I'm trying to prove the following claim:
fn∈Cc, Cc being the set of continuous functions with compact support, then limn→∞||fn−f||∞=0 implies fn(x)→f(x) uniformly.
So, according to my understanding,
limn→∞||fn−f||∞=0
⇔
∀ε>0∃N:n>N⇒|fn(x)−f(x)|≤||fn−f||∞<ε
Now my problem is that I don't see how uniform convergence follows from pointwise convergence μ almost everywhere. Can someone give me a hint? I guess I have to use the fact that they have compact support but I don't see how to put this together.
Thanks for your help!
Answer
You were not very specific about your hypotheses - I assume you're working on Rn with Lebesgue measure.
Suppose there exists a point x such that |fn(x)−f(x)|>ε. Then there exists an open set U containing x such that for all y∈U you have |fn(y)−f(y)|>ε by continuity. But this contradicts the a.e. statement you gave.
In case you don't know yet that f is continuous (or rather: has a continuous representative in L∞), a similar argument shows that (fn) is a uniform Cauchy sequence (with the sup-norm, not only the essential sup-norm), hence f will be continuous (in fact, uniformly continuous).
Note that I haven't used compact support at all, just continuity.
If you're working in a more general setting (like a locally compact space), you'd have to require that the measure gives positive mass to each non-empty open set.
Finally, note that f need not have compact support. It will only have the property that it will be arbitrarily small outside some compact set ("vanish at infinity" is the technical term). For instance, 11+|x| can easily be uniformly approximated by functions with compact support.
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