Originally Posted by

**steph236** H

**Uniform Boundedness:**

Since each f_k is continuous, and [0,1] is closed and bounded, by the Extreme Value theorem, each f_k attains its max and min on [0,1], ie. they are each bounded. Let c=max{|f_1(x)|, |f_2(x)|, |f_3(x)|,...}. Then |f_n(x)|=<c.

I don't think this last part works, but I'm not sure what else to do. Maybe something like:

If it were unbounded, then in the function space C([0,1], R) for any e>0 sup|f_k(x)-f(x)|>e for some value of x. But that contradicts what uniform convergence tells us, ie. that theres an N s.t k>=N means sup|f_k(x)-f(x)|<e for all x.

Any comments/helpful tips are much appreciated. =)