Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

1) In Rudin's proof towrds the end, he declares that $4M\cdot \sqrt{n}\cdot(1-\delta^2)^n$ goes to zero as n goes to infinity. I just don't see it though -- how is this the case.

2) My second question is that Rudin replaces $f_n(x + t) - f(x) \leq 2\sup(f(x))$. How did he come up with this?

share|improve this question

1 Answer 1

For (1), use $1-\delta^2\leqslant\mathrm e^{-\delta^2}$ and $\delta^2n\leqslant\mathrm e^{\delta^2n}$ to deduce $$ 4M\sqrt{n}(1-\delta^2)^n\leqslant4M\delta^{-1}\sqrt{\delta^2n}\cdot\mathrm e^{-n\delta^2}\leqslant4M\delta^{-1}\sqrt{\mathrm e^{\delta^2n}}\cdot\mathrm e^{-n\delta^2}=C\,r^n, $$ with $$ C=4M\delta^{-1},\qquad r=\mathrm e^{-\delta^2/2}\lt1. $$ And (2) might be a straightforward consequence of the inequalities $$ f_n(x+t)-f(x)\leqslant|f_n(x+t)|+|f(x)|\leqslant\|f_n\|_\infty+\|f\|_\infty.$$

share|improve this answer
    
Hi, thanks for your response! My difficulty with (1), though, was seeing how the limit of 4M*sqrt(n)*(1-delta^2)^n as n goes to infinity is zero. –  AnalysisStudent Jan 20 '13 at 23:22
    
Why though? My answer shows the limit is zero. –  Did Jan 20 '13 at 23:24
    
You're right, it does-- I see it's trivial now. Thank you so much! –  AnalysisStudent Jan 20 '13 at 23:27

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.