Let $(\Omega,\mathcal A,\mu)$ be a probability space, $\kappa$ be a contractive self-adjoint linear operator on $L^2(\mu)$ and $E:\mathbb R\to\mathfrak L(L^2(\mu))$ be the spectral family corresponding to $\kappa$. How can we show that $$\int 1+2\sum_{i=1}^n\frac{n-i}n\lambda^i\:{\rm d}E(\lambda)f\xrightarrow{n\to\infty}\int\frac{1+\lambda}{1-\lambda}\:{\rm d}E(\lambda)f\tag1$$ for all $f\in L^2(\mu)$? This looks like a simple application of Lebesgue's dominated convergence theorem, but since the sum in the integrand on the left-hand side tends to $\infty$, I'm confused.

  • $\begingroup$ Why does the sum go to infinity? Since $\kappa$ is contractive you have that only $|\lambda|≤1$ contribute to the integral. Interpreting the word contractive in a more strict fashion ($\|\kappa\|<1$) will also cut off the integral before the problematic point $\lambda=1$. $\endgroup$ – s.harp Aug 14 at 7:08
  • $\begingroup$ @s.harp Well, it's not clear to me why the integral is $0$ outside $[-1,1]$: math.stackexchange.com/q/3322373/47771. $\endgroup$ – 0xbadf00d Aug 14 at 7:10
  • $\begingroup$ Do you know that $\sigma(\kappa)\subseteq [-\|\kappa\|,\kappa\|]$? (For self-adjoint $\kappa$.) $\endgroup$ – s.harp Aug 14 at 7:13
  • $\begingroup$ @s.harp Yes. By self-adjointness, even $1\ge\left\|\kappa\right\|=\sup_{\lambda\in\sigma(\kappa)}|\lambda|$. $\endgroup$ – 0xbadf00d Aug 14 at 7:15
  • $\begingroup$ The relevant statement is that $E(\lambda)$ is an increasing family of projections, that becomes equal to $\Bbb 1$ at $\sup_{\lambda\in\sigma(\kappa)}\lambda$. You would need to compare with the definition of the space associated to $E(\lambda)$ to note this. $\endgroup$ – s.harp Aug 14 at 7:19

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