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## Hot answers tagged operator-theory

3

If $A$ is any unital C*-algebra, the hypotheses are satisfied with $C=A$ and $B=\mathbb C1$.

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You can reduce integral identities to the scalar integrals by applying the operators to vectors and then applying a bounded linear functional $x^{\star}$ to the corresponding vector expressions. Because the bounded linear functionals separate points, you can later remove them from your expressions to obtain a vector identity; finally the vectors are removed ...

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If $t\gt0$, then we want $a\ge2|A|$ and $$\gamma_r=a+ir[-1,1]\cup[a,-r]+ir\cup-r+ir[1,-1]\cup[-r,a]-ir$$ If $t\lt0$, then we want $a\le-2|A|$ and $$\gamma_r=a+ir[1,-1]\cup[a,r]-ir\cup r+ir[-1,1]\cup[r,a]+ir$$ In each case, $\gamma_r$ is counterclockwise. When $t\gt0$, the finite part of $\gamma_r$ is $[a-i\infty,a+i\infty]$. When $t\lt0$, the finite ...

2

Note that if $\mathcal{H}$ is real then the notion of a unitary operator does not make any sense. However, you can do that with five orthogonal operators instead. See e.g. Theorem 4.3 in A. Böttcher, A. Pietsch, Orthogonal and Skew-Symmetric Operators in Real Hilbert Space, Integral Equations and Operator Theory 74 (2012), 497-511. It should be added ...

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Assume $T=T^{\star}\in\mathcal{L}(H)$, where $H$ is a complex Hilbert Space. Implication 1: Show $(Tx,x) \ge 0$ for all $x \in H$ implies $\sigma(T)\subseteq [0,\infty)$. To do this, assume that $(Tx,x) \ge 0$ for all $x \in H$, and let $\lambda < 0$. Then $$0 \le -\lambda(x,x) \le ((T-\lambda I)x,x)$$ implies  |\lambda|\|x\|^{2} \le ...

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If $\widehat{\mu}$ is non-negative, then its square root defines a bounded Fourier multiplier operator $S$ on $L^2$. In this case $S$ is self-adjoint, so $T=S^*S$. Conversely, any linear and shift invariant $S$ is a Fourier multiplier operator with an $L^\infty$ symbol. The adjoint $S^*$ would have a symbol that is the complex conjugate of the symbol for ...

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An isometric operator on a (complex) Hilbert space is a linear operator that preserves distances. That is, $T$ is an isometry if (by definition) $\|Tx-Ty\|=\|x-y\|$ for all $x$ and $y$ in the space. By linearity, this is equivalent to $\|Tx\|=\|x\|$ for all $x$. Because of the definition of the norm in terms of the inner product and the definition of ...

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As you suspected $i($ker $T)$ needn't be dense in ker $T'$. Take $E=\ell^2$ and $E'=\lbrace (x_n)_{n\in\mathbb N}: (x_n/n)\in \ell^2\rbrace$ endowed with the obvious Hilbert norm. The inclusion $E\hookrightarrow E'$ is dense and compact. Define $T':E'\to E'$, $(x_n)_{n\in\mathbb N} \mapsto (x_n-x_{n+1})_{n\in\mathbb N}$. This is a continuous linear operator ...

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