Tuesday, 25 October 2011

fa.functional analysis - Show a linear operator is not compact

Anton already gave a very clean answer. Another way to see it is to work backwards: start from a sequence of functions Fj in L2 that does is non-compact, and define fj(x)=fracddx(xFj(x)).



For example, let phi(x) be an arbitrary smooth bump function supported in [1/4,1/4], then the sequence of functions Fj(x)=2jphi(4jx1) all have disjoint support, but all have the same L2 norm, so obviously does not have a converging subsequence in L2.



Now set fj=(xFj)=Fj(x)+8jxphi(4jx1). Since phi has support only in [1/4,1/4], on the support of fj we can bound 4jx absolutely by, say, 2. So we have that fj is a bounded sequence in L2, whose corresponding Fj=Tfj cannot have a Cauchy subsequence.



Edit: I should also provide some motivation: observe that the scaling argument also works the other way (replace j by j, so that you can dilate). The Hardy-type inequality that you are using is a scaling invariant inequality: you estimate f/x in L2 by its derivative f. If we treat x as having units of distance, then the two objects have the same units regardless of what units f has. This gives scaling invariance of the estimate. In other words, the estimate is invariant under the natural scaling action of mathbbR+ on L2(mathbbR+), where the group operation for mathbbR+ is multiplication.



Observe that (mathbbR+,times) is a non-compact Lie group. Generally, if you have an inequality/operator that is invariant under the action of a non-compact Lie group, the inequality/operator cannot be compact. You just need to start with some test function and act on it by the Lie group action to generate a bounded sequence that runs off non-compactly in the "infinity dimension" direction. Terry summarised it in his Buzz http://www.google.com/buzz/114134834346472219368/9UseDXTJN74/There-are-three-ways-that-sequential-compactness a short while back.



This is, of course, closely related to the notion of concentration compactness.

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