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From what I recall, every $C^{\ast}$-seminorm arises from a $\ast$-homomorphism. ie. $\exists$ a $\ast$-homomorphism $\varphi : A\to B$ between $C^{\ast}$-algebras such that $$p(a) = \|\varphi(a)\|$$ Your result follows from this because a $\ast$-homomorphism is norm-decreasing. To prove the result I mention, the argument (I think) is as follows: Take $N ... 3 The problem is that your circle is unnecessarily big, and you are hitting$0$where the square root is not analytic. If you use the circle$1+e^{it}/2$and the analytic expression for the square root in the disk of radius 1 around 1 $$f(z)=\sum_{k=0}^\infty {1/2 \choose k}\,(z-1)^k,$$ you will get the right values. The uniform convergence will allow you ... 2 Well, nothing special: write$C \in M_{mn}(\mathbb C)$. Let$C_{ab} \in M_{mn} (\mathbb C)$, where$a,b \in \{1, 2, \cdots, n\}$so that $$(C_{ab})_{ij} = \begin{cases} C_{ij} & \text{if } (a-1)m +1 \le i\le am, (b-1)m+1\le j\le bm,\\ 0 & \text{otherwise.}\end{cases}$$ Abusing notations, we also consider$C_{ab} \in M_m(\mathbb C)$. Then $$C = ... 2 If g is an element of order p (this always exists by Cauchy's Theorem), let$$ Q=\frac1p\,\sum_{j=0}^{p-1} g^j. $$Then \text{tr}(Q)=1/p and$$ Q^*=\frac1p\,\left(\sum_{j=0}^{p-1} g^j\right)^*=\frac1p\,\sum_{j=0}^{p-1} g^{-j} =\frac1p\,\sum_{j=0}^{p-1} g^{p-j}=\frac1p\,\sum_{k=1}^{p} g^k=\frac1p\,\sum_{k=0}^{p-1} g^k=Q. $$Note that ... 2 Marten, this is true by an observation due to Kaplansky, I believe, which asserts that an infinite-dimensional C*-algebra contains a self-adjoint element with infinite spectrum. See Ex. 4.6.12 in R. V. Kadison and J. R. Ringrose, Fundamentals of the Theory of Operator Algebras, Vol. I, Elementary Theory, Pure and Applied Math., Vol. 100 Academic Press, ... 1 Let X\in M . Then, for any Y\in M you have XY\omega\in M\omega . So XP=PXP . If you do this for X selfadjoint and take adjoints on the equality, you get XP=PX . As selfadjoints span M , you get that P\in M'. 1 These terms come from applying the product rule when multiplying out the operators: For example, the outer terms read as$$\left(\cos\theta\frac{\partial}{\partial r}\right)\left(-\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right)=-\cos\theta\frac{\partial}{\partial r}\left(\frac{\sin\theta}{r}\frac{\partial}{\partial \theta}\right)$\$ You must then ...