Stone Weierstrass application?

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SUMMARY

The discussion focuses on proving that the span of the set $\{x^{2n}:n \geq0\}$ is dense in the space of continuous functions $C([0,1])$, utilizing the Stone-Weierstrass theorem. The user also seeks to demonstrate that the closure of the span of $\{x^{2n+1}:n \geq0\}$ corresponds to the set of continuous functions that vanish at zero, specifically $\{f \in C([0,1]):f(0) = 0\}$. Key steps include establishing that the span forms an algebra of functions, separating points in the interval, and confirming that the functions do not vanish at any point in the compact set [0,1].

PREREQUISITES
  • Understanding of the Stone-Weierstrass theorem
  • Knowledge of continuous functions on compact intervals
  • Familiarity with the concept of function algebras
  • Basic principles of polynomial density in function spaces
NEXT STEPS
  • Study the implications of the Stone-Weierstrass theorem in functional analysis
  • Explore the properties of continuous functions on compact sets
  • Investigate the concept of function algebras and their applications
  • Learn about polynomial approximation and its role in density arguments
USEFUL FOR

Mathematics students, particularly those studying real analysis, functional analysis, or anyone interested in the applications of the Stone-Weierstrass theorem in proving function properties.

nalgas
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Homework Statement



I want to prove that the span of $\{x^{2n}:n \geq0\}$ is dense in $C([0,1])$.

Furthermore, that the closure of the span of $\{x^{2n+1}:n \geq0\}$ is $\{f \in C([0,1]):f(0) = 0\}$.

Homework Equations



Is my solution correct?

Now I do not know how to tackle the second part. Any help?

Thank you for helping the community.

The Attempt at a Solution



My work:

I guess I can use the Stone-Weierstrass theorem.

Define $A =$ span of $\{x^{2n}:n \geq0\}$.
I need to prove A is an algebra of real continuous functions on a compact set I = [0,1]. The set I is compact since it is continuous and bounded.

A is an algebra since if, for example I multiply $fg = x^2x^4 = x^8 \in A$.

Next, need to show A separates points on I. So, points $x_1,x_2 \in [0,1] \Rightarrow f(x_1) \neq f(x_2), f(x_1), f(x_2) \in \{x^{2n}:n \geq0\}$.

Finally, A vanishes at no point of I, i.e. $\forall x \in [0,1]$ elements of $\{x^{2n}:n \geq0\}$ are $\neq0$.
 
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Yes, you can certainly do that, it will work fine. Or, if you're lazy, you can just recycle the fact that polynomials are dense (presumably you've already proved that since you're using Weierstrass), by using a simple transformation (homeomorphism of [0,1] with itself), do you see which one ?
 

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