Thread: General Proofs of Uniform Convergence

Let S subset R and f, fn, g, gn : S -> R. Suppose fn -> uniformly on S and gn -> g uniformly on S. Then,

1. Show that, for any real number c, cfn -> cf uniformly on S.

2. Show that fn +gn -> f+g uniformly on S.

3. Let S = R, fn(x) = x, f(x) = , gn(x) = 1/n and g(x) = 0. Show that fngn does not converge to fg uniformly on S.

1. Since fn converges uniformly lim[sup{|f(x)-fn(x)|: x ele S}] = 0. Then multiply both sides by c so c*lim[sup{|f(x)-fn(x)|: x ele S}] = c*0. You can move the c inside the limit and sup (not sure exactly why?) so lim[sup{|cf(x)-cfn(x)|: x ele S}] = 0. Hence cfn converges uniformly.

2. So for each eps1 > 0 there exists a number N such that |fn(x)-f(x)|< eps1 for all x ele S and all n > N. Also for each eps2 > 0 there exists a number N such that |gn(x)-g(x)|< eps2 for all x ele S and all n > N. Let eps1+eps2=eps and N>n. Since |fn + gn - f - g|<= |fn-f|+|gn-g|< eps1+eps2 = eps.

3. |x/n - 0|< 1. Since n>N, a particular n is N+1. The first statement fails to be true when x>N+1. So not unifromly convergent.

Do these make sense and does any rigor need to be added.

THanks

and
Use these to complete the proof.

I think this is better

1. Since fn is uniform convergence on S, then |fn - f|< eps/|c|. Then, since |cfn - cf|< |c||fn-f| < eps. For c = 0, |cfn - cf| = |0|< eps since eps>0.