For n N , let fn (x) = nx( 1 - x2)n and gn(x) = xn/1 + xn, x [0,1]. Prove that b

Question

For n N , let fn (x) = nx( 1 - x2)n and gn(x) = xn/1 + xn, x [0,1]. Prove that both of the sequences {fn} and {gn} do not converge uniformly on [0, 1].

Explanation / Answer

lim fn(x) = lim (nx*(1-x^2)^n) n->infinity n->infinity as 0infinity therefore fn(x)->0 f(x)=0 for uniform convergence Mn test M=|fn(x)-f(x)| M=|nx(1-x^2)^n| to find max value of M differentiate wrt x maximum of M occurs at x=(1/n)^0.5 maximum value of M(Mmax) is ((n-1)^n/(n^(n-0.5)) we see that as n->infinity Mmax does not tend to zero therefore fn does not converge uniformly b)gn(x)=(x^n/(1+x^n)) gn(x)->1 as n-> infinity Mn=gn(x)-g(x) =-1/((x^n)+1) as x lies between 0 and 1 max of Mn occurs at x =1 which is M=-1/2 as M->(-0.5) as n-> infinity gn(x) does not converge uniformly

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