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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 ...


4

Suppose $T$ is not bounded. So, there is a sequence of elements $(x_n)$ of A with norm $1$ such that $\|T(x_n)\| > n^2$. But then the sequence $y_n = \frac{1}{n} x_n$ is a Cauchy sequence, because $\|y_n - y_{n+k}\| < \left |\frac{1}{n} + \frac{1}{n+k} \right| < \frac{2}{n}$ and $\|T(y_n)\| = \frac{1}{n} \|T(x_n)\| > n$ that is not Cauchy ...


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

If $1<p<q<\infty$, and $f\in L^q(0,1)$, then $$ \int_0^1 \lvert\, f(x)\rvert^p\,dx=\int_0^1 \lvert\, f(x)\rvert^p\cdot 1\,dx \le \left(\int_0^1\lvert\, f(x)\rvert^q\,dx\right)^{p/q}\left(\int_0^1 1^r\,dx\right)^r= \left(\int_0^1\lvert\, f(x)\rvert^q\,dx\right)^{p/q}, $$ where $$ \frac{p}{q}+\frac{1}{r}=1\quad\text{or}\quad r=\frac{q}{q-p}. $$ Hence ...


2

Yes, your argument is correct.


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

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

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

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} ...


2

We have the relationship $\lVert f\rVert_p\leqslant \lVert f\rVert_q$ for each function $f$. Since the embedding is linear, it follows that it is continuous. The image contains $\mathbb L^\infty$, which is dense. It can be written as a countable union of closed sets with empty interior.


2

Yes. Here are two examples. 1) Let $\Omega=\mathbb{N}$, so that $L^\infty(\Omega)=\ell^\infty$. Let $\mathcal{c}$ be the subspace of convergent sequences. The linear functional $f(\{x_n\})=\lim_{n\to\infty}x_n$ is bounded on $\mathcal{c}$. By the Hann-Banach theorem it can me extended to $\ell^\infty$. This extension is not represented by any $\ell^1$ ...


2

It is convinient to use the following notation $\mu_{x, y}(S) := \left<P(S)x, y\right>$ and shortly $\mu_{x}$ for $\mu_{x,x}$. Let $B(\mathbb{R})$ denote the abelian C*-algebra of bounded Borel functions on $\mathbb{R}$. Define a map $\Phi_P \colon B(\mathbb{R}) \rightarrow \mathcal{L}(\mathcal{H})$ by $$\left<\Phi_P(f)x, y\right> = ...


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

There are plenty of non-equivalent norms. Let $X$ be an infinite-dimensional normed space with norm $\|\cdot\|_X$. Let $Y$ be another normed space with norm $\|\cdot\|_Y$. Let $T\in \mathcal L(X,Y)$ be compact and injective. Then $$ \|x\|_T:=\|Tx\|_Y $$ is a norm on $X$. Moreover, $\|x\|_T \le \|T\|_{\mathcal L(X,Y)} \|x\|_X$. However, both norms cannot ...


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

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 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.


1

If I understand your question correctly, then what you want to show is not true. Take $A_n = [n, n + 1/n]$ and let $\mu$ be Lebesgue measure. Then $\mu (A_n) \to 0$, which easily implies your "Cauchy" property, but $$ \mu(\bigcup_{n=1}^M A_n) = \sum_{n=1}^M 1/n \to \infty$$ for $M \to \infty$.


1

Note that $\int_{x_n}^{x} f(z) \ dz = \int_{0}^{x} f(z) \chi_{(x_n,x)} \ dz $ ($ \chi_{(x_n,x)}$ is the caracteristic function of the interval $(x_n , x)$) Clearly $\chi_{(x_n,x)} \in L^{2}(0, \infty)$. Note that $|f \chi_{(x_n,x)}| \leq 2(|f|^2 + |\chi_{(x_n,x)}|^2 )$, then it follows that $f \chi_{(x_n,x)} \in L^{1}(0.\infty)$ because $f$ and $ ...


1

Define $f_n^2(0)=n^\frac{3}{4}$,$f_n^2(x)=0 \quad \forall x\in [\frac{1}{n},1]$ and linear on $[0,\frac{1}{n}]$. Let $f_n=\sqrt{f_n^2} \in C([0,1])$. Note $f_n\to f=0$ in $L^2$ norm since $ \|f_n\|_2 = (\int_0^1|f|^2 \, dx)^{1/2}=(\int_0^1|f^2| \, dx)^{1/2}$ and $f^2\to f=0$ in $L^1$ norm, but $f^2_n$ doesn't converge to $f^2=0$ in $L^2$ norm. Hence the ...


1

Recall that $(v_1,\dots,v_k)$ is a tight frame if and only if the frame operator $$ x\mapsto \sum_{j=1}^k \langle x,v_k\rangle v_k $$ is multiple of identity. Equivalently, the matrix $\sum_{j=1}^k (v_k\otimes v_k)$ is a scalar matrix. Adding a vector $v$ to a frame means adding rank-1 operator $v\otimes v$ to the frame operator. Can this operation make ...


1

HINT: Imagine for a moment that an extension $T$ of $T_0$ exists and take $x\in E\setminus E_0$. Since $E_0$ is dense, you can approximate $x$ with a sequence $x_n\in E_0$. Since our imaginary operator $T$ is continuous, it must hold that $$\tag{1}Tx=\lim_{n\to \infty} T_0 x_n.$$ Now go back to reality, where $T$ does not exist yet. You need to construct it. ...


1

Let $f(x) := 0$ for all $x \in \mathbb{N}$, $v=1$ and define $$f_n(x) := \begin{cases} 0 & x=1 \\ \frac{x-1}{n} & 2 \leq x \leq n+1 \\ 1 & x>n +1\end{cases}.$$ Then $$\begin{align*} \mathcal{F}(f_n-f) + o_v(f_n-f) &= \mathcal{F}(f_n) = \sum_{k=1}^{\infty} (f_n(k+1)-f_n(k))^2 = n \cdot \frac{1}{n^2} = \frac{1}{n} \to 0, \end{align*}$$ ...


1

If you start with any topological space $\Omega $ then $ C_b (\Omega) $, the set of bounded continuous functions on $\Omega $, is a C $^*$-algebra. But the Gelfand transform allows you to show that there exists a locally compact $\Omega'$ with $ C_b (\Omega)\simeq C_b (\Omega') $. So when talking in abstract, you gain nothing by considering ...



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