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 Jan24 awarded Popular Question Jul2 awarded Curious Jun4 comment About inverse matrix in portfolio choice You are right. I had a typo. In the textbook, there is no 1/2 in Equation (*). Thanks. Jun4 accepted About inverse matrix in portfolio choice Jun4 asked About inverse matrix in portfolio choice Mar17 awarded Notable Question Jan13 awarded Popular Question Sep9 accepted The product of integrable random variables need not be integrable Feb18 awarded Popular Question Sep29 awarded Yearling May6 accepted uniform convergence of continuous functions Apr30 comment uniform convergence of continuous functions I finished all the details but one thing: how can I prove uniform convergence. It seems that even on the compact set, pointwise convergence does not imply uniform convergence. By the way, I would accept the comments as the answer (should you post it or them), since they are very helpful. Thank you. Apr30 comment uniform convergence of continuous functions One quick question: is it right to argue that Convex combination of continuous functions is continuous $\Rightarrow C[0,1]$ is convex. Apr30 comment uniform convergence of continuous functions Riesz representation theorem seems to apply to linear functionals. But C[0,1] meas all continuous functions on [0,1]. Am I wrong? Apr30 asked uniform convergence of continuous functions Apr25 accepted Quotient space and $L_p$ space Apr23 asked Quotient space and $L_p$ space Apr22 comment The spectrum of a bounded linear operator Thank you. I basically proved that. Apr22 comment The spectrum of a bounded linear operator I agree with your EDIT''. In fact, $\sigma(T^{n})=[\sigma(T)]^{n}$ for the complex case. That's why the converse is true for the complex. The proof is not trivial. Do you agree? Also @Norbert Apr22 comment The spectrum of a bounded linear operator One simple linear algebra question. Given that $$T^n-\lambda^nI=\left(\sum\limits_{k=0}^{n-1}\lambda^{n-1-k}T^k\right)(T-\lambda I)$$, why $T^n-\lambda^n I$ is invertible $\Rightarrow T-\lambda I$ is invertible? Thank you.