Trying to Prove Uniform Convergence: Analysis II

1. Jan 22, 2012

Abraham

1. The problem statement, all variables and given/known data

I have a solution to the following problem. I feel it is somewhat questionable though

If fn converges uniformly to f, i.e. fn$\rightarrow$f as n$\rightarrow$∞ and
gn converges uniformly to g, i.e. gn$\rightarrow$f as n$\rightarrow$∞ ,

Prove that fngn $\rightarrow$ fg

The domain is ℝ for all functions.

3. The attempt at a solution

If f and g are uniformly continuous, then for any ε>0, there is an N for which n$\geq$N implies | fn - f | < ε. The same for g.

This is my idea:

For any given ε >0, find N such that for n>N, |fn-f| < ε1, and |gn-f| < ε1. This ε1 should be made very small, such that:

|fngn - fg| = | (f+ε1)(g+ε1) - fg|
= | f*g + f*ε1 + g*ε1 + (ε1)2 - fg|
The ε1 terms can be made very small as needed, such that:
= | f*g + [STRIKE]f*ε1[/STRIKE] +[STRIKE] g*ε1[/STRIKE] + [STRIKE](ε1)2[/STRIKE] - fg| < ε

Agree, or disagree? Secondly, is there a shorter, simpler method? Thanks.
--Abraham

Last edited: Jan 22, 2012
2. Jan 22, 2012

LCKurtz

I probably should refrain from answering this because I have to leave momentarily. But my suggestion is that you need to state the question more carefully to begin with. I think you are confusing the ideas of uniform continuity and uniform convergence. You haven't told us what $f_n\rightarrow f$ means. Pointwise convergence? Uniform convergence? Any information about the domain? Problems like this are frequently addressed by adding and subtracting terms:$$|f_ng_n-fg| =|f_ng_n-f_ng+f_ng-fg|$$ and working with that.

3. Jan 22, 2012

Abraham

Sorry, I wrote the problem incorrectly. I meant to write:

fn converges uniformly to f, i.e. fn→f.

I don't know why I wrote "uniformly continuous" instead.....