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Let $T$ be a closed operator on a Hilbert space with discrete spectrum. Then for $\{\lambda_1,...\lambda_n\}\in\sigma(T)$ one can define the spectral projections $$P_{\{\lambda_1,...\lambda_n\}}=\frac{1}{2\pi i}\oint_\gamma(T-z)^{-1}dz,$$ where $\gamma$ is a smooth curve encircling $\{\lambda_1,...\lambda_n\} $ (and no other eigenvalues). One can show that $P^2=P$.

My Question:

Formally, one can split the curve $\gamma$ into $n$ curves encircling only $\lambda_i$, respectively. This suggests that $$P_{\{\lambda_1,...\lambda_n\}}=\sum_{i=1}^nP_{\{\lambda_i\}}.$$ Is this equation true?

If not, is there any sensible relationship between $P_{\{\lambda_1,...\lambda_n\}}$and $\sum_{i=1}^nP_{\{\lambda_i\}}$?

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  • $\begingroup$ maybe these need to be mutually orthogonal? $\endgroup$
    – Suvrit
    Commented Oct 9, 2015 at 13:59
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    $\begingroup$ That's actually my motivation. In general, a sum of projections isn't a projection anymore, unless they are orthogonal. Buy the Cauchy Formula for spectral projections suggests that this is the case and I want to know if that's true. $\endgroup$
    – Frank
    Commented Oct 9, 2015 at 14:01
  • $\begingroup$ This is just the deformation theorem for complex contour integration. $\endgroup$
    – Michael Renardy
    Commented Oct 9, 2015 at 14:25
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    $\begingroup$ I don't see why the spliting of the curves in $n$ smaller curves is not a proof of this ? $\endgroup$
    – Simon Henry
    Commented Oct 9, 2015 at 14:27

1 Answer 1

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Certainly when $T$ is a compact self-adjoint operator we have the spectral decomposition. Say $\sigma(T) = \{\lambda_1,\lambda_2, \cdots\}, \lambda_i\in \mathbb R$, then $\lambda_i \rightarrow 0$. The spectral theorem implies that $$ T = \sum_{i=1}^\infty \lambda_i P_i$$ where $\{P_i\}$ are mutually orthogonal finite-dimensional spectral projections. So in this case your formula would work.

Going the other way, if you have a finite-dimensional $T$ that has mutually orthogonal spectral projections then it will be orthogonally diagonalizable which implies self-adjoint.

For the general case where your spectral projections are infinite dimensional I am not sure.

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    $\begingroup$ For finite sums and a closed operator $T$ acting on a complex Banach space $X$, the formula is valid in general. This is a consequence of the properties of the analytic functional calculus. See Theorem 9.1 in Chapter V of the book A.E. Taylor and D.C. Lay "Introduction to functional analysis, 2nd ed". Wiley 1980. $\endgroup$
    – M.González
    Commented Oct 9, 2015 at 14:36
  • $\begingroup$ Thanks for these answers. The book by Taylor and Lay was a very helpful reference! It really surprised me that non-normal operaors are so well-behaved... $\endgroup$
    – Frank
    Commented Oct 9, 2015 at 15:49
  • $\begingroup$ @Frank: It is rather the holomorphic functional calculus that is well-behaved... This holds in any Banach algebra! $\endgroup$
    – Alain Valette
    Commented Oct 9, 2015 at 20:37
  • $\begingroup$ Hang on... But unbounded operators don't form a Banach Algebra, do they? $\endgroup$
    – Frank
    Commented Oct 10, 2015 at 16:33
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    $\begingroup$ If $T$ is a closed operator with non-empty resolvent set $\rho(T)$ and you take $\alpha\in \rho(T)$, then $A:=(T-\alpha I)^{-1}$ is in the Banach algebra of bounded operator on $X$, and given an analytic function $f$ on a neighborhood of $\sigma(T)$ you can find an analytic function $g$ on a neighborhood of $\sigma(A)$ so that $f(T)=g(A)$. $\endgroup$
    – M.González
    Commented Oct 11, 2015 at 9:55

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