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Let F be the set of all continuous functions with domain [-1,1] and codomain R. Let A be the algebra of all polynomials that contain only terms of even degree (A is a subset of F). Show that the closure of A in F is the set of even functions in F.

The attempt at a solution

I have to show that (i) if f in F is even, then f is in the closure of A and (ii) if f is in the closure of A, then f is even. I don't have problems proving (ii), rather I'm stuck proving (i). Here's what I have so far:

Let f in F be even. By the Weierstrass Approx. Theorem, there is a sequence of polynomials {p_n} that converge uniformly to f. Now let q_n be the polynomial derived from p_n by squaring each term so that all the degrees are even. For x in [0,1], q_n(sqrt(x)) = p_n(x), so {q_n(sqrt(x))} converges to f(x). Since f is even, {q_n(sqrt(x))} converges to f(-x). Now it would be nice to show that {q_n} converges to f, but this is not the case. If anything, {q_n} converges to f(x^2). How do I proceed from here?

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# Even Continuous Functions

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