# Integration with respect to the Borel measure on the unit sphere

Let $$\mathcal{L}(E)$$ the algebra of all bounded linear operators on a complex Hilbert space $$E$$.

Let $$B_n$$ be the open unit ball of $$\mathbb{C}^n$$ and $$\sigma$$ be the normalized positive Borel measure on the unit sphere $$\partial B_n$$.

By this (book), we have $$\int_{\partial B_n} |\lambda_i|^2 d \sigma(\lambda)=\frac1n,$$ and $$\int_{\partial B_n} \lambda_i\overline{\lambda_j} d \sigma(\lambda)=0,\;\;\forall i,j\in\{1,...,n\}\;\text{with}\;i\neq j.$$

I don't understand this notation $$\int_{\partial B_n} |\lambda_i|^2 d \sigma(\lambda)$$ i.e. here is $$\lambda=(\lambda_1,\cdots,\lambda_n)\in \partial B_n$$?

Also

If $$T_1,\cdots,T_n\in \mathcal{L}(E)$$, why $$\sup_{(\lambda_1,\cdots,\lambda_n)\in \partial B_n}\langle (\sum_{i=1}^n \lambda_i T_i)h, (\sum_{i=1}^n \overline{\lambda_i} T_i^*)h \rangle\geq \int_{\partial B_n}\sum_{i,j=1}^n\lambda_i\overline{\lambda_j}\langle T_iT_j^* h, h\rangle d \sigma(\lambda)?$$

Your interpretation of the notation in the integral is correct, $$\lambda = (\lambda_1, \dots, \lambda_n)$$. This is a completely standard notation for elements of $$\mathbb{R}^d$$.
For the other part of the question, notice that \begin{align} \sup_{(\lambda_1,\cdots,\lambda_n)\in \partial B_n}\langle (\sum_{i=1}^n \lambda_i T_i)h, (\sum_{i=1}^n \overline{\lambda_i} T_i^*)h \rangle &= \sup_{(\lambda_1,\cdots,\lambda_n)\in \partial B_n} \sum_{i,j = 1}^n \lambda_i \overline{\lambda_j} \langle T_i h, T_j^*h \rangle \\ \end{align} Now notice that \begin{align} \int_{\partial \mathbb{B}^n} \sum_{i,j=1}^n \mu_i \overline{\mu_j} \langle T_i h, T_j^* h \rangle d \sigma(\mu) &\leq \int_{\partial \mathbb{B}^n} \sup_{(\lambda_1,\cdots,\lambda_n)\in \partial B_n} \sum_{i,j=1}^n \lambda_i \overline{\lambda_j} \langle T_i h, T_j^* h \rangle d \sigma(\mu) \\& = \sup_{(\lambda_1,\cdots,\lambda_n)\in \partial B_n} \sum_{i,j=1}^n \lambda_i \overline{\lambda_j} \langle T_i h, T_j^* h \rangle \end{align} since $$\sigma$$ is normalised such that $$\sigma(\partial \mathbb{B}^n) =1$$. This is the desired inequality.
• Thank you. This is clear. Just I think you have used the notation $\int_{\partial B_n} |\lambda_i|^2 d \sigma(\mu)$ instead of $\int_{\partial B_n} |\lambda_i|^2 d \sigma(\lambda)$ in order to avoid confusion? Aug 16, 2019 at 13:06
• When I write $d\sigma(\mu)$ I am integrating over $\mu = (\mu_1, \dots, \mu_n)$. I did this because in the inequality, I want to take a $\sup$ over $\lambda$ and hence don't want to be integrating over $\lambda$. Aug 16, 2019 at 13:46