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The zero representation of an algebra on a Hilbert space $H$ is the map that sends every element of the algebra to the zero operator on $H$. Murphy's book gives the following definitions: If $A$ is a C*-subalgebra of $B(H)$, it is said to be irreducible, or to act irreducibly on $H$, if the only closed vector subspaces of $H$ that are invariant for ...


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For any complex number $z$ we have $$z=|z|e^{i\theta}$$ where $\theta\in(-\pi,\pi]$ is the principal argument of $z$.


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Obviously, the norm is at least what you anticipate by setting $f = e_n$ where $n$ almost realizes the supremum of $|\lambda_n - \lambda|^{-1}$. Conversely, write $f$ as $f = \sum_n c_n e_n$. Then $$\|(K - \lambda I)^{-1} f\|^2 = \sum_n |c_n|^2 |\lambda_n - \lambda|^{-2} \leq (\sup_n |\lambda_n - \lambda|^{-2}) (\sum_n |c_n|^2).$$ Taking the square root ...


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There is a general method for finding sequences which satisfy linear recurrent relations with constant coefficients. In this case, we have the relation $$ x_{n+1}+x_{n-1} = \mu x_n \text{ for } n\geq 1. \text{ (*)} $$ Here is how we solve it: consider all the sequences (not nesessarily from $l^2$) which satisfy (*). They form a linear space (it's easy to ...


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The process is good. Maybe some wording could be added. For example, the first displayed inequality only holds for $m$ and $n$ large enough. The existence of the limit of the sequence $(x_n)$ has to be justified (by completeness of a Hilbert space).


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Since $g$ is clearly and odd function, and cosine is even and sine is odd, the coefficient of cosine will have to be $0$, and that's what you got. And you can check by subsitution whether your two linear equations are satisfied by the answer you got.


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If you have $H = X\oplus Y$, then the projection onto $X$ is not well-defined without $Y$. The reason for this is that the splitting $H = X\oplus Y$ is not unique. That said, given $X$ there are many spaces $Y$ such that $H=X\oplus Y$. For $H=\mathbb R^2$, $X=span\pmatrix{1\\0}$, $Y_t=span\pmatrix{t \\1}$, we have $H=X\oplus Y_t$ for all $t\in\mathbb R$. ...


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No. Let $V$ be the orthogonal complement of $L^2(X,\mu)$ in $H$ and let $f$ be a function in $V$. Then, $\int fg d \mu =0$ for all $g$ in $L^2(X,\mu)$. This condition easily implies that $f$ is identically zero.


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$p=2$ does the job. To see this, represent the bounded linear form $\Phi$ as $\Phi(x)=\langle x_0,x \rangle$. Bessel's equality gives the wanted convergence. Indeed, we have $$\sum_{n=1} ^\infty|\langle x_0,e_n\rangle|^2=\lVert x_0\rVert^2 $$ and $\langle x_0,e_n\rangle =\Phi(e_n)$. In general, we cannot hope for a better $p$, that is, $p\lt 2$. Indeed, ...



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