Proposition: Let be uniformly convergent on A and take any . Then we have:
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If each is uniformly continuous on A then f is uniformly continuous on A.
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[6.10.4] |
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If each is continuous at a then f is continuous at a.
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[6.10.5] |
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for all n.
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[6.10.6] |
Proof:
1. ► For any we need to find a (see [6.5.2]) such that
for all with . As we firstly find an for such that
for all . [1]
Now, as is uniformly continuous there is a such that
[2]
for all with . Due to [1] and [2] these x now satisfy
2. ► Now we need to find a for each (see [6.5.1]) such that
for all with . But this is actually the proof of 1 if we replace y by a. The argument for [2] now is the continuity of at a.
3. ► is an immediate consequence of 2.
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