Tag Info

6

We have the following characterization of adjoint operators: Suppose that $X$ and $Y$ are normed spaces. If $T:X\rightarrow Y$ is a bounded linear operator, then $T^*$ is weak*-weak* continuous. Conversely, if $S$ is a weak*-weak* continuous linear operator from $Y^*$ to $X^*$, then there is a bounded linear operator $T:X\rightarrow Y$ such that $T^*=S$. ...

5

Fix $\varepsilon>0$ and $g\in L^{\infty}([0,1])$. If $T$ is surjective, then $g=T(f)$ for some $f\in L^1([0,1])$, indeed. By the continuity of $T$ at $f$, there exists some $\delta>0$ such that \begin{align*} \left.\begin{array}{ll}\bullet\,h\in ...

4

In fact, any infinite-dimensional Banach space $X$ has a compact subset that does not lie in any finite-dimensional linear subspace. Namely, define a sequence $x_n$ inductively such that $x_{n+1}$ is not in the linear span of $\{x_1, \ldots, x_n\}$ but $\|x_{n+1}\| < 1/n$. Then $\{0\} \cup \{x_1, x_2, \ldots\}$ is your set.

4

This doesn't look right near $0$. Take for example $y = \frac{x + x^2}{2}$, $0 \leq x \leq 0.25$ - This satisfies everything near $0$. But $(\frac{xy'}{y})' = \frac{1}{1+x^2} > 0$.

4

One can define a meaningful integral as $$\int f \, d\mu = \int f_+ \, d\mu - \int f_- \, d\mu,$$ As long as at least one of the two integrals on the right hand side is finite. Here, $f_+ (x) = \max \{0, f(x)\}$ is the positive part of $f$ and the negative part is defined analogously.

3

Consider the map $$\Phi : L^1 \to L^1, f \mapsto \varphi f.$$ By assumption, this is a well-defined linear map. Use the closed graph theorem to show that it is bounded (there are some details to fill in here). This shows that the functional $$\psi : L^1 \to \Bbb{C}, f \mapsto \int f \varphi d\mu$$ is bounded (why)? Now use the characterisation of ...

3

Every bounded linear operator on a complex Banach space has non-empty spectrum $\sigma(B)$. A simple way to argue this is to assume the contrary, and conclude that $(B-\lambda I)^{-1}$ is an entire function of $\lambda$ which vanishes at $\infty$; then Liouville's theorem gives the contradiction that $(B-\lambda I)^{-1}\equiv 0$ for all ...

3

I'm adding another answer, now that you've changed the question. Assume that $W$ is finite-dimensional with $W^{\perp} \subseteq U$. The goal is to show that $U\oplus U^{\perp} = V$. All '$\oplus$' decompositions are orthogonal in what follows. Because $W$ is finite-dimensional then $V=W\oplus W^{\perp}$ which can be seen by choosing any basis of $W$ and ...

2

Here's a (hopefully right) recipe to construct "bad" incomplete spaces: Start with a complete infinite dimensional space $\mathcal{H}$. Choose a normalized vector $e_0$. Extend it to an ONB $\mathcal{E}$. Extend this then to a Hamel basis $\mathcal{B}$. Rip of $e_0$ to get an orthonormal $\mathcal{S}:=\mathcal{E}\setminus\{e_0\}$ and a linear independent ...

2

As pointed out in the comments, $C([a,b])$ is not reflexive. I will consider only $C([0,1])$, since this space is isomorphic to $C([a,b])$. I assume you know that $c_0(\mathbb{N})$ is not reflexive. Since closed subspaces of reflexive spaces are reflexive, it suffices to show that $c_0(\mathbb{N})$ can be embedded isometrically in $C([0,1])$. This can be ...

2

Perhaps the author was a bit sloppy in writing it up. I would focus on the decompositions rather than the subspaces: Look for a maximal collection of mutually orthogonal $\mathcal{A}$-cyclic subspaces instead, and let $V$ be the sum of this collection.

2

Sure: the image of the unit ball in $\ell^2$ under the map that sends the standard basis element $e_n$ to $e_n/n$ (with $\ge 1$) is Hilbert-Schmidt, so a compact operator, etc. Indeed, any sequence going to $0$ replacing $1/n$ produces (pre-) compact image.

2

Assume that $A$ is bounded. Take $x_n\in A, y_n\in B$ such that $\lim||x_n-y_n||=e$. Since $A$ is bounded - sequence $x_n$ is bounded and so is $y_n$. Thus using Banach-Alaoglu theorem one can choose subsequences $x_{n_k},y_{n_k}$ converging in the weak topology to $x,y$. Now since closed and convex sets are weakly closed (Mazur theorem) one gets that $x\in ... 2 It is not compact by the Khintchine inequality which tells you that the closed span of Rademacher functions is isomorphic to a Hilbert space in every$L_p$-space. The inclusion then takes a copy of a Hilbert space to another copy of a Hilbert, so it cannot be compact. 2 1: On a quick look, I don't immediately see how to show that$\psi$is contractive in general. But note that when$A$is unital so is$\psi$, which together with positivity makes it contractive. I think this idea can be extended to the non-unital case. 2: When you prove that$\psi$is positive, you don't need to use that$A$is$A$, just that it is a ... 2 First, we can show that$\overline{\mathcal B(\Bbb R)\times\mathcal B(\Bbb R)} = \mathcal B(\Bbb R^2)$, as every open disk -hence every open set in$\Bbb R^2$- can be covered by open rectangles (with sides parallel to the axes). Second,$f(x)$can also be considered as a function in two variables (but constant in$y$). More precisely we consider ... 2 Let$(X_n)_{n=1}^\infty$be a sequence of disjoint sets of strictly positive finite measure. Set$g_n = \mathbf{1}_{X_n}$. For each non-empty set$A\subset \mathbb{N}$let $$f_A(x) = \sum_{n\in A}g_n(x).$$ Then$\|f_A - f_B\| = 1$for distinct subsets$A,B\subseteq \mathbb{N}$. The power-set of$\mathbb{N}$has cardinality continuum so$L_\infty(\mu)$is ... 2 Yes, correct and also true in reflexive Banach spaces. Kakutani's theorem (at least the direction you use in the proof) is also a special case of Banach-Alaoglu theorem. In History of Banach Spaces and Linear Operators, Albrecht Pietsch remarks ... the weak* compactness theorem is an elementary corollary of Tychonoff's theorem. Therefore priority ... 2 Let's first check that$P^\perp$has invariant range. Let$a \in A$and let$\eta \in [A\xi_1]^\perp = (A\xi_1)^\perp$. Then, for any$b \in A$, $$\langle a\eta, b\xi_1 \rangle = \langle \eta, a^\ast b \xi_1 \rangle = 0,$$ where$a^\ast b \in A$and hence$a^\ast b \xi_1 \in A\xi_1$precisely because$A$is a self-adjoint algebra of operators in$B(H)$, ... 2 The nearest point projection onto a closed subset$E\subset \mathbb R^n$is single-valued if and only if$E$is convex*. In this case, the Lipschitz constant is equal to$1$. If$E$is not convex, there is at least one point$x\in \mathbb R^n$for which$\min_{y\in E}\|x-y\|$is attained at more than one point. We could try to discuss the continuity of ... 2 The equation$aN+bN=(a+b)N$is true for any convex set$N$in a real vector space when$a$and$b$have the same sign. Clearly$aN+bN\supset(a+b)N$, so only the other direction needs to be shown. Take$x\in aN+bN$. Then there are$y,z\in N$so that$x=ay+bz=(a+b)\left(\frac{a}{a+b}y+\frac{b}{a+b}z\right)$. By convexity$\frac{a}{a+b}y+\frac{b}{a+b}z\in N$, ... 2 For fixed$x \in \mathbb{R}$, define a mapping $$\tau_x: \mathbb{R} \to \mathbb{R}, y \mapsto y-x.$$ Then$\tau_x$is continuous (hence measurable) and$\tau_x^{-1} = \tau_{-x}. By definition, $$\int_{\mathbb{R}} |g(y-x)| \, m(dy) = \int_{\mathbb{R}} |g \circ \tau_x(y)| \, m(dy).$$ We can rewrite the right-hand side using image measures: $$... 2 Physically, Hamiltonian operators in Quantum Mechanics should be semibounded, meaning that (Ax,x) \ge M(x,x) for all x\in\mathcal{D}(A) and for some fixed M. This has to be done with energy considerations. Second order ODES and PDES, in order to be symmetric, are quadratic in nature, and usually end up being semibounded--again, this is related to ... 2 If you have a bounded operator A, then the holomorphic functional calculus is always an option, and it is based on Cauchy's integral representation:$$ f(A) = \frac{1}{2\pi i} \oint_{C} f(\lambda)\frac{1}{\lambda I-A}"\,d\lambda = \frac{1}{2\pi i} \oint_{C} f(\lambda)(\lambda I-A)^{-1}\,d\lambda. $$The contour C is any simple ... 2 By the fundamental theorem of calculus, we have that$$u_\delta(x)=u_\delta (c)+\int_c^xu_\delta'(t)dt,$$Can you conclude now? 2 Let f(t) = \begin{cases}e^{-1/t} & t > 0 \\ 0 & t \le 0\end{cases}, g(t) = \dfrac{f(2-t)}{f(t-1)+f(2-t)}, and \phi(t) = g(|t|). It is well known that f \in C^{\infty}. Since \max\{t-1,2-t\} \ge \frac{1}{2} > 0, one of f(t-1) and f(2-t) is strictly positive. The other is non-negative. Hence f(t-1)+f(2-t) > 0. Thus, g \in ... 2 Ok, let me elaborate on my comment. The point is that we don't need convergence of the derivative, but only a bound on the derivative (which - for differentiable functions - is the same as a Lipschitz estimate). Take any "mollifier" \varphi \in C_c^\infty (\Bbb{R}) with \varphi \geq 0, \int \varphi \, dx = 1 and \rm{supp}(\varphi) \subset (-1,1). ... 2 Assume that A is unital (otherwise we can take A to be zero-dimensional). We need to prove that every non-zero element x is invertible and then apply Gelfand-Mazur theorem. The set of non-invertible elements is closed. If it contains any non-zero element x then its boundary also contains some non-zero element, call it y. Since y is a boundary ... 1 Build on the identity that you found:$$ f\left(\sum_{j=1}^{n}a_{j}e_{j}\right) = \sum_{j=1}^{n}a_{j}f(e_{j}) $$The way you build on this is to choose a_{j} so that a_{j}f(e_{j})=|f(e_{j})|^{q}. Then you get$$ \begin{align} \sum_{j=1}^{n}|f(e_{j})|^{q} & \le \|f\|\left(\sum_{j=1}^{n}|a_{j}|^p\right)^{1/p} \\ & = ... 1 This looks like an "associativity" property. Both algebras live inB(H_B\otimes H_A\otimes \ell^2(\Gamma))$, so the topology is the same. At the pre-closure level, you should convince yourself that$C_c(\Gamma,B\odot A)$and$B\odot C_c(\Gamma,A)\$ are equal, and that their closures give the two algebras you want to consider.

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