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## Hot answers tagged hilbert-spaces

3

Let $\kappa$ be any cardinal. Define $\ell^2(\kappa)$ as follows: As a set, we have $$\ell^2(\kappa) := \left\{x \colon \kappa \to \mathbf K \Biggm| \sum_{i \in \kappa} |x(i)|^2 < \infty \right\}$$ With the scalar product $$(x,y) := \sum_{i \in \kappa} x(i)\overline{y(i)}$$ $\ell^2(\kappa)$ is a Hilbert space, of which the maps $e^i$, $i \in \kappa$, ...

2

I think you misunderstood it: the "set theoretic complement" is $H \setminus E(H)$, not $E^{\dagger}(H)$

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Suppose $\mathcal{H}=L^{2}[0,\pi]$. Suppose $\mathcal{M}$ is a countably dense subset of compactly supported $C^{\infty}$ functions on $(0,\pi)$. Apply Gram-Schmidt in order to obtain an orthonormal basis $\{ f_{k} \}_{k=1}^{\infty}$ of smooth compactly supported functions on $(0,\pi)$. The following operators $L_{\alpha,\beta}$ are selfadjoint $$... 2 If I understood you well, your \ell^0 is the space of complex sequences that are eventually vanishing. If that is right, consider the sequence (c_n) \in (\ell^0)^{\mathbb N} defined by c_n(k) = \frac{1}{k} for 1 \le k \le n and c_n(k) = 0 for k > n. (c_n) is a Cauchy sequence as for n < m$$\Vert c_n - c_m \Vert^ 2=\sum_{k=n+1}^m ...

2

Yes, you are correct. For reference: The Spectral Theorem For a Pair of Commuting Operators http://www.mi.ras.ru/~snovikov/78.pdf

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Yes, since $B(H)$ is a Banach algebra it makes perfect sense to talk about questions of convergence. Unfortunately, the sum in question need not converge. Let $v$ be some unit vector in $H$. We can construct a counterexample as follows: Let $P_v$ denote the orthogonal projection onto the subspace spanned by $v$. Let $a = - P_v$. Then $a^{n} = (-1)^{n} P_v$, ...

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@AdamHughes answer works just fine, I am giving here a complete proof of the next proposition: If $(X, \|\cdot \|$) is a normed space, then any finite dimensional subset is closed. Here is the proof Proof Let $M \subset X$ have dimension $k \in \mathbb{N}$, then there exist $\{ e_1, \cdots, e_k \}$ such that $$M= \text{span}\{ e_1, \cdots, e_k \}$$ Take ...

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Since $M\cong \Bbb F^n$, $M$ is locally compact, hence closed.

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With the constraint that the cardinality $\kappa$ of a Hamel basis shall satisfy $\kappa^{\aleph_0} = \kappa$, the answer is affirmative. For an infinite set $S$, let us denote the family of countably infinite subsets of $S$ by $\mathfrak{P}_\omega(S)$. Then we look at the space $$\ell^2(S) = \biggl\{ f\colon S \to \mathbb{K} : \sum_{s\in S} \lvert ... 1 Consider the POSET:$$\lambda\in\Lambda:\quad\mathcal{S}\in\lambda\implies\mathcal{S}=\mathcal{S}_\varphi\lambda\in\Lambda:\quad\mathcal{S}\in\lambda\implies\mathcal{S}\perp\mathcal{S}'$$It is nonempty:$$\mathcal{S}_0=(0):\quad\{\mathcal{S}_0\}\in\Lambda$$And admits upper bounds: ... 1 You have trivially that$$C^{\infty}_c(\mathbb{R^+}) \subset C^{\infty}_0(\mathbb{R^+})$$But W^{1,2}_0(\mathbb{R^+}) is usually defined as the closure of C^{\infty}_c(\mathbb{R^+}) for the W^{1,2} norm$$W^{1,2}_0(\mathbb{R^+}) := \overline{C^{\infty}_c(\mathbb{R^+})} $$1 Embedding Consider the embeddings:$$J_\alpha:\mathcal{S}_\alpha\to\mathcal{H}:\quad J_\beta^*J_\alpha=\delta_{\beta\alpha}1_\alpha\quad J_\alpha J_\alpha^*=P_\alpha$$Construct the unitary map:$$U\varphi:=(J_\alpha^*\varphi)_\alpha\quad V(\varphi_\alpha)_\alpha:=\sum_\alpha J_\alpha\varphi_\alpha$$Indeed they are inverses:*$$VU\varphi=\sum_\alpha ...

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Unitary Map Construct the unitary map: $$V:\mathcal{L}^2(\nu_0)\to\mathcal{H}:\quad Vh:=h(N)\varphi_0$$ By measurable calculus: $$h\in\mathcal{L}^2(\nu_0):\quad\|h(N)\varphi_0\|^2=\int|h|^2\mathrm{d}\nu_0$$ Especially one has: $$V\chi_A=E(A)\varphi_0:\quad\mathcal{H}=\overline{\{V\chi_A:A\in\mathcal{B}(\mathbb{C})\}}$$ Concluding unitarity. ...

1

Use the Cauchy criterion. If $m<n$ then $$\Bigl\|\sum_{k=m}^na_k\,h_k\Bigr\|^2=\sum_{k=m}^n|a_k|^2,$$ which can be made smaller than any $\epsilon>0$ because $\sum_{k=1}^\infty|a_k|^2<\infty$. Since a Hilbert space is complete, this implies that $\sum_{k=1}^\infty a_k\,h_k$ is convergent.

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Suppose $T$ is selfadjoint with $\|T\| =1$. Then $|(Tx,x)| \le \|T\|\|x\|=1$ for all unit vectors $x$. If $|(Tx,x)|=1$ for some unit vector $x$, then the above gives $$1=|(Tx,x)|\le \|Tx\|\|x\|\le \|T\|\|x\|^{2}\le 1$$ Therefore, $|(Tx,x)|=\|Tx\|\|x\|$ with $x\ne 0$ which, by Cauchy-Schwarz, forces $Tx=\alpha x$ for some scalar $\alpha$ and unit vector ...

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