# Tag Info

25

No, such a bijection from the unit interval $I$ to the unit square $S$ cannot exist. Since $I$ is compact and $S$ is Hausdorff, a continuous bijection would be a homeomorphism. But in $I$ there are only two non-cut-points, whereas in $S$ each point is a non-cut-point.

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 ...

14

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 ...

13

If $V$ is a Banach space we call $V'$ the dual space (see continuous dual space on wikipedia), i.e. the space of linear continuous functionals $\xi \colon V\to \mathbb R$. Then it is well known that there exists a natural injection $$J \colon V \to V''$$ defined by $$J(v)(\xi) = \xi(v)$$ for all $\xi \in V'$. We know that $J$ is an isometry, in ...

13

In short the list of reasons is: 1) The modern trend in math today is to develop non-commutative analogues of well known theories. Vaguely speaking operator spaces are normed spaces over "non-commutative" scalars, in fact over matricies. 2) There were several long standing problems, that were solved via methods of operator space theory. As soon as a ...

12

Take $$f\left(x\right)=\begin{cases} x^{-1/2}, & x\in\left(0,1\right]\\ 0, & \text{otherwise} \end{cases}$$ This function is obviously in $L^1$; note also that $f\ast f$ is $0$ on $(-\infty,0] \cup [2,+\infty)$, $\pi$ on $(0,1]$, and it decays from $\pi$ to $0$ continuously on $[1,2]$. Therefore, $f \ast f(x)$ is everywhere defined, in the sense ...

12

I submit this counterexample which, in my opinion, proves that the statement is false. In the vein of this MathOverflow post by Gerald Edgar, let $V$ denote the real vector space of all polynomials of one variable and let $$\lVert P\rVert=\max_{x\in[0, 1]} \lvert P(x)\rvert,\qquad \forall P\in V.$$ Moreover, let $$W=\{a_0+a_2x^2+a_4x^4+\dots+a_{2k}x^{2k}\ ... 11 The reason is that we would like to define a norm by the following formula:$$ \|f\|_p:=\left(\int_X |f|^p d\mu \right)^{1/p}. $$Therefore, we need to have triangle inequality (Minkowski's Inequality) which is available only for p\geq 1. See more details in Chapter 6 of Folland's book: Real Analysis. 11 The problem is that an element of a Hamel basis might be an infinite linear combination of the other basis elements. Essentially, linear dependence changes definition. 10 For any f \in L^1 we have$$\lim_{\omega \to \infty} \underbrace{\int_{\mathbb{R}} f(x) \cdot e^{-\imath \, x \omega} \, dx}_{=:\hat{f}(\omega)} = 0, \tag{1}$$this result is known as Riemann-Lebesgue lemma. As f \in L^1 we can choose a sequence (f_n)_{n \in \mathbb{N}} of simple functions such that f_n \stackrel{L^1}{\to} f. Since ... 10 Here's the way I like to think about it. I'll start with the finite dimensional space \Bbb{R}^n because it looks like that's where you are, but I'll give an analogy for infinite dimensional spaces as well. The quantity z^Tx represents a linear functional on \Bbb{R}^n, that is a linear function which eats a vector and spits out a real number:$$ ...

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

Suppose that $R^2=T$. Then $\ker R\subseteq\ker T=\Bbb Re_0$, where $e_0=\langle 1,0,0,\ldots\rangle$. Clearly $\ker R$ is non-trivial, so $\ker R=\ker T$. Moreover, $T$ is surjective, so $R$ must also be surjective. In particular, $e_0=Rx$ for some $x\in\ell^1(\Bbb N)\setminus\ker T$, and therefore $R^2x=Re_0=0\ne Tx$. Now suppose that $R^2=S$, and let ...

9

$\mathbb Q(\sqrt 2)$ has two norms which are equivalent on $\mathbb Q$ derived from the two embeddings into $\mathbb R$

9

Hint: Consider what happens to the connected $[0,1]$ if the point $\frac12$ is removed. What happens to $[0,1]\times[0,1]$ when $f(\frac12)$ is removed?

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

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 $M:=$ $\lbrace ... 8 By the hypothesis the Neumann series$\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 Let T\in J where J is a closed two-sided ideal in B(H). Consider its polar decomposistion T=U|T|, where |T|=(T^*T)^{1/2}. Clearly T^*T\in J, so its square root |T| is in J too (granted, we use some small bit of functional calculus there). Therefore, T^*=|T|U^*\in J as well. Edit: As commented by Martin Argerami, this proof works for norm ... 8 Probably not what you are looking for but one COULD use some functional analysis to answer the question: If \ell^q = \bigcup_{p<q} \ell^p =\bigcup_{n\in\mathbb N} \ell^{q-1/n} Baire's theorem implies that some \ell^{q-1/n} would be of second category in \ell^q and then the open mapping theorem for the inclusion \ell^{q-1/n} \hookrightarrow \ell^q ... 8 Let e \in C_0(X) be the unit. For each x \in X% there is a f \in C_0(X) with f(x) \ne 0. As e(x)f(x) = f(x) we must have e(x) = 1. So e is the constant function e=1. But 1 "vanishes at infinity" only if X as compact: By definition there is a compact K such that 1 = |1(x)| \le \frac 12 outside K, i. e. on X \setminus K, so we must ... 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 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 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 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 ... 7 This is the proof that I know. But it's not easy, but it is well known in the literature. L^1 has a property called cotype 2, that is, there is a constant C>0 such that for any x_1,\dots,x_n \in L^1$$ E \left\|\sum_{k=1}^n \epsilon_k x_k \right\|_1 \ge C \left(\sum_{k=1}^n \|x_k\|_1^2\right)^{1/2} ,$$where$\epsilon_k$is a sequence of ... 7 No, consider$J\oplus_2 J^*$, where$J$is a James space 7 This is not true in general. In fact, totally bounded metric spaces are unions of finitely many arbitrarily fine open balls, but are certainly closed in themselves. Nor is the whole space the only such counterexample. For instance, consider the subset$X=(0,1)\cup(2,3)$of$\Bbb R.$Restricting the usual metric of$\Bbb R$to$X$makes$X$a metric space, ... 7 Yes, this can be done without making the$C^*$-algebra concrete in$B(H)$. All we need is that$z=0$iff$z^*z=0$, which follows from the$C^*$-identity$\|z^*z\|=\|z\|^2$. Lemma Let$a\in A$. The following assertions are equivalent. 1 -$p=a^*a$is idempotent, i.e.$p^2=p$. 2 -$aa^*a=a$3 -$a^*aa^*=a^*$4 -$q=aa^*\$ is idempotent. Proof ...

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