Prove that if fn are uniformly continuous on I and fn f uni

Prove that if {f_n} are uniformly continuous on I, and f_n -> f uniformly, then f is uniformly continuous.

Solution

We need to show the following:
For any ? > 0, there exists ? > 0 such that
|x - y| < ? ==> |f(x) - f(y)| < ? for any x, y in [a,?).
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Let ? > 0 be given. Since lim(x-->?) f(x) = L for some finite number L
(by hypothesis), there exists some positive integer N such that
|f(x) - L| < ?/4 for all x >= N. (We may assume WLOG that N > a.)

Since [a, N] is compact, f is uniformly continuous on [a, N].
In other words, there exists ?1 > 0 such that
|x - y| < ?1 ==> |f(x) - f(y)| < ?/4 for any x, y in [a, N].
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Now, onto the proof...
(i) If both x and y are in [a, N], we are done (by the second part of the argument above).

(ii) If both x and y are in [N, ?), then we use the first part:
There exists ?2 > 0 such that for all x, y in [N, ?),
|x - y| < ?2 ==> |f(x) - f(y)| <= |f(x) - L| + |L - f(y)| < ?/4 + ?/4 = ?/2 ( < ?).

(iii) If x is in [a, N], but y is in (N, ?) [or vice versa...]
note that |f(x) - f(y)| <= |f(x) - f(N)| + |f(N) - f(y)|.
Now, apply (i) and (ii) to find ?3 = min {?1, ?2}.
so that for all x, y (as above) with |x - y| < ?3, we have
|f(x) - f(y)| <= |f(x) - f(N)| + |f(N) - f(y)| < ?/4 + ?/2 = 3?/4 < ?.

In summary, given ? > 0, set ? = min{?1, ?2}. (Putting ?3 in this would be redundant.)
Then, for all x, y in [a,?) with |x - y| < ?, we have |f(x) - f(y)| < ?.