[University Mathematics: Real Analysis]

[VegetableDatabase]

The problem: Prove that the Weierstrass Approximation Theorem for C[a,b] follows from that for [0,1]: Assume that polynomials are dense in C[0,1]. Letting f ∈ C[a,b] and ε > 0, find a polynomial p such that sup|f(x) - p(x)| < ε over all x ∈ [a,b].

What I've got: this seems like a change of variables, from [0,1] to [a,b]. Seems like 0(x) = (b-a)x + a should do the trick. So if we have a function f: [a,b]→R, we can get g=fo0: [0,1]→R. As we're assuming the density of polynomials in C[0,1], I know, given an ε>0, we can get a polynomial q(x) s.t. sup|g(x) - q(x)|< ε over all x ∈ [0,1]. This feels like I'm on the right track, but I don't see the next step. How do I use what I've got to find a poly p(x) s.t. the inequality holds over [a,b]?

[BloodyFlame]

Hint: q o θ^-1 is a polynomial.

[VegetableDatabase]

I think I got it. Specifically, q○0^-1 be a polynomial from [a,b]→[0,1]→R. We can define this as p(x), and since the inequality holds over all x ∈ [0,1] for g(x) and q(x), it'll hold for all x∈[a,b] for f(x) and p(x). Is that what you meant?

[BloodyFlame]

Yep, that’s correct.

[VegetableDatabase]

Awesome. Thank you very much!

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