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4

In general, the inequality $$\mathbb{E}\|X-Y\|^2 \leq \mathbb{E}\|X-Z\|^2 + \mathbb{E}\|Z-Y\|^2$$ does not hold. Simply consider arbitrary $X$, $Z := \frac{1}{2} X$ and $Y = \frac{1}{4}X$. Then $$\mathbb{E}\|X-Y\|^2 = \frac{9}{16} \mathbb{E}\|X\|^2$$ whereas $$\mathbb{E}\|X-Z\|^2 + \mathbb{E}\|Z-Y\|^2 = \left( \frac{1}{4} + \frac{1}{16} \right) ...


4

You can prove that $2 | \langle a, b \rangle | \leq \| a \|^2 + \| b \|^2$ as I believe you have proven to attain $\| a + b \|^2 \leq 2\| a \|^2 + 2\| b \|^2$. Then you just expand: $\| a + b + c \|^2 = \langle a, a \rangle + \langle b, b \rangle + \langle c, c \rangle + 2\langle a, b \rangle + 2\langle a, c \rangle + 2\langle b, c \rangle \leq \| a \|^2 + ...


3

You can expand the LHS and use $\left<x,y\right>+\left<y,x\right>\leq ||x||^2+||y||^2$ 3 times. This latter inequality follows by expanding the non-negative entity $\left<x-y,x-y\right>$.


3

$U$ need not be unitary. Let $U_n$ be the map on $\ell^2$ that sends $e_j\mapsto e_{j+1}$ for $j=1,\ldots, n$, $e_{n+1}\mapsto e_1$, and $U_n$ is the identity on $L(\{e_j: j>n+1\})$. Then $U_n$ is unitary (it sends an ONB to an ONB again), but $U=S$ (the shift) is only an isometry because $e_1$ is not in the range.


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 ...


2

Let $(e_i)_{i\in I}$ be an an orthonormal basis, and $x =\sum_{i\in I} y_i e_i$ with norm $1$, i.e. $\sum_{i\in I} |y_i|^2 = 1$, then $$\|Tx\| \leq \sum_{i\in I}|y_i|\|Te_i\| \leq \sqrt{\sum_{i\in I} |y_i|^2}\sqrt{\sum_{i\in I}\|Te_i\|^2} = \|T\|_{HS}$$


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 $H$ be a separable Hilbert space as desired. For simplicity assuming we are working over the real field. Then being an integral operator, we have $$ K: H\rightarrow H, (Kf)(x)=\int K(x,y)f(y)dy $$ Since $K$ is a symmetric compact operator, it is normal and can be diagonalized. Let us write its eigenvectors as $\phi_{j}$ with eigenvalue $\lambda_{j}$. ...


2

Let $V$ be the span of $\phi_1,\dots,\phi_n$, and let $V^\perp$ be its orthogonal complement. Introduce new inner product on $V$ so that $\langle \phi_i,\phi_j \rangle_V = \delta_{ij}$ (this formula defines it for basis vectors; extend by linearity). Every vector $u$ in $H$ is the sum of vectors $u_1\in V$ and $u_2\in V^\perp$. Define $$\langle ...


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

If $\Im\lambda \ne 0$, and $x \in X$, then $$ \Im\lambda \|x\|^{2} = \Im((A-\lambda I)x,x),\\ |\Im\lambda|\|x\|^{2} \le |((A-\lambda I)x,x)|\le \|(A-\lambda I)x\|\|x\|,\\ |\Im\lambda|\|x\| \le \|(A-\lambda I)x\|. $$ So $A-\lambda I$ is injective for all $\lambda\notin\mathbb{R}$. The above inequality can be used to show that ...


2

(This answer makes the additional assumption that multiplication is associative in your number system. There are difficulties even with this restriction. For more generality, you might want to look at composition algebras, for example.) The number system with addition and multiplication as you describe it is called a ring. The conjugation you refer to is ...


2

Generally, people tend to think of distances between normed spaces in multiplicative terms, because it fits the way how composition of operators works. That is, the smallest value of the distance is $1$ and the triangle inequality has multiplication instead of addition. If you don't like this, take the logarithm. The Banach-Mazur distance does not directly ...


1

Partial answer to side question: $G$ is a nonassociative ring if we assume multiplication is also left distributive (it also does not assume the existence of 1). Also, I am reasonably sure that calling the operation conjugation implies all of the properties (before the side question) besides the linear order. You could also call the conjugate an adjoint ...


1

Yes, it is valid for any Hilbert space. The key fact you need is that in a complete inner-product space $H$, for each non-void, convex, closed set $C$ and every $x\in H\setminus C$ there is a unique element $c\in C$ such that $$\mbox{dist}(x,C) = \|x-c\|.$$ In the real case your isomorphism is, of course, linear.


1

First consider the case where $f(x)=e^{i\lambda x}$ and $g(x)=e^{i\lambda' x}$. In this case, $f(x)g(x)=\exp(i(\lambda-\lambda')x)$ and we can compute explicitely the integral. The result holds by linearity when $f$ and $g$ are linear combinations of $e^{i\lambda \cdot}$. Then we use an approximation argument: let $f_N$, $g_N$ of this form such that ...


1

Use polar coordinates: $x=z+\rho e^{i\theta}$, $$ \int_{B(z,r)}(x - z)^{n}dx = 2\pi \int_0^r \int_0^{2\pi}\rho^n e^{in\theta}\,\rho\,d\theta\,d\rho $$ Here $\int_0^{2\pi} e^{in\theta}\,d\theta=0$ when $n\ne 0$, because an antiderivative of $e^{in\theta}$ is $\frac{1}{in}e^{in\theta}$, which is $2\pi$-periodic.


1

$||AB-BA||=||AB-A_{\alpha}B+A_{\alpha}B-BA|| \leq ||AB-A_{\alpha}B||+||A_{\alpha}B-BA||$. But $A_{\alpha}B=BA_{\alpha}$, and $||AB-A_{\alpha}B|| \leq ||B||||A-A_{\alpha}||$. So $S'$ is closed in the norm topology as well. A "high-level" explanation can be given: $S'$ is a von-neumann algebra, and every von-neumann algebra is a $C^*$ algebra- and so $S'$ ...


1

I think very likely the question you might wish to be asking includes more structure than the question you literally asked... based on your example of a Laplacian. That is, your Hilbert space $H$ is really a Sobolev space $H^1$ on some compact Riemannian manifold. Then, yes, the Laplacian maps $H^1$ to $H^{-1}$ continuously, and $H^{-1}$ is the Hilbert space ...


1

You can see it directly using the orthogonal decomposition $$ X = \mathcal{N}(T)\oplus\mathcal{N}(T)^{\perp}. $$ If $Tx=y$, then $x=x'+x''$ for unique $x'\in\mathcal{N}(T)$, $x''\in\mathcal{N}(T)^{\perp}$. And $Tx=Tx''$. If $Tw=y$ also holds then $w''=x''$ because $w''-x''\in\mathcal{N}(T)^{\perp}\bigcap\mathcal{N}(T)$. So, with a small ...


1

The representation of the elements of $H^*$ is not needed. Let $\lambda$ be in the point spectrum of $A$. Then there is $0\neq x\in H=(H,(\cdot,\cdot))$ such that $Ax=\lambda x$ and by the self-adjointness of $A$: $\lambda(x,x)=(\lambda x,x)=(Ax,x)=(x,Ax)=(x,\lambda x)=\overline{\lambda}(x,x)$. Hence $\lambda=\overline{\lambda}$, thus ...


1

Define $T_{N}h = \sum_{n=1}^{N}h_{n}Te_{n}$. Because $\sum_{n}\|Te_{n}\| < \infty$, then there is a constant $M$ such that $\|Te_{n}\| \le M$, or $\|Te_{n}\|^{1/2} \le M^{1/2}$ for all $n \ge 1$. Then you use the Cauchy-Schwarz inequality to get what you want: $$ \begin{align} \|T_{N}h-Th\| & \le \sum_{n=N+1}^{\infty}|h_{n}|\|Te_{n}\| \\ ...


1

You are given that $$ (-2\pi i)Tf = \int_{-\pi}^{\pi}K(x-y)f(y)\,dy \\ =\int_{-\pi}^{\pi}(\pi\,\mbox{sgn}(x-y)-(x-y))f(y)\,dy\\ =\int_{-\pi}^{x}(\pi\,-(x-y))f(y)\,dy\\ +\int_{x}^{\pi}(-\pi-(x-y))f(y)\,dy\\ =(\pi-x)\int_{-\pi}^{x}f(y)\,dy+\int_{-\pi}^{x}yf(y)\,dy \\ -(\pi+x)\int_{x}^{\pi}f(y)\,dy+\int_{x}^{\pi}yf(y)\,dy. $$ ...


1

Yes to both questions. First, recall that a closed convex set in a Hilbert space is weakly closed, and that a weakly closed and norm bounded set is weakly compact (Alaoglu's theorem). Let $\epsilon := \inf\{\|x-y\| : x \in A, y \in B\}$. We will show there exist $x \in A$, $y \in B$ with $\|x-y\| = \epsilon$ which proves both statements. Choose a sequence ...


1

First observation: For all $u,v\in\mathcal H$ and $j\ne k$ $$ \langle T_ju,T_k v\rangle=\langle T_j^*u,T_k^* v\rangle=0. $$ Thus the linear subspaces $T_j^*\mathcal H$, $j\in\mathbb Z$, are perpendicular to each other, and so are their closures. Set $$ Y=\bigoplus_{j\in\mathbb Z}\overline{T_j^*\mathcal H}. $$ This infinite direct sum contains elements of ...


1

Let $P_{k}$ be the orthogonal projection onto the closure of the range of $T_{k}$. Then $P_{k}P_{k'}=P_{k'}P_{k}=0$ for $k\ne k'$ because $(T_{k}x,T_{k'}y)=(T_{k'}^{\star}T_{k}x,y)=0$ for all $x,y \in H$. Similarly, if $Q_{k}$ is the orthogonal projection onto the closure of the range of $T_{k}^{\star}$, then $Q_{k}Q_{k'}=0$ for $k\ne k'$. Furthermore, $$ ...


1

Note that at first your definition is only suitable for $\phi\in L^1(\mathbb T) \cap L^2(\mathbb T)$, but can be continuously extended to $\phi\in L^2(\mathbb T)$. The Operator taking $\phi$ to $f$ is precisely the fourier transform $\mathcal F$. It is a fundamental result of complex analysis that $\mathcal F: L^2(\mathbb T) \to L^2(\mathbb T)$ is an ...


1

The space of Hilbert–Schmidt operators is a Hilbert space with the inner product $\langle A,B\rangle = \operatorname{tr}(A^*B)$. And I suppose that sequences are considered as a special kind of functions. Then... I don't know of any other source of Hilbert spaces.


1

Let $\mathcal{A}$ be an uncountable family of infinite subsets of $\mathbb{N}$ which have finite intersection. Consider the Boolean subalgebra of the power-set algebra $\wp(\mathbb{N})$ generated by $\mathcal{A}$ and all finite subsets. The Stone space $K$ of this algebra is scattered (actually of Cantor-Bandixon height 3), which is separable as ...



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