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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 in $B(H_B\otimes H_A\otimes \ell^2(\Gamma))$, 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.



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