Proving Uni.Conv. of Fn to F on (0,1): Questions & Answers

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SUMMARY

The discussion centers on the proof of uniform continuity of a function \( f \) on the interval \( (0,1) \) given that a sequence of functions \( f_n \) is uniformly convergent to \( f \) on the same interval. The user proposes a method involving the continuous extension theorem, asserting that if \( f_n \) is uniformly continuous on \( (0,1) \), then \( F_n \) is continuous on the closed interval \([0,1]\). The conclusion drawn is that if \( F_n \) converges uniformly to \( F \) on \([0,1]\), then \( f \) is uniformly continuous on \( (0,1) \).

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  • Understanding of uniform continuity and uniform convergence
  • Familiarity with the continuous extension theorem
  • Knowledge of the Triangle Inequality in analysis
  • Basic proficiency in using the \( \varepsilon/3 \) argument in proofs
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  • Study the properties of uniform continuity in detail
  • Learn about the continuous extension theorem and its applications
  • Explore the Triangle Inequality and its role in mathematical proofs
  • Practice using \( \varepsilon \)-\( \delta \) definitions in convergence proofs
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Mathematics students, particularly those studying real analysis, educators teaching uniform convergence concepts, and researchers exploring function continuity properties.

bw0young0math
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Hello, today I have a question.

If uni.cont. function sequence fn on (0,1) is uni.conv. to f on (0,1), then f is uni.cont. on (0,1).

The above is true.
The wonder I have is...May I prove the above by this way?

This way: fn is uniform continuous on (0,1). So Fn is continuous on [0,1] (by continuous extention theorem.)
If I can prove Fn is uni. convergent to F on [0,1], (since Fn is cont. on [0,1]) F is cont. on [0,1] and f is uni. cont. on (0,1).

is this way true?
IF it is true, how can I prove Fn is uni.conv. to F on [0,1]?
IF not, could you show me some counter example?Please...(Sadface)(Sadface)(Sadface)
 
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I think you can use a standard $\varepsilon/3$ argument to prove this. Write out what uniform continuity and uniform convergence mean, using $\epsilon/3$ as the RHS. Then use the Triangle Inequality to finish.
 

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