# Tagged Questions

For question involving Hilbert spaces, that is, complete normed spaces whose norm comes from an inner product.

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### Proving that the set is closed.

We use the sequential definition to prove a set is closed. So no continuity or closure or anything related to the topology of the set is allowed. Show $A = \{ x \in \ell^2: |x_n| \leq 1/n \}$ is ...
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### Existence of adjoint operator on a Hilbert space

friends! I read that the algebra $\mathscr{L}(H,H)$ of the bounded operators on a Hilbert space $H$ is a $B^\ast$-algebra in the sense defined here . I easily verify all the properties except for the ...
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Let $A$ be a compact, self-adjoint operator, $A \geq 0$. We need to prove that for any orthonormal system $\{e_i\}_1^{\infty}$ and for any $N$, $$\sum_1^N \langle Ae_i,e_i \rangle \leq \sum_1^N \... 0answers 58 views ### Two Hilbert spaces V \subset H, a basis for both spaces? Let V \subset H be a pair of Hilbert spaces (with different inner products). The embedding is continuous and dense, and both spaces are separable. Is it always the case that one can I find a ... 1answer 85 views ### Hilbert Spaces; eigenvalues of PBP vs. B for B compact selfadjoint and P orthoprojection. An exercise I have come upon while studying Hilbert Spaces: Let A be a compact operator, and P \in L(H) be an orthoprojection. Prove that$$\lambda_n (PA^*AP) \leq \lambda_n (A^*A)$$(Where \... 3answers 22 views ### Describing a Subset of a Hilbert Space H Let H be a Hilbert space. How can we describe the set \{ x \in H \mid \|x-y\| = a \|x-z\| \}, where y, z \in H are fixed and a > 0? Geometrically how does it look like? 2answers 90 views ### Spectral Measures: Concentration Given a Hilbert space \mathcal{H}. Consider spectral measures:$$E:\mathcal{B}(\mathbb{C})\to\mathcal{B}(\mathcal{H}):\quad E(\mathbb{C})=1$$Define its support:$$\operatorname{supp}(E):=\bigg(\...
Given a Hilbert space $\mathcal{H}$. Consider a spectral measure: $$E:\Sigma(\Omega)\to\mathcal{B}(\mathcal{H})$$ Can you give me a hint for: $$E(A)E(B)=E(A\cap B)$$ So far for disjoints I checked: ...