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I might be missing something trivial- but this result seems to be false. Let's take the $C^*$ algebra to be $M_2(\mathbb{C})$, $a=\begin{pmatrix} 1 & 0\\ 0 & 2 \end{pmatrix}$, $b=\begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}$. Then $b^*b=I$, and $b^*ab=\begin{pmatrix} 2 & 0\\ 0 & 1 \end{pmatrix}$, and it's not true that $b^*ab \leq ... 4 Assuming that the operator norm is taken with respect to the Euclidean norm on$\mathbb{R}^n$, we have $$\left\lVert \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix}\right\rVert = 1 = \left\lVert \begin{pmatrix} 0 & 0\\ 0 & 1\end{pmatrix}\right\rVert,$$ hence $$2\left\lVert\begin{pmatrix}1 & 0\\0&0 \end{pmatrix}\right\rVert^2 + 2 ... 3 Observe that for p<\infty$$ l_p\cap k = l_p \cap k_0 = l_p. $$Hence these spaces are complete with respect to the l_p norm. To see these inclusions: If x\in l^p, then \sum_{n=1}^\infty |x_n|^p converges, hence \lim_{n\to\infty}x_n=0 and x\in k_0. These spaces are not complete with respect to the l_\infty norm: Take the sequence x_n ... 3 The answer to your question is no, the spectrum can be infinite, and pairs of self-adjoint involutions are actually a good class of counterexamples because they can be fully described using the spectral theorem. The irreducible pairs of involutions occur in dimensions 1 and 2, and the rest are direct integrals of irreducibles. So every such pair is ... 2 Any normed space is also a vector space. However, you do not need to define a norm on a vector space, meaning that all vector spaces are a union of the "normed space" rectangle and another rectangle that is outside the big "topological space" rectangle. 2 Let x,y>0$$f(x+y)-f(y)=\int _0^xf'(t+y)dt<\int_0^xf'(t)dt=f(x)-f(0)$$as f(0)=0, f is subadditive. So$$f(d(x,z))\le f(d(x,y)+d(y,z))\le f(d(x,y))+f(d(y,z)))$$2 Yes, your argument is correct. 2 It is often difficult to prove that an operator is self-adjoint, which seems reasonable since you know a lot about an operator when it is self-adjoint. For instance it is unitarily equivalent to a multiplication operator by a real function. There are many possible approaches, one which is sometimes useful is the following Show that D is symmetric, i.e. ... 2 False. Take f_n zero except in a very narrow (area^2 < 1/n) and tall (height >n) spike. The sequence f_n\to 0 in L^2 but is unbounded. 2 I) We have an answer up to an overall sign. From the AdS_4 Poincare bulk metric$$ ds^2~=~ \frac{\ell^2}{r^2}dr^2 +\frac{r^2}{\ell^2}\left(-dt^2+dx^2_1+dx^2_2 \right) \tag{44}$$in Ref. 1, we deduce that the 3D \gamma_{\mu\nu} boundary metric is diagonal and$$-\gamma_{tt}~=~ \gamma_{x_1x_1}~=~ \gamma_{x_2x_2}~=~\frac{r^2}{\ell^2}. \tag{A} $$[This ... 2 Unfortunately, the multipliers of the Dirichlet space are not as easy to describe as of Hardy and Bergman spaces. The characterization was obtained by Stegenga in Multipliers of the Dirichlet space (free access). In addition to boundedness, it requires a Carleson-type condition in terms of capacity:$$ \iint_{\bigcup S(I_j)} |f'|^2 \le A ... 2 This follows directly from the Orlicz–Pettis theorem (applied to the$p^{{\rm th}}$power of your expression). You can however avoid using this theorem by showing that the function $$f(\phi) = \left( \sum_{n=1}^{\infty} \lvert \phi(x(n))\rvert^p \right)^{1/p}\quad (\phi\in B_{E^*})$$ is continuous with respect to the weak*-topology. Then it will be ... 2 The simplest example would be a nonzero algebra with any involution and zero multiplication. 2$1_{[a,b]}$is the$L_2$function which is 1 between a and b and 0 otherwise. Presumably, your definition of T should have x in place of s on the right hand side. Applying T to$1_{[a,b]}$will give a function f(x). This function will have an$L_2$norm I guess (square f(x), integrate over x then take the square root). I get $$T({{1}_{[a,b]}})(s)=\left\{ ... 2 The integral$$ \int_X f\,d\mu, $$is by definition equal to$$ \int_X f^+\,d\mu-\int_X f^-\,d\mu, \tag{1} $$where f^+(x)=\max\{f(x),0\} and f^-(x)=\max\{-f(x),0\}, and this definition makes sense, as a real number, if and only if \int_X \lvert\,f\lvert\,d\mu<\infty. Why? If \int_X \lvert\,f\lvert\,d\mu<\infty, then \int_X f^+\,d\mu,\int_X ... 2 The fact that \varphi is a *-homomorphism implies that \bar\varphi:A/\ker\varphi\to\varphi (A) , given by \bar\varphi (a+\ker\varphi)=\varphi (a) , is isometric and onto. So any Cauchy sequence in \varphi (A) comes from a Cauchy sequence in A/\ker\varphi . As the latter is closed (in general the quotient of a Banach space by a Banach subspace ... 2 Any normal operator T gives rise to some spectral measure E:Bor(\sigma(T))\to\mathcal{P}(H) which maps Borel subsets of the spectrum of T into orthogonal projections in H. If you take Borel subset A\subset\sigma(T), then E(A) is called a spectral projection. Search spectral theorem on this site. 2 Consider T_n f=f'(1/n). Since f\in C^1, |T_n f|\leq \sup |f'(x)|<\infty. On the other hand, f_t(x)=e^{tx} is a C^1 function and T_nf_t=te^{t/n}\rightarrow t as n\rightarrow\infty. Edit: I just noticed that you want the functionals to be bounded. In this case, consider instead$$ T_n f=n(f(1/n)-f(0)). $$We see that T_nf\rightarrow f'(0), ... 2 First we prove the theorem for the case when \alpha=1/2 and \beta=1/2. In this case, we have$$Q^n (P_1-P_2)=\frac{1}{2^n}\sum_{k=0}^n\binom{n}{k}P_1^kP_2^{n-k}(P_1-P_2)\\=\frac{1}{2^n}\sum_{k=1}^n[\binom{n}{k-1}-\binom{n}{k}]P_1^kP_2^{n-k-1}+\frac{1}{2^n}P_1^{n+1}-\frac{1}{2^n}P_2^{n+1}$$thus we have that$$||Q^n (P_1-P_2)|| \leq ... 2 What involution do you consider? If just complex conjugation, then$x\mapsto \overline{x}$is not even differentiable. If you consider$f\mapsto f^*$where$f^*(z) = \overline{f(\overline{z})}$then it does not satisfy the C*-identity. 2 Based on your statement of the problem, you don't assume that$X$is closed. So I don't think it is clear on the front end that the limit of a sequence is an element of$X$. Instead here is a hint. Assume$(y_k)_{k=1}^\infty$is a cauchy sequence. Express each$y_k = a_{1,k}e_1 + \cdots a_{n,k}e_n$using your basis. Use the definition of cauchy to prove ... 2 Assuming that$S$is hermitian we have that $$A=S+i(I-S^2)^{1/2}\,\,\Longrightarrow\,\,A^*=S-i(I-S^2)^{1/2},$$ and hence $$A^*A=\big(S-i(I-S^2)^{1/2}\big)\big(S+i(I-S^2)^{1/2}\big)=S^2+(1-S^2)=I,$$ as$S(I-S^2)^{1/2}=(I-S^2)^{1/2}S$, which means that$A$is unitary. 2 The Calculus distance scale factors in spherical coordinates are$s_{r}=1$,$s_{\theta}=r$,$s_{\phi}=r\sin\theta$. So the Laplacian is $$\Delta=\frac{1}{s_{r}s_{\theta}s_{\phi}}\left[ \frac{\partial}{\partial r}\frac{s_{\theta}s_{\phi}}{s_{r}}\frac{\partial}{\partial r} ... 1 That would depend on what you mean by "applied." To give you an example, consider quantitative finance, a field of applied math that studies things like pricing of derivatives/contingent claims, hedging & portfolio risk management, statistical arbitrage, etc. and more academic concepts like completeness of markets, models for evolution of asset prices, ... 1 Both norms are equivalent: As all norms on \mathbb R^n are equivalent, there is a constant c>0 such that for all v\in \mathbb R^n$$ c \left( \sum_{i=1}^n |v_i|^p \right)^{1/p} \le \left( \sum_{i=1}^n |v_i|^2 \right)^{1/2} \le c^{-1} \left( \sum_{i=1}^n |v_i|^p \right)^{1/p}.$$Now replace$v_i$by$f_{x_i}(x) = (\frac\partial{\partial x_i}f)(x)$, ... 1 Let$z\in X$. we define$f_{z}(x)=<x,z>$. f is linear:$(f_{z}(ax+by)=<ax+by,z>= a<x,z>+b<y,z>=af_{z}(x)+bf_{z}(y)$f is bounded:$|f_{z}(x)|=|<x,z>|\leq\|x\|\|z\|\Longrightarrow \|f\|\leq\|z\|$(where the first inequality is the Cauchy-swartz inequality). Let$T: X\longrightarrow X^*$with$T(z)=f_{z}$. Observe that$T$is ... 1 Let$T_nx = \langle x_n , x \rangle$. We have$\sup_n |T_n x| < \infty$, hence by the Banach Steinhaus theorem we have$\sup_n \|T_n\| < \infty$. Since$\|T_n\| = \|x_n\|$, we have the desired result. 1 I now looked up the definition of a core of an operator and it could be that the answer actually depends on the exact definition. In Wikipedia, a core (of a closed operator) is defined as a subset$D$of the domain of$A$such that$A$is the closure of$A|D$. If we replace this by requiring that$D$be a subspace, I can prove your claim. By symmetry, it ... 1 For any orthogonal projection$P$, a vector$x$is in the range of$P$iff$\|Px\|=\|x\|$. The condition$P \le Q$for orthogonal projections is equivalent to$\|Px\|\le \|Qx\|$. Because$\|Qx\|\le \|x\|$it follows that if$x \in\mathcal{R}(P)$, then$x\in\mathcal{R}(Q)$. 1 The isomorphism$H \to H^*\$ is always there, whether you pay attention to it or not. Making it explicit would be annoying; it would be akin to disallowing yourself to treat a rational number as if it were a real number, and having to add extra notation whenever you want to convert a rational number to its corresponding real number.