# Convergence and continuity question

1. Nov 21, 2006

### buzzmath

Can anyone help with this question?

Let (f_n) be a sequence of continous functions on D a subset of R^p to R^q s.t. (f_n) converges uniformly to f on D, and let (x_n) be a sequence of elements in D which converges to x in D. Does it follow that (f_n(x_n)) converges to f(x)?

My proof goes like this:
Since (f_n) are all continous and uniformly converge on D to f we know that f is continous on D. By a thm. So f is continous at all the values on (x_n) and is continous at x. Since (f_n) is uniformly convergent to f there exist a natural number k(e) s.t. for all e>0 n>=k(e) x an element of D that ||f_n(x)-f(x)|| < e but since (x_k) and x are both elements of D we know ||f_n(x_k)-f(x_k)||<e for all natural numbers k. So we know f_n(x_k) converges to f(x_k) (By a thm that says if f is continous at a then if (x_n) is any sequences in D(f) that converges to a,k the the sequence (f(x_n)) converges to f(a)) We know that (x_n) an element of D is a sequence which converges to x Thus f(x_n) converges to f(x). Thus f_n(x_n) converges to f(x_n) which converges to f(x). Thus (f_n(x_n)) converges to f(x).

My problem is I can see that if (f_n(x)) converges to f(x) and if f(x_n) converges to f(x) then the result seems clear that (f_n(x_n)) converges to f(x) but in my proof it doesn't seem clear to just say that. Does this make sense or is there a more precise way to say it?

thanks

2. Nov 21, 2006

### StatusX

If you want to be rigorous you should use the epsilon delta definition. You can get |f(x)-f_n(x_n)|=|f(x)-f(x_n)+f(x_n)-f_n(x_n)| <=|f(x)-f(x_n)|+|f(x_n)-f_n(x_n)| by the triangle inequality. Can you see what to do from here?

3. Nov 22, 2006

### buzzmath

I got it now. thanks