Stone-Weierstrass, uniform convergence

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SUMMARY

The discussion centers on the Stone-Weierstrass theorem and the concept of uniform convergence in the context of continuous functions. It establishes that there are continuous functions g: [-1,1] → R for which no sequence of polynomials Q_n can satisfy the condition Q_n(x^2) → g(x) uniformly on the interval as n approaches infinity. The proof demonstrates that assuming such a sequence exists leads to a contradiction, specifically when evaluating the limits at the endpoints of the interval, resulting in an impossibility. The conclusion is that uniform convergence cannot be achieved for the specified functions.

PREREQUISITES
  • Understanding of the Stone-Weierstrass theorem
  • Knowledge of uniform convergence and pointwise convergence
  • Familiarity with polynomial functions and their properties
  • Basic concepts of limits and epsilon-delta definitions in analysis
NEXT STEPS
  • Study the implications of the Stone-Weierstrass theorem in functional analysis
  • Explore examples of continuous functions that fail to meet uniform convergence criteria
  • Learn about the differences between uniform and pointwise convergence in detail
  • Investigate advanced topics in approximation theory related to polynomial convergence
USEFUL FOR

Mathematics students, particularly those studying real analysis, functional analysis, or approximation theory, will benefit from this discussion. It is also relevant for educators and researchers focusing on convergence properties of functions.

holomorphic
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Homework Statement


Show that there are continuous functions [tex]g:[-1,1]\to R[/tex] such that no sequence of polynomials [tex]Q_n[/tex] satisfies [tex]Q_n(x^2)\to g(x)[/tex] uniformly on [-1,1] as [tex]n\to\infty[/tex]

The Attempt at a Solution



Suppose there is a sequence [tex]Q_n[/tex] such that [tex]Q_n(x^2)\to g(x)[/tex] uniformly for [tex]g(x)=x[/tex].

Then [tex]\forall \epsilon > 0 \forall x \in [-1,1] \exists N:(n\geq N\Rightarrow |Q_n(x^2) - g(x)| \leq \epsilon)[/tex]

Take [tex]\epsilon = 1/2[/tex]. Then [tex]\exists N_1, N_2 : ( n \geq max\{N_1,N_2\} \Rightarrow |Q_n(1^2) - g(1)| \leq 1/2[/tex] and [tex]|Q_n((-1)^2) - g(-1)| \leq 1/2)[/tex].

Then for [tex]n \geq max\{N_1,N_2\}[/tex] we have
[tex]1 = 1/2 + 1/2 \geq |Q_n(1^2) - g(1)| + |Q_n((-1)^2) - g(-1)|[/tex]
[tex]=|Q_n(1) - g(1)| + |Q_n(1) - g(-1)|[/tex]
[tex]\geq |g(1)-g(-1)| = |1+1| = 2[/tex], which is false. Therefore there is no such [tex]Q_n[/tex].

Does this solution make sense?
 
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holomorphic said:
Then [tex]\forall \epsilon > 0 \forall x \in [-1,1] \exists N:(n\geq N\Rightarrow |Q_n(x^2) - g(x)| \leq \epsilon)[/tex]

Take [tex]\epsilon = 1/2[/tex]. Then [tex]\exists N_1, N_2 : ( n \geq max\{N_1,N_2\} \Rightarrow |Q_n(1^2) - g(1)| \leq 1/2[/tex] and [tex]|Q_n((-1)^2) - g(-1)| \leq 1/2)[/tex].

Then for [tex]n \geq max\{N_1,N_2\}[/tex] we have
[tex]1 = 1/2 + 1/2 \geq |Q_n(1^2) - g(1)| + |Q_n((-1)^2) - g(-1)|[/tex]
[tex]=|Q_n(1) - g(1)| + |Q_n(1) - g(-1)|[/tex]
[tex]\geq |g(1)-g(-1)| = |1+1| = 2[/tex], which is false. Therefore there is no such [tex]Q_n[/tex].

Oops. I used pointwise convergence instead of uniform. Doesn't change much but this should have read:
Then [tex]\forall \epsilon > 0 \exists N: \forall x \in [-1,1], (n\geq N\Rightarrow |Q_n(x^2) - g(x)| \leq \epsilon)[/tex]

Take [tex]\epsilon = 1/2[/tex]. Then [tex]\exists N : ( n \geq N \Rightarrow |Q_n(1^2) - g(1)| \leq 1/2[/tex] and [tex]|Q_n((-1)^2) - g(-1)| \leq 1/2)[/tex].

Then for [tex]n \geq N[/tex] we have
[tex]1 = 1/2 + 1/2 \geq |Q_n(1^2) - g(1)| + |Q_n((-1)^2) - g(-1)|[/tex]
[tex]=|Q_n(1) - g(1)| + |Q_n(1) - g(-1)|[/tex]
[tex]\geq |g(1)-g(-1)| = |1+1| = 2[/tex], which is false. Therefore there is no such [tex]Q_n[/tex].
 
It certainly makes sense. I haven't gone through all the details, but the main idea is surely correct.
 

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