# Weierstrass Approximation— Rudin's Proof

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?

-

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 $$f_n(x+t)-f(x)\leqslant|f_n(x+t)|+|f(x)|\leqslant\|f_n\|_\infty+\|f\|_\infty.$$

-
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