3.2.2 Consequences of Uniform Convergence
And now we will see that the lies become true if fn → f is replaced by fn → f. (Well, most of them...)
Theorem 3.2.2. Let {fn} be a sequence of functions with common domain D, and let c be a point of I. Suppose that for all n ∈ Z+ we have
lim x→c fn(x) = Ln.
Suppose moreover that fn → f. Then the sequence {Ln} is convergent, lim x→c f(x) exists and we have equality:
lim n→∑ Ln = lim n→∑ lim x→c fn(x) = lim x→c f(x) = lim x→c lim n→∑ fn(x).
Proof. Step 1: We show that the sequence {Ln} is convergent. Since we don't yet have a real number to show that it converges to, it is natural to try to use the Cauchy criterion, hence to try to bound |Lm − Ln|. Now comes the trick: for all x ∈ I we have
|Lm − Ln| ≤ |Lm − fm(x)| + |fm(x) − fn(x)| + |fn(x) − Ln|.
By the Cauchy criterion for uniform convergence, for any ̄̄̄ > 0 there exists N ∈ Z+ such that for all m, n ≥ N and all x ∈ I we have |fm(x) − fn(x)| < ̄̄̄/3. Moreover, the fact that fm(x) → Lm and fn(x) → Ln give us bounds on the first and last terms: there exists ̄̄̄ > 0 such that if 0 < |x − c| < ̄̄̄ then
|Ln − fn(x)| < ̄̄̄/3 and |Lm − fm(x)| < ̄̄̄/3. Combining these three estimates, we find that by taking x ∈ (c − ̄̄̄, c + ̄̄̄), x ≠ c and m, n ≥ N, we have
|Lm − Ln| ≤ ̄̄̄/3 + ̄̄̄/3 + ̄̄̄/3 = ̄̄̄.
So the sequence {Ln} is Cauchy and hence convergent, say to L ∈ R.
Step 2: We show that lim x→c f(x) = L (so in particular the limit exists!). Actually the argument for this is very similar to that of Step 1:
|f(x) − L| ≤ |f(x) − fn(x)| + |fn(x) − Ln| + |Ln − L|.
Since Ln → L and fn(x) → f(x), the first and last term will each be less than ̄̄̄/3 for sufficiently large n. Since fn(x) → Ln, the middle term will be less than ̄̄̄/3 for x sufficiently close to c. Overall we find that by taking x sufficiently close to (but not equal to) c, we get |f(x) − L| < ̄̄̄ and thus lim x→c f(x) = L.