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### Sets, Maps and Symmetry Groups Course Discussion List

## group project 3

Posted by Kevin Hall on January 08, 19100 at 13:21:56:
I'm hoping for a hint on project 3, #12. To prove that the integration operator is a contraction, we have to show that if f: B(r)-->B(r) and g: B(r)-->B(r) are polynomials in z, and r<1, then d(If,Ig)<_ C*d(f,g) for some C<1. Let f(z) = 1 + (a1)z + (a2)z^2 + ... + (an)z^n and let g(z) = 1 + (b1)z + (b2)z^2 + ... + (bn)z^n. Then d(f,g)= sup{(a1-b1)z + (a2-b2)z^2 +...+(an-bn)z^n for z in B(r)}. There should be absolute value signs in there, but I can't make those on the computer. We will also have d(If,Ig) = sup { [z]*[(1/2)(a1-b1)z + (1/3)(a2-b2)z^2 +...+(1/n+1)(an-bn)z^n], for z in B(r)}. In this expression for d(If,Ig), we've factored out a z from the sum. Now the modulus of z is less than one, since f and g are functions from B(r) to B(r) and r<1. So it would seem that d(If,Ig) could be less than d(f,g) by a factor of r. But I can't figure out what to do with the 1/2, 1/3,...,1/n+1, the numbers that appear in the denominators of the terms in d(If,Ig). If it weren't for these terms, we could let C = r and then it would be true that d(If,Ig) <_ d(f,g), but with these terms I'm not sure it's still true. Any ideas on how to deal with these numbers in the denominators?

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