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A Hilbert space $H$ is a real or complex vector space endowed with an inner product such that $H$ is a complete metric space when endowed with the norm induced by this inner product.
0
votes
A bounded operator $T$ is compact if and only if $\sigma_{\mathrm{ess}}(T)=\{0\}$
Suppose that $\sigma_{\mathrm{ess}}(T)=\{0\}$. So, zero is the only one possible accumulation point.
Thus if $\sigma_d(T)$ is finite then $E(\{\lambda\in\mathbb{R}:|\lambda|>\varepsilon\})$ has finite …
1
vote
1
answer
752
views
A bounded operator $T$ is compact if and only if $\sigma_{\mathrm{ess}}(T)=\{0\}$ [closed]
Theorem: Let $T$ be a bounded self-adjoint operator on a complex infinite-dimensional Hilbert space $H$. Then $T$ is compact if and only if $\sigma_{\mathrm{ess}}(T)=\{0\}$.
Proof: If $T$ is compact t …
2
votes
1
answer
91
views
On the dimension of the range of the resolution of the identity
I want to prove the following: Let $A,B$ be bounded self-adjoint operators in a complex-Hilbert space and $E_A(\lambda)$, $E_B(\lambda)$ its corresponding spectral resolutions, i.e.,
$$A=\int_{[m_A,M_ …
2
votes
1
answer
157
views
If $\|S\|<\sin\frac{\pi}{2n}$ then $\|P(I-S)^ku\|\neq 0$ for all $k=0,\ldots,n$
I want to show the following:
Let $H$ be a Hilbert space and let $S:H\to H$ be a bounded operator such that
$$\|S\|<\sin\frac{\pi}{2n}.$$
Let $\mathcal{L}$ be a closed subspace of $H$ and $$u_k:=(I …