Another uniform convergence question

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SUMMARY

The discussion centers on proving the uniform convergence of the series \( t(x) = \sum^{\infty}_{n=0} x^n h(x^n) \) on the interval \([0,s]\) where \( 0 < s < 1 \). The Weierstrass M-test is identified as a key tool for establishing this convergence, leveraging the continuity of the function \( h \) and the boundedness of \( h \) on the compact set. The extreme value theorem is employed to find bounds for \( h(x^n) \), confirming that the series converges uniformly due to the geometric nature of \( s^n \).

PREREQUISITES
  • Understanding of uniform convergence and the Weierstrass M-test.
  • Familiarity with continuity and the extreme value theorem.
  • Knowledge of geometric series and their convergence properties.
  • Basic concepts of series and sequences in real analysis.
NEXT STEPS
  • Study the Weierstrass M-test in detail to understand its application in proving uniform convergence.
  • Review the extreme value theorem and its implications for continuous functions on compact intervals.
  • Explore geometric series and their convergence criteria to solidify understanding of series behavior.
  • Investigate additional examples of uniform convergence in real analysis to reinforce concepts.
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Students and educators in real analysis, mathematicians focusing on series convergence, and anyone seeking to deepen their understanding of uniform convergence in the context of continuous functions.

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Homework Statement



h:[0,1] -> R is continuous
Prove that t(x) = [tex]\sum[/tex][tex]^{infinity}_{n=0}[/tex] xnh(xn) is uniformly convergent on [0,s] where 0<s<1

Homework Equations





The Attempt at a Solution



I have the definition of h being continuous but after this I am pretty clueless about how to tackle this problem. I could use the Weierstrass M-test. I know the series xn converges uniformly on this interval as xn < sn but I don't know how to use the fact that h is continuous to find a sequence of real numbers that xnh(xn) is always less than.
 
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Use the extreme value theorem to bound h(x^n), and then find small bounds for x^n h(x^n).
 
h is continuous (it is a continuous function of a polynomial which is continuous) on a compact set therefore h is bounded for x in the interval [0,s]. the sequence of h's for all n is uniformly bounded by a single M. as you said the s^n is a geometric series so it converges. hence uniform convergence.
 

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