Thursday, February 12, 2009
12/2 lecture: correction of argument
During the lecture I gave an incomplete argument why the inequality |T(u(t))-T(v(t))|<= Ka d(u,v) implies that T is a contraction if aK<1. (As correctly pointed out to me by one of you.)
Clearly, d(T(u),T(v))>=|T(u(t))-T(v(t))| for any t, but this observation does not lead to the required conclusion, since the inequality is in the wrong direction.
An argument that leads to the desired result is that since |T(u(t))-T(v(t))| <= Ka d(u,v) FOR ALL t in J, we have in particular d(Tu,Tv)=sup_t|T(u(t))-T(v(t))|<= Ka d(u,v).
Clearly, d(T(u),T(v))>=|T(u(t))-T(v(t))| for any t, but this observation does not lead to the required conclusion, since the inequality is in the wrong direction.
An argument that leads to the desired result is that since |T(u(t))-T(v(t))|
# posted by Jeroen Lamb @ 11:24 AM 
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