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By Carothers N.L.

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Weierstrass's Second Theorem, 1885) Let f 2 C 2 . Then, for every " > 0, there exists a trig polynomial T such that kf ; T k < ". Ultimately, we will give several di erent proofs of this theorem. Weierstrass gave a separate proof of this result in the same paper containing his theorem on approximation by algebraic polynomials, but it was later pointed out by Lebesgue (1898) that the two theorems are, in fact, equivalent. Lebesgue's proof is based on several elementary observations. We will outline these elementary facts as \exercises with hints," supplying a few proofs here and there, but leaving full details to the reader.

If f Best Approximation 50 Now consider the (uniformly) continuous function ' = f ; p. We may partition a b ] by way of a = t0 < t1 < < tn = b into su ciently small intervals so that j'(x) ; '(y)j < E=2 whenever x y 2 ti ti+1 ]: Here's why we'd want to do such a thing: If ti ti+1 ] contains a (+) point for ' = f ; p, then ' is positive on all of ti ti+1 ]. Indeed, x y 2 ti ti+1 ] and '(x) = E =) '(y) > E=2 > 0: Similarly, if ti ti+1 ] contains a (;) point for ', then ' is negative on all of ti ti+1 ].

A b ] by (t) = a + t(b ; a) for 0 t 1, and de ne a transformation T : C a b ] ! C 0 1 ] by (T (f ))(t) = f ( (t)). Prove that T satis es: (a) T (f + g) = T (f ) + T (g) and T (cf ) = c T (f ) for c 2 R. (b) T (fg) = T (f ) T (g). In particular, T maps polynomials to polynomials. (c) T (f ) T (g) if and only if f g. (d) kT (f )k = kf k. (e) T is both one-to-one and onto. Moreover, (T );1 = T ;1 . The point to exercise 18 is that C a b ] and C 0 1 ] are identical as vector spaces, metric spaces, algebras, and lattices.

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