=\sin^2 nx=\frac{1}{2}-\frac{\cos (2nx)}{2}\rightharpoonup \frac{1}{2}=v,$$ and $u^2\ne v$. Note. If $\{u_n\}$ is bounded and ...

8

Suppose that the unit sphere $S_X$ of $(X, \| . \|_X)$ is compact. Then the unit ball of $X$ is the image of the compact set $[0,1] \times S_X$ by the continuous map $(t, v) \mapsto tv$, and hence is compact.

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