# Another sequence problem

1. Oct 5, 2010

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

For some reason, although it looks simple, it's giving me trouble.

Let X be a topological space, and Y a metric space. Let fn : X --> Y be a sequence of continuous functions, and let xn be a sequence of points in X converging to x. Show that if fn converges uniformly to f, then (fn(xn)) converges to f(x).

3. The attempt at a solution

The facts I know:

i) since fn is a sequence of continuous finctions which converges uniformly to f, f is continuous
ii) since f is continuous, f(xn) converges to f(x).

Now, since fn converges uniformly to f, for every ε > 0 there exists some N such that for all x in X and for any n >= N, d(fn(x), f(x)) < ε holds. (iii)

I need to show that for any ε > 0, there exists some integer N such that, for all n >=N d(fn(xn), f(x)) < ε holds.

Let ε > o be given. Since, because of uniform convergence (iii) holds for any x, it holds for the members of the sequence xn, too. So, there exists N such that for n >= N, d(fn(xn), f(xn)) < ε holds. Now I'm stuck. Could this mean that the sequences fn(xn) and f(xn) converge to the same limit? Since then fn(xn) would definitely converge to f(x). But I can't find a theorem or result which says anything about that right now.

Perhaps I'm not on the right track at all. Thanks for any replies.

2. Oct 5, 2010

### ystael

You're almost there -- you have all the pieces, you just need to put them together.

Would it illuminate anything if I suggested that you choose $$N$$ so that for $$n > N$$ and every $$x' \in X$$ (thus, in particular, for the $$x_n$$) you have $$d(f_n(x'), f(x')) < \frac\epsilon2$$ ?

3. Oct 5, 2010

ystael, just before I read your answer, and just before I think about it, I came up with a Lemma which I proved (hopefully correct - if it works, I proved what I need to prove.)

Lemma. Let xn and yn be sequences in the metric space (X, d), with xn --> x and yn --> y. If for every ε > 0 there is a positive integer N such that for all n >= N, d(xn, yn) < ε holds, then x = y.

Proof. Let ε > 0 be given. Since xn and yn converge, for ε/2 there is some N1 such that n>= N1 implies d(xn, x) < ε/2, and some N2 such that n>= N2 implies d(yn, y) < ε/2. Let N3 be the integer associated with ε. Let N = max{N1, N2, N3}. Then d(x, y) <= d(x, xn) + d(xn, yn) + d(yn, y) < 2ε, for all n >= N. Since this holds for any ε > 0, we conclude that d(x, y) = 0, i.e. x = y.

4. Oct 6, 2010

### ystael

This lemma and its proof are correct.

5. Oct 6, 2010

OK, but there's one step more I'm not sure about in the proof of the original problem - we know that f(xn) --> f(x), but I don't know if fn(xn) converges (which is partially what I need to show) - without this, I can't apply the Lemma above, it just occured to me.

I know that fn(xn) --> fn(x), since fn is continuous for every n, but that's a different thing. I'm interested in the convergence of the sequence f1(x1), f2(x2), ... and not fn(x1), fn(x2), ...

6. Oct 6, 2010

OK, let ε > 0 be given. Then, for ε/2 there exist some N1 such shat for n >= N1, d(fn(xn), f(xn)) < ε/2 and some N2 such that for n >= n2, d(f(xn), f(x)) < ε/2. If N = max {N1, N2}, we have d(fn(xn), f(x)) <= d(fn(xn), f(xn)) + d(f(xn), f(x)) < ε, for any n >= N. Hence, fn(xn) converges to f(x).

7. Oct 6, 2010

### ystael

This is correct. Make sure you are clear on how this construction requires uniform convergence of $$(f_n)$$ to $$f$$. When you say "there exists some $$N_1$$ such that for $$n \geq N_1$$, $$d(f_n(x_n), f(x_n)) < \frac\epsilon2$$", the point $$x_n$$ is not constant in the limit expression, so this statement is not true if $$(f_n)$$ only converges to $$f$$ pointwise.

8. Oct 6, 2010

Yes, I'm aware of that. It allows us to "fit in" members of the sequence xn.

Thanks a lot for your help.

9. Oct 6, 2010

By the way, the way I tried to prove it, along with the Lemma in one of the posts above, it doesn't seem to work, right? Because of what I wrote in post #5..?