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Unformatted text preview: MATH 220: DISTRIBUTIONS AND WEAK DERIVATIVES ANDRAS VASY Suppose V is a vector space over F = R or F = C . The algebraic dual of V is the vector space L ( V, F ) consisting of linear functionals from V to F . That is elements of f L ( V, F ) are linear maps f : V F satisfying f ( v + w ) = f ( v ) + f ( w ) , f ( cv ) = cf ( v ) , v,w V, c F . When V is infinite dimensional, we need additional information, namely con tinuity. So if V is a topological space with the compatible with the vector space structure, i.e. if V is a topological vector space, we define the dual space V as the space of continuous linear maps f : V F . For us, V is the class of very nice objects, and V will be the class of bad objects. Of course, normally there is no way of comparing elements of V with those of V , so we will also need an injection : V V so that elements of V can be regarded as elements of V (by identifying v V with ( V ). As we want to differentiate functions, as much as we desire, V will consist of infinitely differentiable functions. As we need to control behavior at infinity to integrate, the elements of V will be compactly supported. So we define V = C c ( R n ) to be the space of infinitely continuously differentiable functions of compact support. Here recall that the support supp of continuous function is the closure of the set where negationslash = 0; so C c ( R n ) means that there is a compact (i.e. closed and bounded) subset K of R n such that 0 outside K . As C c ( R n ) is infinite dimensional, we also need to put a topology on this. Tech nically this means that we should define what open sets are. Rather than doing this (to avoid complexity) we define what convergence of a sequence j of functions in C c ( R n ) means. Definition 1. Suppose { j } j =1 , 2 ,... is a sequence in C c ( R n ), and C c ( R n ). We say that lim j j = if (i) there is a compact set K such that j 0 outside K for all j , (ii) and all derivatives of j converge uniformly to , i.e. for all multiindices , sup R n  D ( j )  0 as j . Explicitly, for all N n and > exists N such that for j N , sup R n  D ( j )  < . Lemma 0.1. For all x R n and > there is a function C c ( R n ) such that ( x ) > , and supp { x :  x x  < } . Proof. First one checks that the function defined by ( t ) = e 1 /t , t > 0; ( t ) = 0 , t , is in C ( R ). Then we let ( x ) = parenleftbigg 2 2  x x  2 parenrightbigg ....
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 Fall '10
 Vasy

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