# If $E$ is the spectral measure coresponding to a self-adjoint operator, show $\int1+2\sum_{i=1}^n\frac{n-i}nλ^idE(λ)f→\int\frac{1+λ}{1-λ}dE(λ)f$

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.

• 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 $\lambda=1$. – s.harp Aug 14 at 7:08
• @s.harp Well, it's not clear to me why the integral is $0$ outside $[-1,1]$: math.stackexchange.com/q/3322373/47771. – 0xbadf00d Aug 14 at 7:10
• Do you know that $\sigma(\kappa)\subseteq [-\|\kappa\|,\kappa\|]$? (For self-adjoint $\kappa$.) – s.harp Aug 14 at 7:13
• @s.harp Yes. By self-adjointness, even $1\ge\left\|\kappa\right\|=\sup_{\lambda\in\sigma(\kappa)}|\lambda|$. – 0xbadf00d Aug 14 at 7:15
• 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. – s.harp Aug 14 at 7:19